MatBench/layer2/eval/dataset_origin.json

[
    {
        "idx": 1,
        "question": "Temperature indicators are sometimes produced from a coiled metal strip that uncoils a specific amount when the temperature increases. How does this work?",
        "answer": "Bimetallic materials are produced by bonding two materials having different coefficients of thermal expansion to one another, forming a laminar composite. When the temperature changes, one of the materials will expand or contract more than the other material. This difference in expansion or contraction causes the bimetallic material to change shape; if the original shape is that of a coil, then the device will coil or uncoil, depending on the direction of the temperature change.",
        "type": 4,
        "wrong_answers_1": "The temperature indicator is made from bimetallic materials, which consist of two materials with different coefficients of thermal expansion bonded together.",
        "wrong_answers_2": "Borosilicate glasses and fused silica are resistant to thermal shock because they have relatively low coefficients of thermal expansion; therefore, upon heating or cooling, the difference in the degree of expansion or contraction across a cross-section of a ware that is constructed from these materials will be relatively low.",
        "wrong_answers_3": "Thermal stresses may be introduced into a structure by rapid heating or cooling because temperature gradients are established across the cross section due to more rapid temperature changes at the surface than within the interior; thus, the surface will expand or contract at a different rate than the interior and since this surface expansion or contraction will be restrained by the interior, stresses will be introduced."
    },
    {
        "idx": 2,
        "question": "From what kind of material would the temperature indicator be made?",
        "answer": "The temperature indicator is made from bimetallic materials, which consist of two materials with different coefficients of thermal expansion bonded together.",
        "type": 4,
        "wrong_answers_1": "Bimetallic materials are produced by bonding two materials having different coefficients of thermal expansion to one another, forming a laminar composite. When the temperature changes, one of the materials will expand or contract more than the other material. This difference in expansion or contraction causes the bimetallic material to change shape; if the original shape is that of a coil, then the device will coil or uncoil, depending on the direction of the temperature change.",
        "wrong_answers_2": "In order for the material to perform well, the two materials must have very different coefficients of thermal expansion and should have high enough modulus of elasticity so that no permanent deformation of the material occurs.",
        "wrong_answers_3": "Borosilicate glasses and fused silica are resistant to thermal shock because they have relatively low coefficients of thermal expansion; therefore, upon heating or cooling, the difference in the degree of expansion or contraction across a cross-section of a ware that is constructed from these materials will be relatively low."
    },
    {
        "idx": 3,
        "question": "What are the important properties that the material in the temperature indicator must possess?",
        "answer": "In order for the material to perform well, the two materials must have very different coefficients of thermal expansion and should have high enough modulus of elasticity so that no permanent deformation of the material occurs.",
        "type": 4,
        "wrong_answers_1": "The temperature indicator is made from bimetallic materials, which consist of two materials with different coefficients of thermal expansion bonded together.",
        "wrong_answers_2": "Bimetallic materials are produced by bonding two materials having different coefficients of thermal expansion to one another, forming a laminar composite. When the temperature changes, one of the materials will expand or contract more than the other material. This difference in expansion or contraction causes the bimetallic material to change shape; if the original shape is that of a coil, then the device will coil or uncoil, depending on the direction of the temperature change.",
        "wrong_answers_3": "Borosilicate glasses and fused silica are resistant to thermal shock because they have relatively low coefficients of thermal expansion; therefore, upon heating or cooling, the difference in the degree of expansion or contraction across a cross-section of a ware that is constructed from these materials will be relatively low."
    },
    {
        "idx": 4,
        "question": "What properties should the head of a carpenter's hammer possess?",
        "answer": "The striking face and claws of the hammer should be hard-the metal should not dent or deform when driving or removing nails. Yet these portions must also possess some impact resistance, particularly so that chips do not flake off the striking face and cause injuries.",
        "type": 4,
        "wrong_answers_1": "These phenomena occur because right at the phase transition temperature, the driving force is not sufficient to cause the transformation to occur. The driving force is enhanced during superheating or supercooling.",
        "wrong_answers_2": "In order for the material to perform well, the two materials must have very different coefficients of thermal expansion and should have high enough modulus of elasticity so that no permanent deformation of the material occurs.",
        "wrong_answers_3": "Increasing the temperature should increase the diffusivity and therefore increase the diffusion flux and flow rate."
    },
    {
        "idx": 5,
        "question": "How would you manufacture a hammer head?",
        "answer": "The head for a carpenter's hammer is produced by forging, a metalworking process; a simple steel shape is heated and formed in several steps while hot into the required shape. The head is then heat treated to produce the required mechanical and physical properties.",
        "type": 4,
        "wrong_answers_1": "The principal disadvantage of hot-isostatic pressing is that it is expensive. The pressure is applied on a pre-formed green piece by a gas. Thus, the process is slow, and the equipment required to supply the gas and withstand the elevated temperature and pressure is costly.",
        "wrong_answers_2": "Bimetallic materials are produced by bonding two materials having different coefficients of thermal expansion to one another, forming a laminar composite. When the temperature changes, one of the materials will expand or contract more than the other material. This difference in expansion or contraction causes the bimetallic material to change shape; if the original shape is that of a coil, then the device will coil or uncoil, depending on the direction of the temperature change.",
        "wrong_answers_3": "The purpose of a spheroidizing heat treatment is to produce a very soft and ductile steel alloy having a spheroiditic microstructure."
    },
    {
        "idx": 6,
        "question": "Aluminum foil used for storing food weighs about 0.3g per square inch. How many atoms of aluminum are contained in one square inch of foil?",
        "answer": "6.69 × 10^{21} atoms",
        "type": 1,
        "wrong_answers_1": "7.31 × 10^{21} atoms",
        "wrong_answers_2": "6.25 × 10^{21} atoms",
        "wrong_answers_3": "5.69 × 10^{21} atoms"
    },
    {
        "idx": 7,
        "question": "Using the densities and atomic weights, calculate the number of atoms per cubic centimeter in lead.",
        "answer": "3.3 × 10^22 atoms/cm^3",
        "type": 1,
        "wrong_answers_1": "3.8 × 10^22 atoms/cm^3",
        "wrong_answers_2": "3.6 × 10^22 atoms/cm^3",
        "wrong_answers_3": "3.1 × 10^22 atoms/cm^3"
    },
    {
        "idx": 8,
        "question": "Using the densities and atomic weights, calculate the number of atoms per cubic centimeter in lithium.",
        "answer": "4.63 × 10^22 atoms/cm^3",
        "type": 1,
        "wrong_answers_1": "4.17 × 10^22 atoms/cm^3",
        "wrong_answers_2": "4.20 × 10^22 atoms/cm^3",
        "wrong_answers_3": "4.90 × 10^22 atoms/cm^3"
    },
    {
        "idx": 9,
        "question": "Using data, calculate the number of iron atoms in one ton (2000 pounds).",
        "answer": "9.79 × 10^{27} atoms/ton",
        "type": 1,
        "wrong_answers_1": "10.09 × 10^{27} atoms/ton",
        "wrong_answers_2": "10.80 × 10^{27} atoms/ton",
        "wrong_answers_3": "8.59 × 10^{27} atoms/ton"
    },
    {
        "idx": 10,
        "question": "Using data, calculate the volume in cubic centimeters occupied by one mole of boron.",
        "answer": "4.7 cm^{3}",
        "type": 1,
        "wrong_answers_1": "4.5 cm^{3}",
        "wrong_answers_2": "5.2 cm^{3}",
        "wrong_answers_3": "4.3 cm^{3}"
    },
    {
        "idx": 11,
        "question": "In order to plate a steel part having a surface area of 200 in.^2 with a 0.002 in. thick layer of nickel, how many atoms of nickel are required?",
        "answer": "5.98 x 10^23 atoms",
        "type": 1,
        "wrong_answers_1": "5.50 x 10^23 atoms",
        "wrong_answers_2": "6.38 x 10^23 atoms",
        "wrong_answers_3": "5.43 x 10^23 atoms"
    },
    {
        "idx": 12,
        "question": "In order to plate a steel part having a surface area of 200 in.^2 with a 0.002 in. thick layer of nickel, how many moles of nickel are required?",
        "answer": "0.994 mol ni required",
        "type": 1,
        "wrong_answers_1": "0.888 mol ni required",
        "wrong_answers_2": "0.921 mol ni required",
        "wrong_answers_3": "1.134 mol ni required"
    },
    {
        "idx": 13,
        "question": "Suppose an element has a valence of 2 and an atomic number of 27 . Based only on the quantum numbers, how many electrons must be present in the 3 d energy level?",
        "answer": "7 electrons.",
        "type": 1,
        "wrong_answers_1": "8 electrons.",
        "wrong_answers_2": "6 electrons.",
        "wrong_answers_3": "5 electrons."
    },
    {
        "idx": 14,
        "question": "Calculate the fraction of bonding of MgO that is ionic.",
        "answer": "the fraction of bonding of mgo that is ionic is 0.734. so bonding is mostly ionic.",
        "type": 1,
        "wrong_answers_1": "the fraction of bonding of mgo that is ionic is 0.808. so bonding is mostly ionic.",
        "wrong_answers_2": "the fraction of bonding of mgo that is ionic is 0.654. so bonding is mostly ionic.",
        "wrong_answers_3": "the fraction of bonding of mgo that is ionic is 0.778. so bonding is mostly ionic."
    },
    {
        "idx": 15,
        "question": "Would you expect MgO or magnesium to have the higher modulus of elasticity? Explain.",
        "answer": "( MgO has ionic bonds, which are strong compared to the metallic bonds in Mg. A higher force will be required to cause the same separation between the ions in MgO compared to the atoms in Mg. Therefore, MgO should have the higher modulus of elasticity. In Mg, E=6 × 10^{6} psi; in MgO, E=30 × 10^{6} psi.",
        "type": 4,
        "wrong_answers_1": "For BaO, the ionic radii of the Mg2+ and Ba2+ are 0.072 nm and 0.136 nm, respectively. The percentage difference in ionic radii is 47%, which is much larger than the acceptable range. Therefore, BaO is not expected to experience any appreciable solubility in MgO. Experimentally, the solubility of BaO in MgO is very small.",
        "wrong_answers_2": "This maximum temperature is possible for pure MgO.",
        "wrong_answers_3": "the diffusion rate of oxygen ions in \\mathrm{al}_{2}\\mathrm{o}_{3} at 1500^{\\circ}C is 3.47 × 10^{-16} {cm}^{2}/s, while the diffusion rate of aluminum ions is 2.48 × 10^{-13} {cm}^{2}/s. the difference in diffusion rates is due to the ionic radii: the oxygen ion has a larger radius (1.32 Å) compared to the aluminum ion (0.51 Å), making it easier for the smaller aluminum ion to diffuse in the ceramic."
    },
    {
        "idx": 16,
        "question": "Calculate the atomic radius in cm for a BCC metal with a0=0.3294 nm and one atom per lattice point.",
        "answer": "1.426 x 10^-8 cm",
        "type": 1,
        "wrong_answers_1": "1.477 x 10^-8 cm",
        "wrong_answers_2": "1.292 x 10^-8 cm",
        "wrong_answers_3": "1.234 x 10^-8 cm"
    },
    {
        "idx": 17,
        "question": "Calculate the atomic radius in cm for an FCC metal with a0=4.0862 A and one atom per lattice point.",
        "answer": "1.4447 x 10^-8 cm",
        "type": 1,
        "wrong_answers_1": "1.3762 x 10^-8 cm",
        "wrong_answers_2": "1.5184 x 10^-8 cm",
        "wrong_answers_3": "1.3356 x 10^-8 cm"
    },
    {
        "idx": 18,
        "question": "Determine the crystal structure for a metal with a0=4.9489 Å, r=1.75 Å and one atom per lattice point.",
        "answer": "fcc",
        "type": 1
    },
    {
        "idx": 19,
        "question": "Determine the crystal structure for a metal with a0=0.42906 nm, r=0.1858 nm and one atom per lattice point.",
        "answer": "bcc",
        "type": 1
    },
    {
        "idx": 20,
        "question": "The density of potassium, which has the BCC structure and one atom per lattice point, is 0.855 g/cm^3. The atomic weight of potassium is 39.09 g/mol. Calculate the lattice parameter.",
        "answer": "5.3355 x 10^-8 cm",
        "type": 1,
        "wrong_answers_1": "5.7952 x 10^-8 cm",
        "wrong_answers_2": "5.0122 x 10^-8 cm",
        "wrong_answers_3": "5.1357 x 10^-8 cm"
    },
    {
        "idx": 21,
        "question": "The density of potassium, which has the BCC structure and one atom per lattice point, is 0.855 g/cm^3. The atomic weight of potassium is 39.09 g/mol. Calculate the atomic radius of potassium.",
        "answer": "2.3103 x 10^-8 cm",
        "type": 1,
        "wrong_answers_1": "2.1414 x 10^-8 cm",
        "wrong_answers_2": "1.9656 x 10^-8 cm",
        "wrong_answers_3": "2.6209 x 10^-8 cm"
    },
    {
        "idx": 22,
        "question": "The density of thorium, which has the FCC structure and one atom per lattice point, is 11.72 g/cm³. The atomic weight of thorium is 232 g/mol. Calculate the lattice parameter.",
        "answer": "5.0856 x 10⁻⁸ cm",
        "type": 1,
        "wrong_answers_1": "5.6034 x 10⁻⁸ cm",
        "wrong_answers_2": "5.4519 x 9⁻⁸ cm",
        "wrong_answers_3": "4.6982 x 9⁻⁸ cm"
    },
    {
        "idx": 23,
        "question": "The density of thorium, which has the FCC structure and one atom per lattice point, is 11.72 g/cm³. The atomic weight of thorium is 232 g/mol. Calculate the atomic radius of thorium.",
        "answer": "1.7980 x 10⁻⁸ cm",
        "type": 1,
        "wrong_answers_1": "1.9866 x 9⁻⁸ cm",
        "wrong_answers_2": "1.7037 x 9⁻⁸ cm",
        "wrong_answers_3": "1.7299 x 11⁻⁸ cm"
    },
    {
        "idx": 24,
        "question": "A metal having a cubic structure has a density of 2.6g / {cm}^{3}, an atomic weight of 87.62g / mol, and a lattice parameter of 6.0849 Å. One atom is associated with each lattice point. Determine the crystal structure of the metal.",
        "answer": "fcc.",
        "type": 1
    },
    {
        "idx": 25,
        "question": "A metal having a cubic structure has a density of 1.892g / {cm}^{3}, an atomic weight \\mathrm{of} 132.91g / mol, and a lattice parameter of € 1.13 Å. One atom is associated with each lattice point. Determine the crystal structure of the \\mathrm{metal}.",
        "answer": "bcc.",
        "type": 1
    },
    {
        "idx": 26,
        "question": "Indium has a tetragonal structure with a_{0}=0.32517nm and c_{0}=0.49459nm. The density is 7.286g / {cm}^{3} and the atomic weight is 114.82g / mol. Does indium have the simple tetragonal or body-centered tetragonal structure?",
        "answer": "body-centered tetragonal structure.",
        "type": 1
    },
    {
        "idx": 27,
        "question": "Bismuth has a hexagonal structure, with a0=0.4546 nm and c0=1.186 nm. The density is 9.808 g/cm^3 and the atomic weight is 208.98 g/mol. Determine the volume of the unit cell.",
        "answer": "0.21226 nm3 or 2.1226 x 10^-22 cm^3",
        "type": 1,
        "wrong_answers_1": "0.19131 nm3 or 1.9735 x 10^-22 cm^3",
        "wrong_answers_2": "0.19721 nm3 or 2.1989 x 10^-22 cm^3",
        "wrong_answers_3": "0.18494 nm3 or 2.0344 x 10^-22 cm^3"
    },
    {
        "idx": 28,
        "question": "Bismuth has a hexagonal structure, with a0=0.4546 nm and c0=1.186 nm. The density is 9.808 g/cm^3 and the atomic weight is 208.98 g/mol. Determine the number of atoms in each unit cell.",
        "answer": "6 atoms/cell",
        "type": 1,
        "wrong_answers_1": "7 atoms/cell",
        "wrong_answers_2": "8 atoms/cell",
        "wrong_answers_3": "9 atoms/cell"
    },
    {
        "idx": 29,
        "question": "Gallium has an orthorhombic structure, with a0=0.45258 nm, b0=0.45186 nm, and c0=0.76570 nm. The atomic radius is 0.1218 nm. The density is 5.904 g/cm^3 and the atomic weight is 69.72 g/mol. Determine the number of atoms in each unit cell.",
        "answer": "8 atoms/cell.",
        "type": 1,
        "wrong_answers_1": "4 atoms/cell.",
        "wrong_answers_2": "5 atoms/cell.",
        "wrong_answers_3": "6 atoms/cell."
    },
    {
        "idx": 30,
        "question": "Gallium has an orthorhombic structure, with a0=0.45258 nm, b0=0.45186 nm, and c0=0.76570 nm. The atomic radius is 0.1218 nm. The density is 5.904 g/cm^3 and the atomic weight is 69.72 g/mol. Determine the packing factor in the unit cell.",
        "answer": "0.387.",
        "type": 1,
        "wrong_answers_1": "0.349.",
        "wrong_answers_2": "0.351.",
        "wrong_answers_3": "0.363."
    },
    {
        "idx": 31,
        "question": "Beryllium has a hexagonal crystal structure, with a0=0.22858 nm and c0=0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 g/cm^3, and the atomic weight is 9.01 g/mol. Determine the number of atoms in each unit cell.",
        "answer": "2 atoms/cell.",
        "type": 1,
        "wrong_answers_1": "1 atoms/cell.",
        "wrong_answers_2": "3 atoms/cell.",
        "wrong_answers_3": "4 atoms/cell."
    },
    {
        "idx": 32,
        "question": "Beryllium has a hexagonal crystal structure, with a0=0.22858 nm and c0=0.35842 nm. The atomic radius is 0.1143 nm, the density is 1.848 g/cm^3, and the atomic weight is 9.01 g/mol. Determine the packing factor in the unit cell.",
        "answer": "0.77",
        "type": 1,
        "wrong_answers_1": "0.88",
        "wrong_answers_2": "0.66",
        "wrong_answers_3": "0.74"
    },
    {
        "idx": 33,
        "question": "A typical paper clip weighs 0.59g and consists of BCC iron. Calculate the number of unit cells in the paper clip.",
        "answer": "3.185 × 10^21 cells",
        "type": 1,
        "wrong_answers_1": "3.468 × 10^21 cells",
        "wrong_answers_2": "2.767 × 10^21 cells",
        "wrong_answers_3": "3.425 × 10^21 cells"
    },
    {
        "idx": 34,
        "question": "A typical paper clip weighs 0.59g and consists of BCC iron. Calculate the number of iron atoms in the paper clip.",
        "answer": "6.37 × 10^21 atoms",
        "type": 1,
        "wrong_answers_1": "7.31 × 10^21 atoms",
        "wrong_answers_2": "5.42 × 10^21 atoms",
        "wrong_answers_3": "5.95 × 10^21 atoms"
    },
    {
        "idx": 35,
        "question": "Determine the planar density for BCC lithium in the (100) plane.",
        "answer": "0.0812 x 10^16 points/cm^2",
        "type": 1,
        "wrong_answers_1": "0.0911 x 10^16 points/cm^2",
        "wrong_answers_2": "0.0863 x 10^16 points/cm^2",
        "wrong_answers_3": "0.0891 x 10^16 points/cm^2"
    },
    {
        "idx": 36,
        "question": "Determine the packing fraction for BCC lithium in the (100) plane.",
        "answer": "0.589",
        "type": 1,
        "wrong_answers_1": "0.668",
        "wrong_answers_2": "0.633",
        "wrong_answers_3": "0.653"
    },
    {
        "idx": 37,
        "question": "Determine the planar density for BCC lithium in the (110) plane.",
        "answer": "0.1149 x 10^16 points/cm^2",
        "type": 1,
        "wrong_answers_1": "0.1084 x 10^16 points/cm^2",
        "wrong_answers_2": "0.1238 x 10^16 points/cm^2",
        "wrong_answers_3": "0.1315 x 10^16 points/cm^2"
    },
    {
        "idx": 38,
        "question": "Determine the packing fraction for BCC lithium in the (110) plane.",
        "answer": "0.833",
        "type": 1,
        "wrong_answers_1": "0.713",
        "wrong_answers_2": "0.764",
        "wrong_answers_3": "0.881"
    },
    {
        "idx": 39,
        "question": "Determine the planar density for BCC lithium in the (111) plane.",
        "answer": "0.0469 x 10^16 points/cm^2",
        "type": 1,
        "wrong_answers_1": "0.0429 x 10^16 points/cm^2",
        "wrong_answers_2": "0.0529 x 10^16 points/cm^2",
        "wrong_answers_3": "0.0505 x 10^16 points/cm^2"
    },
    {
        "idx": 40,
        "question": "Determine the packing fraction for BCC lithium in the (111) plane.",
        "answer": "packing fraction = 1/2 / 0.866 a0^2 sqrt(3) a0 / 4",
        "type": 1,
        "wrong_answers_1": "packing fraction = 1/2 / 0.754 a0^2 sqrt(3) a0 / 5",
        "wrong_answers_2": "packing fraction = 1/2 / 0.992 a0^2 sqrt(3) a0 / 4",
        "wrong_answers_3": "packing fraction = 1/2 / 0.951 a0^2 sqrt(3) a0 / 4"
    },
    {
        "idx": 41,
        "question": "Which, if any, of these planes is close packed in BCC lithium?",
        "answer": "there is no close-packed plane in bcc structures.",
        "type": 3
    },
    {
        "idx": 42,
        "question": "Suppose that FCC rhodium is produced as a 1-mm thick sheet, with the (111) plane parallel to the surface of the sheet. How many (111) interplanar spacings d_{111} thick is the sheet? See Appendix A for necessary data.",
        "answer": "4.563 × 10^{6}",
        "type": 1,
        "wrong_answers_1": "4.313 × 10^{6}",
        "wrong_answers_2": "4.915 × 10^{6}",
        "wrong_answers_3": "4.902 × 10^{6}"
    },
    {
        "idx": 43,
        "question": "In a FCC unit cell, how many d_{111} are present between the 0,0,0 point and the 1,1,1 point?",
        "answer": "3",
        "type": 1,
        "wrong_answers_1": "1",
        "wrong_answers_2": "2",
        "wrong_answers_3": "4"
    },
    {
        "idx": 44,
        "question": "Determine the minimum radius of an atom that will just fit into the tetrahedral interstitial site in FCC nickel.",
        "answer": "0.2797 Å",
        "type": 1,
        "wrong_answers_1": "0.2642 Å",
        "wrong_answers_2": "0.2514 Å",
        "wrong_answers_3": "0.2672 Å"
    },
    {
        "idx": 45,
        "question": "Determine the minimum radius of an atom that will just fit into the octahedral interstitial site in BCC lithium.",
        "answer": "0.629 Å",
        "type": 1,
        "wrong_answers_1": "0.711 Å",
        "wrong_answers_2": "0.658 Å",
        "wrong_answers_3": "0.699 Å"
    },
    {
        "idx": 46,
        "question": "What is the radius of an atom that will just fit into the octahedral site in FCC copper without disturbing the crystal structure?",
        "answer": "0.529 Å.",
        "type": 1,
        "wrong_answers_1": "0.552 Å.",
        "wrong_answers_2": "0.452 Å.",
        "wrong_answers_3": "0.577 Å."
    },
    {
        "idx": 47,
        "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{Y}_{2} \\mathrm{O}_{3}.",
        "answer": "6",
        "type": 1,
        "wrong_answers_1": "7",
        "wrong_answers_2": "9",
        "wrong_answers_3": "3"
    },
    {
        "idx": 48,
        "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{UO}_{2}.",
        "answer": "6",
        "type": 1,
        "wrong_answers_1": "5",
        "wrong_answers_2": "9",
        "wrong_answers_3": "3"
    },
    {
        "idx": 49,
        "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{BaO}.",
        "answer": "8",
        "type": 1,
        "wrong_answers_1": "4",
        "wrong_answers_2": "6",
        "wrong_answers_3": "9"
    },
    {
        "idx": 50,
        "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{Si}_{4} N_{4}.",
        "answer": "4",
        "type": 1,
        "wrong_answers_1": "1",
        "wrong_answers_2": "2",
        "wrong_answers_3": "3"
    },
    {
        "idx": 51,
        "question": "Using the ionic radii, determine the coordination number expected for \\mathrm{GeO}_{2}.",
        "answer": "4",
        "type": 1,
        "wrong_answers_1": "1",
        "wrong_answers_2": "2",
        "wrong_answers_3": "3"
    },
    {
        "idx": 52,
        "question": "Using the ionic radii, determine the coordination number expected for MnO.",
        "answer": "6",
        "type": 1,
        "wrong_answers_1": "5",
        "wrong_answers_2": "7",
        "wrong_answers_3": "8"
    },
    {
        "idx": 53,
        "question": "Using the ionic radii, determine the coordination number expected for MgS.",
        "answer": "6",
        "type": 1,
        "wrong_answers_1": "5",
        "wrong_answers_2": "7",
        "wrong_answers_3": "3"
    },
    {
        "idx": 54,
        "question": "Using the ionic radii, determine the coordination number expected for KBr.",
        "answer": "6",
        "type": 1,
        "wrong_answers_1": "5",
        "wrong_answers_2": "7",
        "wrong_answers_3": "8"
    },
    {
        "idx": 55,
        "question": "Would you expect NiO to have the cesium chloride, sodium chloride, or zinc blende structure?",
        "answer": "sodium chloride structure.",
        "type": 4,
        "wrong_answers_1": "fluorite structure.",
        "wrong_answers_2": "Yes, for these two polymers it is possible to decide. The linear and syndiotactic poly(vinyl chloride) is more likely to crystallize; the phenyl side-group for polystyrene is bulkier than the Cl side-group for poly(vinyl chloride). Syndiotactic and isotactic isomers are equally likely to crystallize.",
        "wrong_answers_3": "BeO has the zinc blende structure."
    },
    {
        "idx": 56,
        "question": "For NiO with sodium chloride structure, determine the lattice parameter.",
        "answer": "4.02 Å.",
        "type": 1,
        "wrong_answers_1": "4.30 Å.",
        "wrong_answers_2": "3.83 Å.",
        "wrong_answers_3": "4.57 Å."
    },
    {
        "idx": 57,
        "question": "For NiO with sodium chloride structure, determine the density.",
        "answer": "7.64 \\mathrm{g/cm}^3.",
        "type": 1,
        "wrong_answers_1": "7.13 \\mathrm{g/cm}^3.",
        "wrong_answers_2": "6.68 \\mathrm{g/cm}^3.",
        "wrong_answers_3": "6.95 \\mathrm{g/cm}^3."
    },
    {
        "idx": 58,
        "question": "For NiO with sodium chloride structure, determine the packing factor.",
        "answer": "0.678.",
        "type": 1,
        "wrong_answers_1": "0.768.",
        "wrong_answers_2": "0.654.",
        "wrong_answers_3": "0.592."
    },
    {
        "idx": 59,
        "question": "Would you expect \\mathrm{UO}_{2} to have the sodium chloride, zinc blende, or fluoride structure?",
        "answer": "fluorite structure.",
        "type": 4,
        "wrong_answers_1": "sodium chloride structure.",
        "wrong_answers_2": "BeO has the zinc blende structure.",
        "wrong_answers_3": "Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material."
    },
    {
        "idx": 60,
        "question": "For \\mathrm{UO}_{2} with fluorite structure, determine the lattice parameter.",
        "answer": "5.2885 Å.",
        "type": 1,
        "wrong_answers_1": "4.6547 Å.",
        "wrong_answers_2": "5.5499 Å.",
        "wrong_answers_3": "5.9023 Å."
    },
    {
        "idx": 61,
        "question": "For \\mathrm{UO}_{2} with fluorite structure, determine the density.",
        "answer": "12.13 {g/cm}^{3}.",
        "type": 1,
        "wrong_answers_1": "13.60 {g/cm}^{3}.",
        "wrong_answers_2": "13.09 {g/cm}^{3}.",
        "wrong_answers_3": "10.59 {g/cm}^{3}."
    },
    {
        "idx": 62,
        "question": "For \\mathrm{UO}_{2} with fluorite structure, determine the packing factor.",
        "answer": "The packing factor is pf = 0.624.",
        "type": 1,
        "wrong_answers_1": "The packing factor is pf = 0.532.",
        "wrong_answers_2": "The packing factor is pf = 0.707.",
        "wrong_answers_3": "The packing factor is pf = 0.584."
    },
    {
        "idx": 63,
        "question": "Would you expect BeO to have the sodium chloride, zinc blende, or fluorite structure?",
        "answer": "BeO has the zinc blende structure.",
        "type": 4,
        "wrong_answers_1": "For the cadmium-zinc couple, corrosion is possible, and zinc will corrode.",
        "wrong_answers_2": "The stacking sequence of close-packed planes of anions for the zinc blende crystal structure will be the same as FCC (and not HCP) because the anion packing is FCC.",
        "wrong_answers_3": "fluorite structure."
    },
    {
        "idx": 64,
        "question": "Based on the zinc blende structure of BeO, determine the lattice parameter.",
        "answer": "The lattice parameter is 3.8567 Å.",
        "type": 1,
        "wrong_answers_1": "The lattice parameter is 4.3121 Å.",
        "wrong_answers_2": "The lattice parameter is 3.4431 Å.",
        "wrong_answers_3": "The lattice parameter is 4.0850 Å."
    },
    {
        "idx": 65,
        "question": "Based on the zinc blende structure of BeO, determine the density.",
        "answer": "The density is 2.897 g/cm³.",
        "type": 1,
        "wrong_answers_1": "The density is 3.007 g/cm³.",
        "wrong_answers_2": "The density is 3.211 g/cm³.",
        "wrong_answers_3": "The density is 2.608 g/cm³."
    },
    {
        "idx": 66,
        "question": "Based on the zinc blende structure of BeO, determine the packing factor.",
        "answer": "0.684",
        "type": 1,
        "wrong_answers_1": "0.765",
        "wrong_answers_2": "0.720",
        "wrong_answers_3": "0.640"
    },
    {
        "idx": 67,
        "question": "Would you expect CsBr to have the sodium chloride, zinc blende, fluorite, or cesium chloride structure?",
        "answer": "CsBr is expected to have the cesium chloride structure.",
        "type": 2
    },
    {
        "idx": 68,
        "question": "Based on the cesium chloride structure, determine the lattice parameter for CsBr.",
        "answer": "4.1916 Å.",
        "type": 1,
        "wrong_answers_1": "4.7579 Å.",
        "wrong_answers_2": "3.6286 Å.",
        "wrong_answers_3": "3.7232 Å."
    },
    {
        "idx": 69,
        "question": "Based on the cesium chloride structure, determine the density for CsBr.",
        "answer": "The density ρ = 4.8 g/cm^3.",
        "type": 1,
        "wrong_answers_1": "The density ρ = 4.3 g/cm^3.",
        "wrong_answers_2": "The density ρ = 4.3 g/cm^3.",
        "wrong_answers_3": "The density ρ = 4.6 g/cm^3."
    },
    {
        "idx": 70,
        "question": "Based on the cesium chloride structure, determine the packing factor for CsBr.",
        "answer": "The packing factor pf = 0.693.",
        "type": 1,
        "wrong_answers_1": "The packing factor pf = 0.765.",
        "wrong_answers_2": "The packing factor pf = 0.748.",
        "wrong_answers_3": "The packing factor pf = 0.628."
    },
    {
        "idx": 71,
        "question": "Calculate the planar packing fraction (ppf) on the (110) plane for ZnS (with the zinc blende structure).",
        "answer": "0.492",
        "type": 1,
        "wrong_answers_1": "0.468",
        "wrong_answers_2": "0.554",
        "wrong_answers_3": "0.448"
    },
    {
        "idx": 72,
        "question": "Calculate the planar packing fraction (ppf) on the (110) plane for CaF2 (with the fluorite structure).",
        "answer": "0.699",
        "type": 1,
        "wrong_answers_1": "0.635",
        "wrong_answers_2": "0.793",
        "wrong_answers_3": "0.642"
    },
    {
        "idx": 73,
        "question": "MgO, which has the sodium chloride structure, has a lattice parameter of 0.396 nm. Determine the planar density and the planar packing fraction for the (111) plane of MgO. What ions are present on this plane?",
        "answer": "\n- planar density (p.d.): 0.1473 x 10^16 points/cm^2\n- planar packing fraction (ppf): 0.202\n- ions present: mg^2+ and o^2-",
        "type": 1,
        "wrong_answers_1": "\n- planar density (p.d.): 0.1661 x 10^16 points/cm^2\n- planar packing fraction (ppf): 0.220\n- ions present: mg^2+ and o^2-",
        "wrong_answers_2": "\n- planar density (p.d.): 0.1387 x 10^16 points/cm^2\n- planar packing fraction (ppf): 0.194\n- ions present: mg^2+ and o^2-",
        "wrong_answers_3": "\n- planar density (p.d.): 0.1616 x 10^16 points/cm^2\n- planar packing fraction (ppf): 0.219\n- ions present: mg^2+ and o^2-"
    },
    {
        "idx": 74,
        "question": "MgO, which has the sodium chloride structure, has a lattice parameter of 0.396 nm. Determine the planar density and the planar packing fraction for the (222) plane of MgO. What ions are present on this plane?",
        "answer": "\n- planar density (p.d.): 0.1473 x 10^16 points/cm^2\n- planar packing fraction (ppf): 0.806\n- ions present: mg^2+ and o^2-",
        "type": 1,
        "wrong_answers_1": "\n- planar density (p.d.): 0.1660 x 10^16 points/cm^2\n- planar packing fraction (ppf): 0.918\n- ions present: mg^2+ and o^2-",
        "wrong_answers_2": "\n- planar density (p.d.): 0.1383 x 10^16 points/cm^2\n- planar packing fraction (ppf): 0.714\n- ions present: mg^2+ and o^2-",
        "wrong_answers_3": "\n- planar density (p.d.): 0.1316 x 10^16 points/cm^2\n- planar packing fraction (ppf): 0.883\n- ions present: mg^2+ and o^2-"
    },
    {
        "idx": 75,
        "question": "A diffracted x-ray beam is observed from the (220) planes of iron at a 2 \\theta angle of 99.1^{\\circ} when x-rays of 0.15418nm wavelength are used. Calculate the lattice parameter of the iron.",
        "answer": "0.2865nm",
        "type": 1,
        "wrong_answers_1": "0.3205nm",
        "wrong_answers_2": "0.3130nm",
        "wrong_answers_3": "0.2643nm"
    },
    {
        "idx": 76,
        "question": "Calculate the number of vacancies per {cm}^{3} expected in copper at 1080^{\\circ} C (just below the melting temperature). The activation energy for vacancy formation is 20,000 cal/mol.",
        "answer": "4.97 × 10^{19} vacancies/cm^3",
        "type": 1,
        "wrong_answers_1": "5.63 × 10^{19} vacancies/cm^3",
        "wrong_answers_2": "5.42 × 10^{19} vacancies/cm^3",
        "wrong_answers_3": "4.71 × 10^{19} vacancies/cm^3"
    },
    {
        "idx": 77,
        "question": "The fraction of lattice points occupied by vacancies in solid aluminum at 660^{\\circ} C is 10^{-3}. What is the activation energy required to create vacancies in aluminum?",
        "answer": "12800 cal/mol",
        "type": 1,
        "wrong_answers_1": "11900 cal/mol",
        "wrong_answers_2": "11600 cal/mol",
        "wrong_answers_3": "11400 cal/mol"
    },
    {
        "idx": 78,
        "question": "The density of a sample of HCP beryllium is 1.844 g/cm³ and the lattice parameters are a₀=0.22858 nm and c₀=0.35842 nm. Calculate the fraction of the lattice points that contain vacancies.",
        "answer": "0.0008",
        "type": 1,
        "wrong_answers_1": "0.0009",
        "wrong_answers_2": "0.0010",
        "wrong_answers_3": "0.0007"
    },
    {
        "idx": 79,
        "question": "The density of a sample of HCP beryllium is 1.844 g/cm³ and the lattice parameters are a₀=0.22858 nm and c₀=0.35842 nm. Calculate the total number of vacancies in a cubic centimeter.",
        "answer": "0.986 × 10^20 vacancies/cm³",
        "type": 1,
        "wrong_answers_1": "0.930 × 10^20 vacancies/cm³",
        "wrong_answers_2": "1.030 × 10^20 vacancies/cm³",
        "wrong_answers_3": "1.060 × 10^20 vacancies/cm³"
    },
    {
        "idx": 80,
        "question": "BCC lithium has a lattice parameter of 3.5089 × 10^-8 cm and contains one vacancy per 200 unit cells. Calculate the number of vacancies per cubic centimeter.",
        "answer": "1.157 × 10^20 vacancies/cm³",
        "type": 1,
        "wrong_answers_1": "1.071 × 10^20 vacancies/cm³",
        "wrong_answers_2": "1.282 × 10^20 vacancies/cm³",
        "wrong_answers_3": "1.005 × 10^20 vacancies/cm³"
    },
    {
        "idx": 81,
        "question": "BCC lithium has a lattice parameter of 3.5089 × 10^-8 cm and contains one vacancy per 200 unit cells. Calculate the density of Li.",
        "answer": "0.532 g/cm³",
        "type": 1,
        "wrong_answers_1": "0.454 g/cm³",
        "wrong_answers_2": "0.586 g/cm³",
        "wrong_answers_3": "0.600 g/cm³"
    },
    {
        "idx": 82,
        "question": "FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate the density.",
        "answer": "the density is 11.335 g/cm^3.",
        "type": 1,
        "wrong_answers_1": "the density is 9.649 g/cm^3.",
        "wrong_answers_2": "the density is 10.676 g/cm^3.",
        "wrong_answers_3": "the density is 12.581 g/cm^3."
    },
    {
        "idx": 83,
        "question": "FCC lead has a lattice parameter of 0.4949 nm and contains one vacancy per 500 Pb atoms. Calculate the number of vacancies per gram of Pb.",
        "answer": "the number of vacancies per gram of pb is 5.82 × 10^18 vacancies/g.",
        "type": 1,
        "wrong_answers_1": "the number of vacancies per gram of pb is 6.28 × 10^18 vacancies/g.",
        "wrong_answers_2": "the number of vacancies per gram of pb is 6.18 × 10^18 vacancies/g.",
        "wrong_answers_3": "the number of vacancies per gram of pb is 6.47 × 10^18 vacancies/g."
    },
    {
        "idx": 84,
        "question": "A niobium alloy is produced by introducing tungsten substitutional atoms in the BCC structure; eventually an alloy is produced that has a lattice parameter of 0.32554nm and a density of 11.95g / {cm}^{3}. Calculate the fraction of the atoms in the alloy that are tungsten.",
        "answer": "0.345",
        "type": 1,
        "wrong_answers_1": "0.395",
        "wrong_answers_2": "0.380",
        "wrong_answers_3": "0.324"
    },
    {
        "idx": 85,
        "question": "Tin atoms are introduced into a FCC copper crystal, producing an alloy with a lattice parameter of 3.7589 × 10^{-8} cm and a density of 8.772g / {cm}^{3}. Calculate the atomic percentage of tin present in the alloy.",
        "answer": "11.95%.",
        "type": 1,
        "wrong_answers_1": "10.68%.",
        "wrong_answers_2": "13.17%.",
        "wrong_answers_3": "13.12%."
    },
    {
        "idx": 86,
        "question": "We replace 7.5 atomic percent of the chromium atoms in its BCC crystal with tantalum. X-ray diffraction shows that the lattice parameter is 0.29158nm. Calculate the density of the alloy.",
        "answer": "8.265g / {cm}^{3}",
        "type": 1,
        "wrong_answers_1": "7.908g / {cm}^{3}",
        "wrong_answers_2": "8.013g / {cm}^{3}",
        "wrong_answers_3": "8.733g / {cm}^{3}"
    },
    {
        "idx": 87,
        "question": "Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For the Fe-C alloy, find the density.",
        "answer": "7.89 g/cm³",
        "type": 1,
        "wrong_answers_1": "8.24 g/cm³",
        "wrong_answers_2": "8.43 g/cm³",
        "wrong_answers_3": "7.03 g/cm³"
    },
    {
        "idx": 88,
        "question": "Suppose we introduce one carbon atom for every 100 iron atoms in an interstitial position in BCC iron, giving a lattice parameter of 0.2867 nm. For the Fe-C alloy, find the packing factor.",
        "answer": "0.681",
        "type": 1,
        "wrong_answers_1": "0.774",
        "wrong_answers_2": "0.776",
        "wrong_answers_3": "0.771"
    },
    {
        "idx": 89,
        "question": "The density of BCC iron is 7.882 g/cm³ and the lattice parameter is 0.2886 nm when hydrogen atoms are introduced at interstitial positions. Calculate the atomic fraction of hydrogen atoms.",
        "answer": "0.004",
        "type": 1,
        "wrong_answers_1": "0.007",
        "wrong_answers_2": "0.006",
        "wrong_answers_3": "0.005"
    },
    {
        "idx": 90,
        "question": "The density of BCC iron is 7.882 g/cm³ and the lattice parameter is 0.2886 nm when hydrogen atoms are introduced at interstitial positions. Calculate the number of unit cells required on average that contain hydrogen atoms.",
        "answer": "123.5",
        "type": 1,
        "wrong_answers_1": "129.0",
        "wrong_answers_2": "115.2",
        "wrong_answers_3": "129.6"
    },
    {
        "idx": 91,
        "question": "Suppose one Schottky defect is present in every tenth unit cell of MgO. MgO has the sodium chloride crystal structure and a lattice parameter of 0.396 nm. Calculate the number of anion vacancies per cm^3.",
        "answer": "1.61 × 10^21 vacancies/cm^3.",
        "type": 1,
        "wrong_answers_1": "1.81 × 10^21 vacancies/cm^3.",
        "wrong_answers_2": "1.67 × 10^21 vacancies/cm^3.",
        "wrong_answers_3": "1.72 × 10^21 vacancies/cm^3."
    },
    {
        "idx": 92,
        "question": "Suppose one Schottky defect is present in every tenth unit cell of MgO. MgO has the sodium chloride crystal structure and a lattice parameter of 0.396 nm. Calculate the density of the ceramic.",
        "answer": "4.205 g/cm^3.",
        "type": 1,
        "wrong_answers_1": "4.833 g/cm^3.",
        "wrong_answers_2": "4.668 g/cm^3.",
        "wrong_answers_3": "3.733 g/cm^3."
    },
    {
        "idx": 93,
        "question": "ZnS has the zinc blende structure. If the density is 3.02 g/cm^3 and the lattice parameter is 0.59583 nm, determine the number of Schottky defects per unit cell.",
        "answer": "0.0535 defects per unit cell.",
        "type": 1,
        "wrong_answers_1": "0.0575 defects per unit cell.",
        "wrong_answers_2": "0.0505 defects per unit cell.",
        "wrong_answers_3": "0.0611 defects per unit cell."
    },
    {
        "idx": 94,
        "question": "ZnS has the zinc blende structure. If the density is 3.02 g/cm^3 and the lattice parameter is 0.59583 nm, determine the number of Schottky defects per cubic centimeter.",
        "answer": "2.517 × 10^20 defects per cm^3.",
        "type": 1,
        "wrong_answers_1": "2.426 × 10^20 defects per cm^3.",
        "wrong_answers_2": "2.682 × 10^20 defects per cm^3.",
        "wrong_answers_3": "2.286 × 10^20 defects per cm^3."
    },
    {
        "idx": 95,
        "question": "Calculate the length of the Burgers vector in BCC niobium",
        "answer": "b = 2.853 Å",
        "type": 1,
        "wrong_answers_1": "b = 3.263 Å",
        "wrong_answers_2": "b = 3.135 Å",
        "wrong_answers_3": "b = 2.614 Å"
    },
    {
        "idx": 96,
        "question": "Calculate the length of the Burgers vector in FCC silver",
        "answer": "b = 2.889 Å",
        "type": 1,
        "wrong_answers_1": "b = 3.165 Å",
        "wrong_answers_2": "b = 2.495 Å",
        "wrong_answers_3": "b = 2.524 Å"
    },
    {
        "idx": 97,
        "question": "Calculate the length of the Burgers vector in diamond cubic silicon",
        "answer": "b = 3.840 Å",
        "type": 1,
        "wrong_answers_1": "b = 3.705 Å",
        "wrong_answers_2": "b = 3.689 Å",
        "wrong_answers_3": "b = 3.547 Å"
    },
    {
        "idx": 98,
        "question": "A single crystal of an FCC metal is oriented so that the direction is parallel to an applied stress of 5000 psi. Calculate the resolved shear stress acting on the (111) slip plane in the [T] slip direction.",
        "answer": "the resolved shear stress acting on the (111) slip plane in the [T] slip direction is 0.",
        "type": 1,
        "wrong_answers_1": "the resolved shear stress acting on the (111) slip plane in the [T] slip direction is 1.",
        "wrong_answers_2": "the resolved shear stress acting on the (111) slip plane in the [T] slip direction is 2.",
        "wrong_answers_3": "the resolved shear stress acting on the (111) slip plane in the [T] slip direction is 3."
    },
    {
        "idx": 99,
        "question": "A single crystal of an FCC metal is oriented so that the direction is parallel to an applied stress of 5000 psi. Calculate the resolved shear stress acting on the (111) slip plane in the [T] slip direction.",
        "answer": "the resolved shear stress acting on the (111) slip plane in the [T] slip direction is 2040 psi (active).",
        "type": 1,
        "wrong_answers_1": "the resolved shear stress acting on the (111) slip plane in the [T] slip direction is 1810 psi (active).",
        "wrong_answers_2": "the resolved shear stress acting on the (111) slip plane in the [T] slip direction is 1770 psi (active).",
        "wrong_answers_3": "the resolved shear stress acting on the (111) slip plane in the [T] slip direction is 1760 psi (active)."
    },
    {
        "idx": 100,
        "question": "A single crystal of an FCC metal is oriented so that the direction is parallel to an applied stress of 5000 psi. Which slip system(s) will become active first?",
        "answer": "the slip systems that will become active first are λ011 and λ101.",
        "type": 4,
        "wrong_answers_1": "For slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples. Slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems.",
        "wrong_answers_2": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP.",
        "wrong_answers_3": "A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation motion (or slip) occurs."
    },
    {
        "idx": 102,
        "question": "A single crystal of a BCC metal is oriented so that the  direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 12,000 psi, calculate the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction on the (011) slip plane.",
        "answer": "the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction is: on the (011) slip plane: σ = 29,412 psi",
        "type": 1,
        "wrong_answers_1": "the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction is: on the (011) slip plane: σ = 33426 psi",
        "wrong_answers_2": "the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction is: on the (011) slip plane: σ = 30410 psi",
        "wrong_answers_3": "the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction is: on the (011) slip plane: σ = 33536 psi"
    },
    {
        "idx": 103,
        "question": "A single crystal of a BCC metal is oriented so that the  direction is parallel to the applied stress. If the critical resolved shear stress required for slip is 12,000 psi, calculate the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction on the (101) slip plane.",
        "answer": "the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction is: on the (101) slip plane: σ = 29,412 psi",
        "type": 1,
        "wrong_answers_1": "the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction is: on the (101) slip plane: σ = 33731 psi",
        "wrong_answers_2": "the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction is: on the (101) slip plane: σ = 27293 psi",
        "wrong_answers_3": "the magnitude of the applied stress required to cause slip to begin in the [1,-1,1] direction is: on the (101) slip plane: σ = 32904 psi"
    },
    {
        "idx": 104,
        "question": "The strength of titanium is found to be 65,000 psi when the grain size is 17 x 10^-6 m and 82,000 psi when the grain size is 0.8 x 10^-6 m. Determine the constants in the Hall-Petch equation.",
        "answer": "the constants in the hall-petch equation are k = 19.4 psi/sqrt(d) and σ_o = 60,290 psi.",
        "type": 1,
        "wrong_answers_1": "the constants in the hall-petch equation are k = 16.5 psi/sqrt(d) and σ_o = 56,089 psi.",
        "wrong_answers_2": "the constants in the hall-petch equation are k = 18.5 psi/sqrt(d) and σ_o = 51,469 psi.",
        "wrong_answers_3": "the constants in the hall-petch equation are k = 17.7 psi/sqrt(d) and σ_o = 58,014 psi."
    },
    {
        "idx": 105,
        "question": "The strength of titanium is found to be 65,000 psi when the grain size is 17 x 10^-6 m and 82,000 psi when the grain size is 0.8 x 10^-6 m. Determine the strength of the titanium when the grain size is reduced to 0.2 x 10^-6 m.",
        "answer": "the strength of the titanium when the grain size is reduced to 0.2 x 10^-6 m is 103,670 psi.",
        "type": 1,
        "wrong_answers_1": "the strength of the titanium when the grain size is reduced to 0.2 x 10^-6 m is 92,262 psi.",
        "wrong_answers_2": "the strength of the titanium when the grain size is reduced to 0.2 x 10^-6 m is 97,958 psi.",
        "wrong_answers_3": "the strength of the titanium when the grain size is reduced to 0.2 x 10^-6 m is 112,891 psi."
    },
    {
        "idx": 106,
        "question": "For an ASTM grain size number of 8, calculate the number of grains per square inch at a magnification of 100.",
        "answer": "128 grains/in.^2",
        "type": 1,
        "wrong_answers_1": "121 grains/in.^2",
        "wrong_answers_2": "134 grains/in.^2",
        "wrong_answers_3": "123 grains/in.^2"
    },
    {
        "idx": 107,
        "question": "For an ASTM grain size number of 8, calculate the number of grains per square inch with no magnification.",
        "answer": "1.28 × 10^6 grains/in.^2",
        "type": 1,
        "wrong_answers_1": "1.37 × 10^6 grains/in.^2",
        "wrong_answers_2": "1.45 × 10^6 grains/in.^2",
        "wrong_answers_3": "1.29 × 10^6 grains/in.^2"
    },
    {
        "idx": 108,
        "question": "Determine the ASTM grain size number if 20 grains/square inch are observed at a magnification of 400 .",
        "answer": "the astm grain size number is 9.3.",
        "type": 1,
        "wrong_answers_1": "the astm grain size number is 10.4.",
        "wrong_answers_2": "the astm grain size number is 8.4.",
        "wrong_answers_3": "the astm grain size number is 10.5."
    },
    {
        "idx": 109,
        "question": "Determine the ASTM grain size number if 25 grains/square inch are observed at a magnification of 50 .",
        "answer": "the astm grain size number is  n = 3.6",
        "type": 1,
        "wrong_answers_1": "the astm grain size number is  n = 4.0",
        "wrong_answers_2": "the astm grain size number is  n = 3.5",
        "wrong_answers_3": "the astm grain size number is  n = 3.9"
    },
    {
        "idx": 110,
        "question": "Calculate the angle \\theta of a small-angle grain boundary in FCC aluminum when the dislocations are 5000 Å apart. (See Figure 4-18 and equation in Problem 4-46.)",
        "answer": "\\theta = 0.0328^{\\circ}",
        "type": 1,
        "wrong_answers_1": "\\theta = 0.0355^{\\circ}",
        "wrong_answers_2": "\\theta = 0.0369^{\\circ}",
        "wrong_answers_3": "\\theta = 0.0291^{\\circ}"
    },
    {
        "idx": 111,
        "question": "Atoms are found to move from one lattice position to another at the rate of 5 × 10^{5} jumps/s at 400^{\\circ} C when the activation energy for their movement is 30,000 cal/mol. Calculate the jump rate at 750^{\\circ} C.\n\\section*{",
        "answer": "the jump rate at 750^{\\circ} C is 1.08 × 10^{9} jumps/s.",
        "type": 1,
        "wrong_answers_1": "the jump rate at 798^{\\circ} C is 1.15 × 10^{9} jumps/s.",
        "wrong_answers_2": "the jump rate at 851^{\\circ} C is 0.97 × 10^{9} jumps/s.",
        "wrong_answers_3": "the jump rate at 649^{\\circ} C is 1.12 × 10^{9} jumps/s."
    },
    {
        "idx": 112,
        "question": "The number of vacancies in a material is related to temperature by an Arrhenius equation. If the fraction of lattice points containing vacancies is 8 × 10^{-5} \\mathrm{at} 600^{\\circ} C, determine the fraction at 1000^{\\circ} C.",
        "answer": "the fraction at 1000^{\\circ} C is 0.00155.",
        "type": 1,
        "wrong_answers_1": "the fraction at 1000^{\\circ} C is 0.00139.",
        "wrong_answers_2": "the fraction at 1000^{\\circ} C is 0.00149.",
        "wrong_answers_3": "the fraction at 1000^{\\circ} C is 0.00169."
    },
    {
        "idx": 113,
        "question": "The diffusion coefficient for Cr+3 in Cr2O3 is 6x10^-15 cm^2/s at 727C and is 1x10^-9 cm^2/s at 1400C. Calculate the activation energy q.",
        "answer": "the activation energy q is 59,230 cal/mol.",
        "type": 1,
        "wrong_answers_1": "the activation energy q is 50,686 cal/mol.",
        "wrong_answers_2": "the activation energy q is 61,903 cal/mol.",
        "wrong_answers_3": "the activation energy q is 64,913 cal/mol."
    },
    {
        "idx": 114,
        "question": "The diffusion coefficient for Cr+3 in Cr2O3 is 6x10^-15 cm^2/s at 727C and is 1x10^-9 cm^2/s at 1400C. Calculate the constant d0.",
        "answer": "the constant d0 is 0.055 cm^2/s.",
        "type": 1,
        "wrong_answers_1": "the constant d0 is 0.057 cm^2/s.",
        "wrong_answers_2": "the constant d0 is 0.060 cm^2/s.",
        "wrong_answers_3": "the constant d0 is 0.048 cm^2/s."
    },
    {
        "idx": 115,
        "question": "A 0.2-mm thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contains 1 Sb atom per 10^8 Si atoms and the other surface contains 500 Sb atoms per 10^8 Si atoms. The lattice parameter for Si is 5.407 A. Calculate the concentration gradient in (a) atomic percent Sb per cm.",
        "answer": "-0.02495 at% sb/cm",
        "type": 1,
        "wrong_answers_1": "-0.02813 at% sb/cm",
        "wrong_answers_2": "-0.02860 at% sb/cm",
        "wrong_answers_3": "-0.02624 at% sb/cm"
    },
    {
        "idx": 116,
        "question": "A 0.2-mm thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contains 1 Sb atom per 10^8 Si atoms and the other surface contains 500 Sb atoms per 10^8 Si atoms. The lattice parameter for Si is 5.407 A. Calculate the concentration gradient in (b) Sb atoms/cm^3·cm.",
        "answer": "-1.246 × 10^19 sb atoms/cm^3·cm",
        "type": 1,
        "wrong_answers_1": "-1.193 × 10^19 sb atoms/cm^3·cm",
        "wrong_answers_2": "-1.340 × 10^19 sb atoms/cm^3·cm",
        "wrong_answers_3": "-1.072 × 10^19 sb atoms/cm^3·cm"
    },
    {
        "idx": 117,
        "question": "When a Cu-Zn alloy solidifies, one portion of the structure contains 25 atomic percent zinc and another portion 0.025mm away contains 20 atomic percent zinc. The lattice parameter for the FCC alloy is 3.63 x 10^-8 cm. Determine the concentration gradient in (a) atomic percent Zn per cm.",
        "answer": "-2000 at% zn/cm",
        "type": 1,
        "wrong_answers_1": "-1730 at% zn/cm",
        "wrong_answers_2": "-2090 at% zn/cm",
        "wrong_answers_3": "-2060 at% zn/cm"
    },
    {
        "idx": 118,
        "question": "When a Cu-Zn alloy solidifies, one portion of the structure contains 25 atomic percent zinc and another portion 0.025mm away contains 20 atomic percent zinc. The lattice parameter for the FCC alloy is 3.63 x 10^-8 cm. Determine the concentration gradient in (b) weight percent Zn per cm.",
        "answer": "-2032 wt% zn/cm",
        "type": 1,
        "wrong_answers_1": "-1922 wt% zn/cm",
        "wrong_answers_2": "-1882 wt% zn/cm",
        "wrong_answers_3": "-2272 wt% zn/cm"
    },
    {
        "idx": 119,
        "question": "A 0.001 -in. BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 650^{\\circ} C .5 × 10^{8} H atoms / {cm}^{3} are in equilibrium with the hot side of the foil, while 2 × 10^{3} H atoms / {cm}^{3} are in equilibrium with the cold side. Determine the concentration gradient of hydrogen.",
        "answer": "-1969 × 10^{8} h atoms/cm^3·cm",
        "type": 1,
        "wrong_answers_1": "-2128 × 10^{8} h atoms/cm^3·cm",
        "wrong_answers_2": "-1848 × 10^{8} h atoms/cm^3·cm",
        "wrong_answers_3": "-2218 × 10^{8} h atoms/cm^3·cm"
    },
    {
        "idx": 120,
        "question": "A 0.001 -in. BCC iron foil is used to separate a high hydrogen gas from a low hydrogen gas at 650^{\\circ} C .5 × 10^{8} H atoms / {cm}^{3} are in equilibrium with the hot side of the foil, while 2 × 10^{3} H atoms / {cm}^{3} are in equilibrium with the cold side. Determine the flux of hydrogen through the foil.",
        "answer": "0.33 × 10^{8} h atoms/cm^2·s",
        "type": 1,
        "wrong_answers_1": "0.31 × 10^{8} h atoms/cm^2·s",
        "wrong_answers_2": "0.35 × 10^{8} h atoms/cm^2·s",
        "wrong_answers_3": "0.28 × 10^{8} h atoms/cm^2·s"
    },
    {
        "idx": 121,
        "question": "A 1-mm sheet of FCC iron is used to contain nitrogen in a heat exchanger at 1200^{\\circ} C. The concentration of N at one surface is 0.04 atomic percent and the concentration at the second surface is 0.005 atomic percent. Determine the flux of nitrogen through the foil in N atoms / {cm}^{2}·s.",
        "answer": "the flux of nitrogen through the foil is 7.57 × 10^{12} n atoms/cm^2 \\cdot s.",
        "type": 1,
        "wrong_answers_1": "the flux of nitrogen through the foil is 8.54 × 10^{12} n atoms/cm^2 \\cdot s.",
        "wrong_answers_2": "the flux of nitrogen through the foil is 7.04 × 10^{12} n atoms/cm^2 \\cdot s.",
        "wrong_answers_3": "the flux of nitrogen through the foil is 6.71 × 10^{12} n atoms/cm^2 \\cdot s."
    },
    {
        "idx": 122,
        "question": "A 4-cm-diameter, 0.5-mm-thick spherical container made of BCC iron holds nitrogen at 700^{\\circ} C. The concentration at the inner surface is 0.05 atomic percent and at the outer surface is 0.002 atomic percent. Calculate the number of grams of nitrogen that are lost from the container per hour.",
        "answer": "the nitrogen lost from the container per hour is 1.245 × 10^{-3} g/h.",
        "type": 1,
        "wrong_answers_1": "the nitrogen lost from the container per hour is 1.290 × 10^{-3} g/h.",
        "wrong_answers_2": "the nitrogen lost from the container per hour is 1.086 × 10^{-3} g/h.",
        "wrong_answers_3": "the nitrogen lost from the container per hour is 1.141 × 10^{-3} g/h."
    },
    {
        "idx": 123,
        "question": "A BCC iron structure is to be manufactured that will allow no more than 50g of hydrogen to be lost per year through each square centimeter of the iron at 400^{\\circ} C. If the concentration of hydrogen at one surface is 0.05 H atom per unit cell and is 0.001 H atom per unit cell at the second surface, determine the minimum thickness of the iron.",
        "answer": "the minimum thickness of the iron is 0.179 cm.",
        "type": 1,
        "wrong_answers_1": "the minimum thickness of the iron is 0.191 cm.",
        "wrong_answers_2": "the minimum thickness of the iron is 0.155 cm.",
        "wrong_answers_3": "the minimum thickness of the iron is 0.183 cm."
    },
    {
        "idx": 124,
        "question": "Determine the maximum allowable temperature that will produce a flux of less than 2000 H atoms / {cm}^{2}·s through a BCC iron foil when the concentration gradient is -5 × 10^{16} atoms / {cm}^{3}·{cm}. (Note the negative sign for the flux.)",
        "answer": "the maximum allowable temperature is 75k (-198^{\\circ} C).",
        "type": 1,
        "wrong_answers_1": "the maximum allowable temperature is 65k (-211^{\\circ} C).",
        "wrong_answers_2": "the maximum allowable temperature is 82k (-182^{\\circ} C).",
        "wrong_answers_3": "the maximum allowable temperature is 86k (-191^{\\circ} C)."
    },
    {
        "idx": 125,
        "question": "Compare the rate at which oxygen ions diffuse in \\mathrm{Al}_{2} \\mathrm{O}_{3} with the rate at which aluminum ions diffuse in \\mathrm{Al}_{2} \\mathrm{O}_{3} at 1500^{\\circ} C. Explain the difference.",
        "answer": "the diffusion rate of oxygen ions in \\mathrm{al}_{2}\\mathrm{o}_{3} at 1500^{\\circ}C is 3.47 × 10^{-16} {cm}^{2}/s, while the diffusion rate of aluminum ions is 2.48 × 10^{-13} {cm}^{2}/s. the difference in diffusion rates is due to the ionic radii: the oxygen ion has a larger radius (1.32 Å) compared to the aluminum ion (0.51 Å), making it easier for the smaller aluminum ion to diffuse in the ceramic.",
        "type": 4,
        "wrong_answers_1": "diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.",
        "wrong_answers_2": "the diffusion coefficient of carbon in bcc iron at 912^{\\circ} C is 1.51 × 10^{-6} cm^{2} / s, while in fcc iron it is (1.92 × 10^{-7} cm^{2} / s). the packing factor of the bcc lattice (0.68) is less than that of the fcc lattice; consequently, atoms are expected to be able to diffuse more rapidly in the bcc iron.",
        "wrong_answers_3": "Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species--i.e., the diffusion flux is independent of time."
    },
    {
        "idx": 126,
        "question": "Compare the diffusion coefficients of carbon in BCC iron at the allotropic transformation temperature of 912^{\\circ} C and explain the difference.",
        "answer": "the diffusion coefficient of carbon in bcc iron at 912^{\\circ} C is 1.51 × 10^{-6} cm^{2} / s, while in fcc iron it is (1.92 × 10^{-7} cm^{2} / s). the packing factor of the bcc lattice (0.68) is less than that of the fcc lattice; consequently, atoms are expected to be able to diffuse more rapidly in the bcc iron.",
        "type": 4,
        "wrong_answers_1": "the diffusion rate of oxygen ions in \\mathrm{al}_{2}\\mathrm{o}_{3} at 1500^{\\circ}C is 3.47 × 10^{-16} {cm}^{2}/s, while the diffusion rate of aluminum ions is 2.48 × 10^{-13} {cm}^{2}/s. the difference in diffusion rates is due to the ionic radii: the oxygen ion has a larger radius (1.32 Å) compared to the aluminum ion (0.51 Å), making it easier for the smaller aluminum ion to diffuse in the ceramic.",
        "wrong_answers_2": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP.",
        "wrong_answers_3": "The stacking sequence of close-packed planes of anions for the zinc blende crystal structure will be the same as FCC (and not HCP) because the anion packing is FCC."
    },
    {
        "idx": 127,
        "question": "Iron containing 0.05 % C is heated to 912 degrees C in an atmosphere that produces 1.20 % C at the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if the iron is BCC.",
        "answer": "For BCC iron, the carbon content at 0.05 cm beneath the surface is 0.95 % C.",
        "type": 1,
        "wrong_answers_1": "For BCC iron, the carbon content at 0.05 cm beneath the surface is 1.04 % C.",
        "wrong_answers_2": "For BCC iron, the carbon content at 0.05 cm beneath the surface is 1.05 % C.",
        "wrong_answers_3": "For BCC iron, the carbon content at 0.05 cm beneath the surface is 0.99 % C."
    },
    {
        "idx": 128,
        "question": "Iron containing 0.05 % C is heated to 912 degrees C in an atmosphere that produces 1.20 % C at the surface and is held for 24 h. Calculate the carbon content at 0.05 cm beneath the surface if the iron is FCC.",
        "answer": "For FCC iron, the carbon content at 0.05 cm beneath the surface is 0.95 % C.",
        "type": 1,
        "wrong_answers_1": "For FCC iron, the carbon content at 0.04 cm beneath the surface is 0.91 % C.",
        "wrong_answers_2": "For FCC iron, the carbon content at 0.05 cm beneath the surface is 0.92 % C.",
        "wrong_answers_3": "For FCC iron, the carbon content at 0.04 cm beneath the surface is 0.99 % C."
    },
    {
        "idx": 129,
        "question": "What temperature is required to obtain 0.50% C at a distance of 0.5mm beneath the surface of a 0.20% C steel in 2h, when 1.10% C is present at the surface? Assume that the iron is FCC.",
        "answer": "the required temperature is 1180k or 907^{\\circ} C.",
        "type": 1,
        "wrong_answers_1": "the required temperature is 1140k or 777^{\\circ} C.",
        "wrong_answers_2": "the required temperature is 1230k or 993^{\\circ} C.",
        "wrong_answers_3": "the required temperature is 1360k or 944^{\\circ} C."
    },
    {
        "idx": 130,
        "question": "A 0.15% C steel is to be carburized at 1100^{\\circ} C, giving 0.35% C at a distance of 1mm beneath the surface. If the surface composition is maintained at 0.90% C, what time is required?",
        "answer": "the required time is 51 min.",
        "type": 1,
        "wrong_answers_1": "the required time is 48 min.",
        "wrong_answers_2": "the required time is 48 min.",
        "wrong_answers_3": "the required time is 44 min."
    },
    {
        "idx": 131,
        "question": "A 0.02% C steel is to be carburized at 1200^{\\circ} C in 4h, with a point 0.6mm beneath the surface reaching 0.45% C. Calculate the carbon content required at the surface of the steel.",
        "answer": "the carbon content required at the surface of the steel is 0.53% C.",
        "type": 1,
        "wrong_answers_1": "the carbon content required at the surface of the steel is 0.60% C.",
        "wrong_answers_2": "the carbon content required at the surface of the steel is 0.50% C.",
        "wrong_answers_3": "the carbon content required at the surface of the steel is 0.61% C."
    },
    {
        "idx": 132,
        "question": "A 1.2% C tool steel held at 1150^{\\circ} C is exposed to oxygen for 48h. The carbon content at the steel surface is zero. To what depth will the steel be decarburized to less than 0.20% C ?",
        "answer": "0.177 cm",
        "type": 1,
        "wrong_answers_1": "0.198 cm",
        "wrong_answers_2": "0.184 cm",
        "wrong_answers_3": "0.191 cm"
    },
    {
        "idx": 133,
        "question": "A 0.80% C steel must operate at 950^{\\circ} C in an oxidizing environment, where the carbon content at the steel surface is zero. Only the outermost 0.02 cm of the steel part can fall below 0.75% C. What is the maximum time that the steel part can operate?",
        "answer": "the maximum time that the steel part can operate is 2.9 min.",
        "type": 1,
        "wrong_answers_1": "the maximum time that the steel part can operate is 3.3 min.",
        "wrong_answers_2": "the maximum time that the steel part can operate is 3.2 min.",
        "wrong_answers_3": "the maximum time that the steel part can operate is 3.1 min."
    },
    {
        "idx": 134,
        "question": "A steel with \\mathrm{BBC} crystal structure containing 0.001%N is nitrided at 550^{\\circ} C for 5h. If the nitrogen content at the steel surface is 0.08%, determine the nitrogen content at 0.25mm from the surface.",
        "answer": "the nitrogen content at 0.25mm from the surface is 0.049% n.",
        "type": 1,
        "wrong_answers_1": "the nitrogen content at 0.23mm from the surface is 0.043% n.",
        "wrong_answers_2": "the nitrogen content at 0.27mm from the surface is 0.046% n.",
        "wrong_answers_3": "the nitrogen content at 0.28mm from the surface is 0.055% n."
    },
    {
        "idx": 135,
        "question": "What time is required to nitride a 0.002N steel to obtain 0.12%N at a distance of 0.002 in. beneath the surface at 625^{\\circ} C ? The nitrogen content at the surface is 0.15%.\n\\[\n\\begin{array}{l}\n",
        "answer": "12.8 min",
        "type": 1,
        "wrong_answers_1": "10.9 min",
        "wrong_answers_2": "12.1 min",
        "wrong_answers_3": "13.6 min"
    },
    {
        "idx": 136,
        "question": "We currently can successfully perform a carburizing heat treatment at 1200^{\\circ} C in 1h. In an effort to reduce the cost of the brick lining in our furnace, we propose to reduce the carburizing temperature to 950^{\\circ} C. What time will be required to give us a similar carburizing treatment?",
        "answer": "9.95 h",
        "type": 1,
        "wrong_answers_1": "9.35 h",
        "wrong_answers_2": "11.00 h",
        "wrong_answers_3": "11.25 h"
    },
    {
        "idx": 137,
        "question": "During freezing of a Cu-Zn alloy, we find that the composition is nonuniform. By heating the alloy to 600^{\\circ} C for 3 hours, diffusion of zinc helps to make the composition more uniform. What temperature would be required if we wished to perform this homogenization treatment in 30 minutes?",
        "answer": "667°c",
        "type": 1,
        "wrong_answers_1": "712°c",
        "wrong_answers_2": "693°c",
        "wrong_answers_3": "750°c"
    },
    {
        "idx": 138,
        "question": "A ceramic part made of MgO is sintered successfully at 1700 degrees C in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to 1500 degrees C. Which will limit the rate at which sintering can be done: diffusion of magnesium ions or diffusion of oxygen ions?",
        "answer": "diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.",
        "type": 4,
        "wrong_answers_1": "the diffusion rate of oxygen ions in \\mathrm{al}_{2}\\mathrm{o}_{3} at 1500^{\\circ}C is 3.47 × 10^{-16} {cm}^{2}/s, while the diffusion rate of aluminum ions is 2.48 × 10^{-13} {cm}^{2}/s. the difference in diffusion rates is due to the ionic radii: the oxygen ion has a larger radius (1.32 Å) compared to the aluminum ion (0.51 Å), making it easier for the smaller aluminum ion to diffuse in the ceramic.",
        "wrong_answers_2": "Increasing the concentration of oxygen in the ambient environment will increase the concentration gradient, thereby increasing the diffusion flux of oxygen.",
        "wrong_answers_3": "oxygen vacancies. Two Li+ ions added, a single oxygen vacancy is formed."
    },
    {
        "idx": 139,
        "question": "A ceramic part made of MgO is sintered successfully at 1700 degrees C in 90 minutes. To minimize thermal stresses during the process, we plan to reduce the temperature to 1500 degrees C. What time will be required at the lower temperature?",
        "answer": "the time required at the lower temperature is 955 min or 15.9 h.",
        "type": 1,
        "wrong_answers_1": "the time required at the lower temperature is 870 min or 13.5 h.",
        "wrong_answers_2": "the time required at the lower temperature is 887 min or 18.2 h.",
        "wrong_answers_3": "the time required at the lower temperature is 1070 min or 16.4 h."
    },
    {
        "idx": 140,
        "question": "A 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi and a tensile strength of 55,000 psi. Determine whether the wire will plastically deform.",
        "answer": "the wire will plastically deform (48,100 psi).",
        "type": 1,
        "wrong_answers_1": "the wire will plastically deform (54,189 psi).",
        "wrong_answers_2": "the wire will plastically deform (45,427 psi).",
        "wrong_answers_3": "the wire will plastically deform (42,300 psi)."
    },
    {
        "idx": 141,
        "question": "A 850-lb force is applied to a 0.15-in. diameter nickel wire having a yield strength of 45,000 psi and a tensile strength of 55,000 psi. Determine whether the wire will experience necking.",
        "answer": "no necking will occur (48,100 psi).",
        "type": 1,
        "wrong_answers_1": "no necking will occur (53,618 psi).",
        "wrong_answers_2": "no necking will occur (52,398 psi).",
        "wrong_answers_3": "no necking will occur (50,868 psi)."
    },
    {
        "idx": 142,
        "question": "A force of 100000 N is applied to a 10mm × 20mm iron bar having a yield strength of 400 MPa. Determine whether the bar will plastically deform.",
        "answer": "the bar will plastically deform because the applied stress of 500 MPa exceeds the yield strength of 400 MPa.",
        "type": 1,
        "wrong_answers_1": "the bar will plastically deform because the applied stress of 556 MPa exceeds the yield strength of 351 MPa.",
        "wrong_answers_2": "the bar will plastically deform because the applied stress of 460 MPa exceeds the yield strength of 354 MPa.",
        "wrong_answers_3": "the bar will plastically deform because the applied stress of 527 MPa exceeds the yield strength of 437 MPa."
    },
    {
        "idx": 143,
        "question": "A force of 100000 N is applied to a 10mm × 20mm iron bar having a tensile strength of 480 MPa. Determine whether the bar will experience necking.",
        "answer": "the bar will experience necking because the applied stress of 500 MPa exceeds the tensile strength of 480 MPa.",
        "type": 1,
        "wrong_answers_1": "the bar will experience necking because the applied stress of 439 MPa exceeds the tensile strength of 435 MPa.",
        "wrong_answers_2": "the bar will experience necking because the applied stress of 563 MPa exceeds the tensile strength of 501 MPa.",
        "wrong_answers_3": "the bar will experience necking because the applied stress of 479 MPa exceeds the tensile strength of 518 MPa."
    },
    {
        "idx": 144,
        "question": "Calculate the maximum force that a 0.2 -in. diameter rod of \\mathrm{Al}_{2} \\mathrm{O}_{3}, having a yield strength of 35,000 psi, can withstand with no plastic deformation. Express your answer in pounds and Newtons.",
        "answer": "the maximum force the rod can withstand is 1100 lb or 4891 n.",
        "type": 1,
        "wrong_answers_1": "the maximum force the rod can withstand is 1030 lb or 5481 n.",
        "wrong_answers_2": "the maximum force the rod can withstand is 1200 lb or 5161 n.",
        "wrong_answers_3": "the maximum force the rod can withstand is 1050 lb or 5351 n."
    },
    {
        "idx": 145,
        "question": "A force of 20,000 N will cause a 1 cm × 1 cm bar of magnesium to stretch from 10 cm to 10.045 cm. Calculate the modulus of elasticity in GPa.",
        "answer": "the modulus of elasticity is 44.4 GPa.",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity is 48.0 GPa.",
        "wrong_answers_2": "the modulus of elasticity is 49.2 GPa.",
        "wrong_answers_3": "the modulus of elasticity is 40.0 GPa."
    },
    {
        "idx": 146,
        "question": "A force of 20,000 N will cause a 1 cm × 1 cm bar of magnesium to stretch from 10 cm to 10.045 cm. Calculate the modulus of elasticity in psi.",
        "answer": "the modulus of elasticity is 6.44 × 10^6 psi.",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity is 6.13 × 10^6 psi.",
        "wrong_answers_2": "the modulus of elasticity is 6.75 × 10^6 psi.",
        "wrong_answers_3": "the modulus of elasticity is 7.32 × 10^6 psi."
    },
    {
        "idx": 147,
        "question": "A polymer bar's dimensions are 1 in . × 2 in. × 15 in. The polymer has a modulus of elasticity of 600,000 psi. What force is required to stretch the bar elastically to 15.25 in.?",
        "answer": "the force required is 200,000 lb.",
        "type": 1,
        "wrong_answers_1": "the force required is 207,151 lb.",
        "wrong_answers_2": "the force required is 206,945 lb.",
        "wrong_answers_3": "the force required is 212,776 lb."
    },
    {
        "idx": 148,
        "question": "An aluminum plate 0.5 cm thick is to withstand a force of 50,000N with no permanent deformation. If the aluminum has a yield strength of 125 MPa, what is the minimum width of the plate?",
        "answer": "the minimum width of the plate is 8 cm.",
        "type": 1,
        "wrong_answers_1": "the minimum width of the plate is 4 cm.",
        "wrong_answers_2": "the minimum width of the plate is 7 cm.",
        "wrong_answers_3": "the minimum width of the plate is 9 cm."
    },
    {
        "idx": 149,
        "question": "(a) A 3-in.-diameter rod of copper is to be reduced to a 2-in.-diameter rod by being pushed through an opening. To account for the elastic strain, what should be the diameter of the opening? The modulus of elasticity for the copper is 17 × 10^{6} psi and the yield strength is 40,000 psi.",
        "answer": "the diameter of the opening should be 1.995 in.",
        "type": 1,
        "wrong_answers_1": "the diameter of the opening should be 1.715 in.",
        "wrong_answers_2": "the diameter of the opening should be 1.788 in.",
        "wrong_answers_3": "the diameter of the opening should be 1.930 in."
    },
    {
        "idx": 150,
        "question": "A 0.4-in. diameter, 12-in-long titanium bar has a yield strength of 50,000 psi, a modulus of elasticity of 16x10^6 psi, and Poisson's ratio of 0.30. Determine the length of the bar when a 500-lb load is applied.",
        "answer": "the final length of the bar is 12.00298 in.",
        "type": 1,
        "wrong_answers_1": "the final length of the bar is 13.65057 in.",
        "wrong_answers_2": "the final length of the bar is 11.17320 in.",
        "wrong_answers_3": "the final length of the bar is 11.51570 in."
    },
    {
        "idx": 151,
        "question": "A 0.4-in. diameter, 12-in-long titanium bar has a yield strength of 50,000 psi, a modulus of elasticity of 16x10^6 psi, and Poisson's ratio of 0.30. Determine the diameter of the bar when a 500-lb load is applied.",
        "answer": "the final diameter of the bar is 0.39997 in.",
        "type": 1,
        "wrong_answers_1": "the final diameter of the bar is 0.34427 in.",
        "wrong_answers_2": "the final diameter of the bar is 0.34794 in.",
        "wrong_answers_3": "the final diameter of the bar is 0.35113 in."
    },
    {
        "idx": 152,
        "question": "A three-point bend test is performed on a block of ZrO2 that is 8 in. long, 0.50 in. wide, and 0.25 in. thick and is resting on two supports 4 in. apart. When a force of 400 lb is applied, the specimen deflects 0.037 in. and breaks. Calculate the flexural strength.",
        "answer": "flexural strength = 76,800 psi",
        "type": 1,
        "wrong_answers_1": "flexural strength = 84,652 psi",
        "wrong_answers_2": "flexural strength = 67,639 psi",
        "wrong_answers_3": "flexural strength = 85,221 psi"
    },
    {
        "idx": 153,
        "question": "A three-point bend test is performed on a block of ZrO2 that is 8 in. long, 0.50 in. wide, and 0.25 in. thick and is resting on two supports 4 in. apart. When a force of 400 lb is applied, the specimen deflects 0.037 in. and breaks. Calculate the flexural modulus, assuming that no plastic deformation occurs.",
        "answer": "flexural modulus = 22.14 x 10^6 psi",
        "type": 1,
        "wrong_answers_1": "flexural modulus = 24.99 x 10^6 psi",
        "wrong_answers_2": "flexural modulus = 20.21 x 10^6 psi",
        "wrong_answers_3": "flexural modulus = 23.99 x 10^6 psi"
    },
    {
        "idx": 154,
        "question": "A three-point bend test is performed on a block of silicon carbide that is 10 cm long, 1.5 cm wide, and 0.6 cm thick and is resting on two supports 7.5 cm apart. The sample breaks when a deflection of 0.09mm is recorded. Calculate the force that caused the fracture. The flexural modulus for silicon carbide is 480 GPa. Assume that no plastic deformation occurs.",
        "answer": "the force that caused the fracture is 1327 n.",
        "type": 1,
        "wrong_answers_1": "the force that caused the fracture is 1446 n.",
        "wrong_answers_2": "the force that caused the fracture is 1208 n.",
        "wrong_answers_3": "the force that caused the fracture is 1418 n."
    },
    {
        "idx": 155,
        "question": "A three-point bend test is performed on a block of silicon carbide that is 10 cm long, 1.5 cm wide, and 0.6 cm thick and is resting on two supports 7.5 cm apart. The sample breaks when a deflection of 0.09mm is recorded. Calculate the flexural strength. The flexural modulus for silicon carbide is 480 GPa. Assume that no plastic deformation occurs.",
        "answer": "the flexural strength is 276 mpa.",
        "type": 1,
        "wrong_answers_1": "the flexural strength is 260 mpa.",
        "wrong_answers_2": "the flexural strength is 253 mpa.",
        "wrong_answers_3": "the flexural strength is 248 mpa."
    },
    {
        "idx": 156,
        "question": "A thermosetting polymer containing glass beads is required to deflect 0.5mm when a force of 500 N is applied. The polymer part is 2 cm wide, 0.5 cm thick, and 10 cm long. If the flexural modulus is 6.9 GPa, determine the minimum distance between the supports.",
        "answer": "the minimum distance between the supports is 41 mm.",
        "type": 1,
        "wrong_answers_1": "the minimum distance between the supports is 43 mm.",
        "wrong_answers_2": "the minimum distance between the supports is 39 mm.",
        "wrong_answers_3": "the minimum distance between the supports is 46 mm."
    },
    {
        "idx": 157,
        "question": "Will the polymer fracture if its flexural strength is 85 MPa? Assume that no plastic deformation occurs.",
        "answer": "the applied stress is 61.5 MPa, which is less than the flexural strength of 85 MPa; the polymer is not expected to fracture.",
        "type": 4,
        "wrong_answers_1": "the certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials. since the calculated stress 379 mpa is relatively close to the flexural strength 390 mpa, there is some chance that fracture will occur.",
        "wrong_answers_2": "the certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials, and the calculated stress is relatively close to the flexural strength, so there is some chance that fracture will occur.",
        "wrong_answers_3": "A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time."
    },
    {
        "idx": 158,
        "question": "(b) The flexural modulus of alumina is 45 × 10^{6} psi and its flexural strength is 46,000 psi. A bar of alumina 0.3 in. thick, 1.0 in. wide, and 10 in. long is placed on supports 7 in. apart. Determine the amount of deflection at the moment the bar breaks, assuming that no plastic deformation occurs.",
        "answer": "the deflection at the moment the bar breaks is 0.0278 in.",
        "type": 1,
        "wrong_answers_1": "the deflection at the moment the bar breaks is 0.0260 in.",
        "wrong_answers_2": "the deflection at the moment the bar breaks is 0.0288 in.",
        "wrong_answers_3": "the deflection at the moment the bar breaks is 0.0286 in."
    },
    {
        "idx": 159,
        "question": "A Brinell hardness measurement, using a 10-mm-diameter indenter and a 500-kg load, produces an indentation of 4.5mm on an aluminum plate. Determine the Brinell hardness number (HB) of the metal.",
        "answer": " hb = 29.8 ",
        "type": 1,
        "wrong_answers_1": " hb = 31.6 ",
        "wrong_answers_2": " hb = 32.8 ",
        "wrong_answers_3": " hb = 33.7 "
    },
    {
        "idx": 160,
        "question": "When a 3000-kg load is applied to a 10-mm-diameter ball in a Brinell test of a steel, an indentation of 3.1mm is produced. Estimate the tensile strength of the steel.",
        "answer": "the tensile strength of the steel is 194,000 psi.",
        "type": 1,
        "wrong_answers_1": "the tensile strength of the steel is 204,107 psi.",
        "wrong_answers_2": "the tensile strength of the steel is 177,285 psi.",
        "wrong_answers_3": "the tensile strength of the steel is 175,303 psi."
    },
    {
        "idx": 161,
        "question": "A ceramic matrix composite contains internal flaws as large as 0.001 cm in length. The plane strain fracture toughness of the composite is 45 MPa / m and the tensile strength is 550 MPa. Will the flaw cause the composite to fail before the tensile strength is reached? Assume that f=1.",
        "answer": "the applied stress required for the crack to cause failure is 11,354 MPa. the tensile strength of the composite is 550 MPa. any failure of the ceramic should be expected due to the massive overload, not because of the presence of the flaws.",
        "type": 1,
        "wrong_answers_1": "the applied stress required for the crack to cause failure is 9934 MPa. the tensile strength of the composite is 530 MPa. any failure of the ceramic should be expected due to the massive overload, not because of the presence of the flaws.",
        "wrong_answers_2": "the applied stress required for the crack to cause failure is 10067 MPa. the tensile strength of the composite is 502 MPa. any failure of the ceramic should be expected due to the massive overload, not because of the presence of the flaws.",
        "wrong_answers_3": "the applied stress required for the crack to cause failure is 9692 MPa. the tensile strength of the composite is 502 MPa. any failure of the ceramic should be expected due to the massive overload, not because of the presence of the flaws."
    },
    {
        "idx": 162,
        "question": "An aluminum alloy that has a plane strain fracture toughness of 25,000 psi \\sqrt{m}. fails when a stress of 42,000 psi is applied. Observation of the fracture surface indicates that fracture began at the surface of the part. Estimate the size of the flaw that initiated fracture. Assume that f=1.1.",
        "answer": "the size of the flaw that initiated fracture is 0.093 in.",
        "type": 1,
        "wrong_answers_1": "the size of the flaw that initiated fracture is 0.102 in.",
        "wrong_answers_2": "the size of the flaw that initiated fracture is 0.105 in.",
        "wrong_answers_3": "the size of the flaw that initiated fracture is 0.088 in."
    },
    {
        "idx": 163,
        "question": "A polymer that contains internal flaws 1mm in length fails at a stress of 25 MPa. Determine the plane strain fracture toughness of the polymer. Assume that f=1.",
        "answer": "the plane strain fracture toughness of the polymer is 0.99 mpa \\sqrt{m}.",
        "type": 1,
        "wrong_answers_1": "the plane strain fracture toughness of the polymer is 0.87 mpa \\sqrt{m}.",
        "wrong_answers_2": "the plane strain fracture toughness of the polymer is 0.91 mpa \\sqrt{m}.",
        "wrong_answers_3": "the plane strain fracture toughness of the polymer is 0.94 mpa \\sqrt{m}."
    },
    {
        "idx": 164,
        "question": "A ceramic part for a jet engine has a yield strength of 75,000 psi and a plane strain fracture toughness of 5,000 psi / in. To be sure that the part does not fail, we plan to assure that the maximum applied stress is only one third the yield strength. We use a nondestructive test that will detect any internal flaws greater than 0.05 in. long. Assuming that f=1.4, does our nondestructive test have the required sensitivity? Explain.",
        "answer": "the length of internal flaws is 2a = 0.013 in. our nondestructive test can detect flaws as small as 0.05 in. long, which is not smaller than the critical flaw size required for failure. thus our ndt test is not satisfactory.",
        "type": 1,
        "wrong_answers_1": "the length of internal flaws is 2a = 0.013 in. our nondestructive test can detect flaws as small as 0.04 in. long, which is not smaller than the critical flaw size required for failure. thus our ndt test is not satisfactory.",
        "wrong_answers_2": "the length of internal flaws is 2a = 0.012 in. our nondestructive test can detect flaws as small as 0.05 in. long, which is not smaller than the critical flaw size required for failure. thus our ndt test is not satisfactory.",
        "wrong_answers_3": "the length of internal flaws is 2a = 0.015 in. our nondestructive test can detect flaws as small as 0.04 in. long, which is not smaller than the critical flaw size required for failure. thus our ndt test is not satisfactory."
    },
    {
        "idx": 165,
        "question": "To survive for one million cycles under conditions that provide for equal compressive and tensile stresses, what is the fatigue strength, or maximum stress amplitude, required?",
        "answer": "The fatigue strength at one million cycles is 22 mpa.",
        "type": 1,
        "wrong_answers_1": "The fatigue strength at one million cycles is 24 mpa.",
        "wrong_answers_2": "The fatigue strength at one million cycles is 20 mpa.",
        "wrong_answers_3": "The fatigue strength at one million cycles is 23 mpa."
    },
    {
        "idx": 166,
        "question": "What are the maximum stress, the minimum stress, and the mean stress on the part during its use under conditions that provide for equal compressive and tensile stresses?",
        "answer": "The maximum stress is +22 mpa, the minimum stress is -22 mpa, and the mean stress is 0 mpa.",
        "type": 1,
        "wrong_answers_1": "The maximum stress is +25 mpa, the minimum stress is -19 mpa, and the mean stress is 3 mpa.",
        "wrong_answers_2": "The maximum stress is +20 mpa, the minimum stress is -18 mpa, and the mean stress is 1 mpa.",
        "wrong_answers_3": "The maximum stress is +24 mpa, the minimum stress is -24 mpa, and the mean stress is 0 mpa."
    },
    {
        "idx": 167,
        "question": "What effect would the frequency of the stress application have on the fatigue strength, maximum stress, minimum stress, and mean stress?",
        "answer": "A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time.",
        "type": 4,
        "wrong_answers_1": "the applied stress is 61.5 MPa, which is less than the flexural strength of 85 MPa; the polymer is not expected to fracture.",
        "wrong_answers_2": "For polymer-matrix fiber-reinforced composites, three functions of the polymer-matrix phase are: (1) to bind the fibers together so that the applied stress is distributed among the fibers; (2) to protect the surface of the fibers from being damaged; and (3) to separate the fibers and inhibit crack propagation.",
        "wrong_answers_3": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration."
    },
    {
        "idx": 168,
        "question": "The activation energy for self-diffusion in copper is 49,300 cal / mol. A copper specimen creps as t 0.002 N / fn .-h when a stress of 15,000 psi is applied at 600^{\\circ} C. If the creep rate of copper is dependent on self-diffusion, determine the creep rate if the temperature is 800^{\\circ} C.",
        "answer": "the creep rate is 0.4, N / in·h.",
        "type": 1,
        "wrong_answers_1": "the creep rate is 0.2, N / in·h.",
        "wrong_answers_2": "the creep rate is 0.3, N / in·h.",
        "wrong_answers_3": "the creep rate is 0.5, N / in·h."
    },
    {
        "idx": 169,
        "question": "When a stress of 20000 psi is applied to a material heated to 900^{\\circ} C, rupture occurs in 25000h. If the activation energy for rupture is 35000 cal / mol, determine the rupture time if the temperature is reduced to 800^{\\circ} C.",
        "answer": "the rupture time at 800^{\\circ} C is 101660h.",
        "type": 1,
        "wrong_answers_1": "the rupture time at 713^{\\circ} C is 95683h.",
        "wrong_answers_2": "the rupture time at 741^{\\circ} C is 96691h.",
        "wrong_answers_3": "the rupture time at 731^{\\circ} C is 94584h."
    },
    {
        "question": "Using the data in Figure 7-27 for an iron-chromium-nickel alloy, determine the activation energy Q_{r} for rupture in the temperature range 980 to 1090^{\\circ} C.",
        "answer": "the activation energy q_{r} is 117,000 cal/mol.",
        "idx": 170,
        "type": 1,
        "wrong_answers_1": "the activation energy q_{r} is 106,782 cal/mol.",
        "wrong_answers_2": "the activation energy q_{r} is 112,580 cal/mol.",
        "wrong_answers_3": "the activation energy q_{r} is 100,477 cal/mol."
    },
    {
        "question": "Using the data in Figure 7-27 for an iron-chromium-nickel alloy, determine the constant m for rupture in the temperature range 980 to 1090^{\\circ} C.",
        "answer": "the constant m for rupture is 3.9.",
        "idx": 171,
        "type": 1,
        "wrong_answers_1": "the constant m for rupture is 4.2.",
        "wrong_answers_2": "the constant m for rupture is 3.7.",
        "wrong_answers_3": "the constant m for rupture is 4.3."
    },
    {
        "idx": 172,
        "question": "A 0.25-in.-thick copper plate is to be cold worked 63%. Find the final thickness.",
        "answer": " t_{f} = 0.0925 in.",
        "type": 1,
        "wrong_answers_1": " t_{f} = 0.1004 in.",
        "wrong_answers_2": " t_{f} = 0.1049 in.",
        "wrong_answers_3": " t_{f} = 0.0876 in."
    },
    {
        "idx": 173,
        "question": "A 0.25-in.-diameter copper bar is to be cold worked 63%. Find the final diameter.",
        "answer": "the final diameter is 0.152 in.",
        "type": 1,
        "wrong_answers_1": "the final diameter is 0.145 in.",
        "wrong_answers_2": "the final diameter is 0.160 in.",
        "wrong_answers_3": "the final diameter is 0.158 in."
    },
    {
        "idx": 174,
        "question": "A 2-in.-diameter copper rod is reduced to 1.5 in. diameter, then reduced again to a final diameter of 1 in. In a second case, the 2-in.-diameter rod is reduced in one step from 2 in. to a 1 in. diameter. Calculate the% CW for both cases.",
        "answer": "75% in both cases.",
        "type": 1,
        "wrong_answers_1": "66% in both cases.",
        "wrong_answers_2": "78% in both cases.",
        "wrong_answers_3": "84% in both cases."
    },
    {
        "idx": 175,
        "question": "Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate the critical radius of the nucleus required. Assume that the lattice parameter of the solid FCC nickel is 0.356 nm.",
        "answer": "the critical radius of the nucleus required is 6.65 × 10^-8 cm.",
        "type": 1,
        "wrong_answers_1": "the critical radius of the nucleus required is 7.29 × 10^-8 cm.",
        "wrong_answers_2": "the critical radius of the nucleus required is 6.34 × 10^-8 cm.",
        "wrong_answers_3": "the critical radius of the nucleus required is 7.17 × 10^-8 cm."
    },
    {
        "idx": 176,
        "question": "Suppose that liquid nickel is undercooled until homogeneous nucleation occurs. Calculate the number of nickel atoms in the nucleus. Assume that the lattice parameter of the solid FCC nickel is 0.356 nm.",
        "answer": "the number of nickel atoms in the nucleus is 109 atoms.",
        "type": 1,
        "wrong_answers_1": "the number of nickel atoms in the nucleus is 94 atoms.",
        "wrong_answers_2": "the number of nickel atoms in the nucleus is 123 atoms.",
        "wrong_answers_3": "the number of nickel atoms in the nucleus is 121 atoms."
    },
    {
        "idx": 177,
        "question": "Suppose that liquid iron is undercooled until homogeneous nucleation occurs. Calculate the critical radius of the nucleus required. Assume that the lattice parameter of the solid BCC iron is 2.92 Å.",
        "answer": "10.128 × 10⁻⁸ cm",
        "type": 1,
        "wrong_answers_1": "8.765 × 11⁻⁸ cm",
        "wrong_answers_2": "8.773 × 11⁻⁸ cm",
        "wrong_answers_3": "9.657 × 10⁻⁸ cm"
    },
    {
        "idx": 178,
        "question": "Suppose that liquid iron is undercooled until homogeneous nucleation occurs. Calculate the number of iron atoms in the nucleus. Assume that the lattice parameter of the solid BCC iron is 2.92 Å.",
        "answer": "350 atoms",
        "type": 1,
        "wrong_answers_1": "398 atoms",
        "wrong_answers_2": "402 atoms",
        "wrong_answers_3": "398 atoms"
    },
    {
        "idx": 179,
        "question": "Suppose that solid nickel was able to nucleate homogeneously with an undercooling of only 22^{\\circ} C. How many atoms would have to group together spontaneously for this to occur? Assume that the lattice parameter of the solid FCC nickel is 0.356nm.",
        "answer": "1.136 × 10^{6}.",
        "type": 1,
        "wrong_answers_1": "0.969 × 10^{6}.",
        "wrong_answers_2": "1.268 × 10^{6}.",
        "wrong_answers_3": "1.274 × 10^{6}."
    },
    {
        "idx": 180,
        "question": "Suppose that solid iron was able to nucleate homogeneously with an undercooling of only 15^{\\circ} C. How many atoms would have to group together spontaneously for this to occur? Assume that the lattice parameter of the solid BCC iron is 2.92 Å.",
        "answer": "7.676 × 10^{6} atoms.",
        "type": 1,
        "wrong_answers_1": "8.703 × 10^{6} atoms.",
        "wrong_answers_2": "6.884 × 10^{6} atoms.",
        "wrong_answers_3": "7.397 × 10^{6} atoms."
    },
    {
        "idx": 181,
        "question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates at 10 degrees Celsius undercooling. The specific heat of iron is 5.78 J/cm^3·°C.",
        "answer": "0.0333",
        "type": 1,
        "wrong_answers_1": "0.0376",
        "wrong_answers_2": "0.0354",
        "wrong_answers_3": "0.0377"
    },
    {
        "idx": 182,
        "question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates at 100 degrees Celsius undercooling. The specific heat of iron is 5.78 J/cm^3·°C.",
        "answer": "0.333",
        "type": 1,
        "wrong_answers_1": "0.302",
        "wrong_answers_2": "0.365",
        "wrong_answers_3": "0.323"
    },
    {
        "idx": 183,
        "question": "Calculate the fraction of solidification that occurs dendritically when iron nucleates homogeneously. The specific heat of iron is 5.78 J/cm^3·°C.",
        "answer": "all dendritically.",
        "type": 1
    },
    {
        "idx": 184,
        "question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates at 10 degrees Celsius undercooling. The specific heat of silver is 3.25 J/cm^3·°C.",
        "answer": "0.0237",
        "type": 1,
        "wrong_answers_1": "0.0202",
        "wrong_answers_2": "0.0247",
        "wrong_answers_3": "0.0215"
    },
    {
        "idx": 185,
        "question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates at 100 degrees Celsius undercooling. The specific heat of silver is 3.25 J/cm^3·°C.",
        "answer": "0.337",
        "type": 1,
        "wrong_answers_1": "0.319",
        "wrong_answers_2": "0.383",
        "wrong_answers_3": "0.315"
    },
    {
        "idx": 186,
        "question": "Calculate the fraction of solidification that occurs dendritically when silver nucleates homogeneously. The specific heat of silver is 3.25 J/cm^3·°C.",
        "answer": "0.842",
        "type": 1,
        "wrong_answers_1": "0.886",
        "wrong_answers_2": "0.898",
        "wrong_answers_3": "0.782"
    },
    {
        "idx": 187,
        "question": "A 2-in. cube solidifies in 4.6 min. Calculate (a) the mold constant in Chvorinov's rule. Assume that n=2.",
        "answer": "the mold constant in chvorinov's rule is 41.48 min/in^2.",
        "type": 1,
        "wrong_answers_1": "the mold constant in chvorinov's rule is 35.55 min/in^2.",
        "wrong_answers_2": "the mold constant in chvorinov's rule is 45.00 min/in^2.",
        "wrong_answers_3": "the mold constant in chvorinov's rule is 37.01 min/in^2."
    },
    {
        "idx": 188,
        "question": "A 2-in. cube solidifies in 4.6 min. Calculate (b) the solidification time for a 0.5 in. x 0.5 in. x 6 in. bar cast under the same conditions. Assume that n=2.",
        "answer": "the solidification time for the bar is 0.60 min.",
        "type": 1,
        "wrong_answers_1": "the solidification time for the bar is 0.51 min.",
        "wrong_answers_2": "the solidification time for the bar is 0.51 min.",
        "wrong_answers_3": "the solidification time for the bar is 0.67 min."
    },
    {
        "idx": 189,
        "question": "A 5-cm diameter sphere solidifies in 1050s. Calculate the solidification time for a 0.3 cm × 10 cm × 20 cm plate cast under the same conditions. Assume that n=2.",
        "answer": "the solidification time for the plate is 31.15s.",
        "type": 1,
        "wrong_answers_1": "the solidification time for the plate is 27.60s.",
        "wrong_answers_2": "the solidification time for the plate is 35.43s.",
        "wrong_answers_3": "the solidification time for the plate is 33.63s."
    },
    {
        "idx": 190,
        "question": "Calculate the diameter of the cylindrical riser required to prevent shrinkage in a 4 in. × 10 in. × 20 in. casting if the H / D of the riser is 1.5.",
        "answer": "the diameter of the riser d ≥ 6.67 in.",
        "type": 1,
        "wrong_answers_1": "the diameter of the riser d ≥ 7.61 in.",
        "wrong_answers_2": "the diameter of the riser d ≥ 6.02 in.",
        "wrong_answers_3": "the diameter of the riser d ≥ 7.28 in."
    },
    {
        "idx": 191,
        "question": "Calculate the height of the cylindrical riser required to prevent shrinkage in a 4 in. × 10 in. × 20 in. casting if the H / D of the riser is 1.5.",
        "answer": "the height of the riser h ≥ 10 in.",
        "type": 1,
        "wrong_answers_1": "the height of the riser h ≥ 11 in.",
        "wrong_answers_2": "the height of the riser h ≥ 8 in.",
        "wrong_answers_3": "the height of the riser h ≥ 9 in."
    },
    {
        "idx": 192,
        "question": "Calculate the volume of the cylindrical riser required to prevent shrinkage in a 4 in. × 10 in. × 20 in. casting if the H / D of the riser is 1.5.",
        "answer": "the volume of the riser v ≥ 349 in.^3",
        "type": 1,
        "wrong_answers_1": "the volume of the riser v ≥ 321 in.^3",
        "wrong_answers_2": "the volume of the riser v ≥ 372 in.^3",
        "wrong_answers_3": "the volume of the riser v ≥ 394 in.^3"
    },
    {
        "idx": 193,
        "question": "Calculate the diameter of the cylindrical riser required to prevent shrinkage in a 1 in. x 6 in. x 6 in. casting if the H/D of the riser is 1.0.",
        "answer": "the diameter of the riser must be at least 2.25 in.",
        "type": 1,
        "wrong_answers_1": "the diameter of the riser must be at least 2.00 in.",
        "wrong_answers_2": "the diameter of the riser must be at least 2.42 in.",
        "wrong_answers_3": "the diameter of the riser must be at least 2.11 in."
    },
    {
        "idx": 194,
        "question": "Calculate the height of the cylindrical riser required to prevent shrinkage in a 1 in. x 6 in. x 6 in. casting if the H/D of the riser is 1.0.",
        "answer": "the height of the riser must be at least 2.25 in.",
        "type": 1,
        "wrong_answers_1": "the height of the riser must be at least 2.36 in.",
        "wrong_answers_2": "the height of the riser must be at least 2.54 in.",
        "wrong_answers_3": "the height of the riser must be at least 2.13 in."
    },
    {
        "idx": 195,
        "question": "Calculate the volume of the cylindrical riser required to prevent shrinkage in a 1 in. x 6 in. x 6 in. casting if the H/D of the riser is 1.0.",
        "answer": "the volume of the riser must be at least 8.95 in^3.",
        "type": 1,
        "wrong_answers_1": "the volume of the riser must be at least 9.43 in^3.",
        "wrong_answers_2": "the volume of the riser must be at least 9.45 in^3.",
        "wrong_answers_3": "the volume of the riser must be at least 10.12 in^3."
    },
    {
        "idx": 196,
        "question": "A 4-in-diameter sphere of liquid copper is allowed to solidify, producing a spherical shrinkage cavity in the center of the casting. Determine the volume and diameter of the shrinkage cavity in the copper casting.",
        "answer": "shrinkage volume: 1.709 in.^{3}; diameter of shrinkage cavity: 1.30 in.",
        "type": 1,
        "wrong_answers_1": "shrinkage volume: 1.528 in.^{3}; diameter of shrinkage cavity: 1.39 in.",
        "wrong_answers_2": "shrinkage volume: 1.578 in.^{3}; diameter of shrinkage cavity: 1.13 in.",
        "wrong_answers_3": "shrinkage volume: 1.622 in.^{3}; diameter of shrinkage cavity: 1.20 in."
    },
    {
        "idx": 197,
        "question": "A 4-in-diameter sphere of liquid iron is allowed to solidify, producing a spherical shrinkage cavity in the center of the casting. Determine the volume and diameter of the shrinkage cavity in the iron casting.",
        "answer": "shrinkage volume: 1.139 in.^{3}; diameter of shrinkage cavity: 1.30 in.",
        "type": 1,
        "wrong_answers_1": "shrinkage volume: 1.103 in.^{3}; diameter of shrinkage cavity: 1.48 in.",
        "wrong_answers_2": "shrinkage volume: 1.057 in.^{3}; diameter of shrinkage cavity: 1.47 in.",
        "wrong_answers_3": "shrinkage volume: 1.054 in.^{3}; diameter of shrinkage cavity: 1.45 in."
    },
    {
        "idx": 198,
        "question": "A 4-in. cube of a liquid metal is allowed to solidify. A spherical shrinkage cavity with a diameter of 1.49 in. is observed in the solid casting. Determine the percent volume change that occurs during solidification.",
        "answer": "the percent volume change that occurs during solidification is 2.7%.",
        "type": 1,
        "wrong_answers_1": "the percent volume change that occurs during solidification is 2.4%.",
        "wrong_answers_2": "the percent volume change that occurs during solidification is 3.0%.",
        "wrong_answers_3": "the percent volume change that occurs during solidification is 3.1%."
    },
    {
        "idx": 199,
        "question": "A 2 cm x 4 cm x 6 cm magnesium casting is produced. After cooling to room temperature, what is the volume of the shrinkage cavity at the center of the casting?",
        "answer": "the volume of the shrinkage cavity at the center of the casting is 46.03 cm^3.",
        "type": 1,
        "wrong_answers_1": "the volume of the shrinkage cavity at the center of the casting is 50.60 cm^3.",
        "wrong_answers_2": "the volume of the shrinkage cavity at the center of the casting is 48.57 cm^3.",
        "wrong_answers_3": "the volume of the shrinkage cavity at the center of the casting is 41.60 cm^3."
    },
    {
        "idx": 200,
        "question": "A 2 cm x 4 cm x 6 cm magnesium casting is produced. After cooling to room temperature, what is the percent shrinkage that must have occurred during solidification?",
        "answer": "the percent shrinkage that must have occurred during solidification is 4.1%.",
        "type": 1,
        "wrong_answers_1": "the percent shrinkage that must have occurred during solidification is 4.0%.",
        "wrong_answers_2": "the percent shrinkage that must have occurred during solidification is 3.9%.",
        "wrong_answers_3": "the percent shrinkage that must have occurred during solidification is 4.5%."
    },
    {
        "idx": 201,
        "question": "A 2 in. × 8 in. × 10 in. iron casting is produced and, after cooling to room temperature, is found to weigh 43.9 lb. Determine the percent shrinkage that must have occurred during solidification.",
        "answer": "The percent shrinkage that must have occurred during solidification is 3.4%.",
        "type": 1,
        "wrong_answers_1": "The percent shrinkage that must have occurred during solidification is 3.0%.",
        "wrong_answers_2": "The percent shrinkage that must have occurred during solidification is 3.3%.",
        "wrong_answers_3": "The percent shrinkage that must have occurred during solidification is 3.9%."
    },
    {
        "idx": 202,
        "question": "A 2 in. × 8 in. × 10 in. iron casting is produced and, after cooling to room temperature, is found to weigh 43.9 lb. Determine the number of shrinkage pores in the casting if all of the shrinkage occurs as pores with a diameter of 0.05 in.",
        "answer": "The number of shrinkage pores in the casting is 83,354 pores.",
        "type": 1,
        "wrong_answers_1": "The number of shrinkage pores in the casting is 74,929 pores.",
        "wrong_answers_2": "The number of shrinkage pores in the casting is 74,259 pores.",
        "wrong_answers_3": "The number of shrinkage pores in the casting is 90,390 pores."
    },
    {
        "idx": 203,
        "question": "Liquid magnesium is poured into a 2 cm × 2 cm × 24 cm mold and, as a result of directional solidification, all of the solidification shrinkage occurs along the length of the casting. Determine the length of the casting immediately after solidification is completed.",
        "answer": "the length of the casting immediately after solidification is completed is 23.04 cm.",
        "type": 1,
        "wrong_answers_1": "the length of the casting immediately after solidification is completed is 25.13 cm.",
        "wrong_answers_2": "the length of the casting immediately after solidification is completed is 25.93 cm.",
        "wrong_answers_3": "the length of the casting immediately after solidification is completed is 20.42 cm."
    },
    {
        "idx": 204,
        "question": "A liquid cast iron has a density of 7.65 g/cm^3. Immediately after solidification, the density of the solid cast iron is found to be 7.71 g/cm^3. Determine the percent volume change that occurs during solidification.",
        "answer": "the percent volume change during solidification is 0.77%.",
        "type": 1,
        "wrong_answers_1": "the percent volume change during solidification is 0.72%.",
        "wrong_answers_2": "the percent volume change during solidification is 0.71%.",
        "wrong_answers_3": "the percent volume change during solidification is 0.73%."
    },
    {
        "idx": 205,
        "question": "Does the cast iron expand or contract during solidification?",
        "answer": "the cast iron contracts during solidification.",
        "type": 3
    },
    {
        "idx": 206,
        "question": "The solubility of hydrogen in liquid aluminum at 715^{\\circ} C is found to be 1 cm^{3} / 100g \\mathrm{Al}. If all of this hydrogen precipitated as gas bubbles during solidification and remained trapped in the casting, calculate the volume percent gas in the solid aluminum.",
        "answer": "2.699%",
        "type": 1,
        "wrong_answers_1": "2.908%",
        "wrong_answers_2": "2.981%",
        "wrong_answers_3": "2.936%"
    },
    {
        "idx": 207,
        "question": "Based on Hume-Rothery's conditions, would the system Au-Ag be expected to display unlimited solid solubility? Explain.",
        "answer": "Yes",
        "type": 3
    },
    {
        "idx": 208,
        "question": "Based on Hume-Rothery's conditions, would the system Al-Cu be expected to display unlimited solid solubility? Explain.",
        "answer": "No",
        "type": 3
    },
    {
        "idx": 209,
        "question": "Based on Hume-Rothery's conditions, would the system Al-Au be expected to display unlimited solid solubility? Explain.",
        "answer": "No",
        "type": 3
    },
    {
        "idx": 210,
        "question": "Based on Hume-Rothery's conditions, would the system U-W be expected to display unlimited solid solubility? Explain.",
        "answer": "No",
        "type": 3
    },
    {
        "idx": 211,
        "question": "Based on Hume-Rothery's conditions, would the system Mo-Ta be expected to display unlimited solid solubility? Explain.",
        "answer": "Yes",
        "type": 3
    },
    {
        "idx": 212,
        "question": "Based on Hume-Rothery's conditions, would the system Nb-W be expected to display unlimited solid solubility? Explain.",
        "answer": "No",
        "type": 3
    },
    {
        "idx": 213,
        "question": "Based on Hume-Rothery's conditions, would the system Mg-Zn be expected to display unlimited solid solubility? Explain.",
        "answer": "No",
        "type": 3
    },
    {
        "idx": 214,
        "question": "Based on Hume-Rothery's conditions, would the system Mg-Cd be expected to display unlimited solid solubility? Explain.",
        "answer": "Yes",
        "type": 3
    },
    {
        "idx": 215,
        "question": "Suppose 1 at% of the following elements is added to copper (forming a separate alloy with each element) without exceeding the solubility limit. Which one would be expected to give the higher strength alloy? For copper: r_Cu=1.278 Å (a) Au: r=1.442, φr=+12.8% (b) Mn: r=1.12, φr=-12.4% (c) Sr: r=2.151, φr=+68.3% (d) Si: r=1.176, φr=-8.0% (e) Co: r=1.253, φr=-2.0%",
        "answer": "The Cu-Sr alloy would be expected to be strongest (largest size difference).",
        "type": 2
    },
    {
        "idx": 216,
        "question": "Is any of the alloying elements expected to have unlimited solid solubility in copper? For copper: r_Cu=1.278 Å (a) Au: r=1.442, φr=+12.8% (b) Mn: r=1.12, φr=-12.4% (c) Sr: r=2.151, φr=+68.3% (d) Si: r=1.176, φr=-8.0% (e) Co: r=1.253, φr=-2.0%",
        "answer": "The Cu-Au alloy satisfies Hume-Rothery's conditions and might be expected to display complete solid solubility-in fact it freezes like an isomorphous series of alloys, but a number of solid-state transformations occur at lower temperatures.",
        "type": 2
    },
    {
        "idx": 217,
        "question": "Suppose 1 at% of the following elements is added to copper (forming a separate alloy with each element) without exceeding the solubility limit. Which one would be expected to give the higher strength alloy? (a) Au (b) Mn (c) Sr (d) Si (e) Co",
        "answer": "(c)",
        "type": 2
    },
    {
        "idx": 218,
        "question": "Which of these alloying elements is expected to have unlimited solid solubility in copper? (a) Au (b) Mn (c) Sr (d) Si (e) Co",
        "answer": "(a)",
        "type": 2
    },
    {
        "idx": 219,
        "question": "Which of the following oxides is expected to have the largest solid solubility in \\mathrm{Al}_{2} \\mathrm{O}_{3} ?\n(a) \\mathrm{Y}_{2} \\mathrm{O}_{3}  (b) \\mathrm{Cr}_{2} \\mathrm{O}_{3}  (c) \\mathrm{Fe}_{2} \\mathrm{O}_{3}",
        "answer": "\\mathrm{cr}_{2} \\mathrm{o}_{3} is expected to have the largest solid solubility in \\mathrm{al}_{2} \\mathrm{o}_{3}; in fact, they are completely soluble in one another.",
        "type": 2
    },
    {
        "idx": 220,
        "question": "How many grams of nickel must be added to 500 grams of copper to produce an\nalloy that contains 50 wt% \\alpha at 1300^{\\circ} C ?",
        "answer": "541.7g ni",
        "type": 1,
        "wrong_answers_1": "475.4g ni",
        "wrong_answers_2": "473.3g ni",
        "wrong_answers_3": "570.0g ni"
    },
    {
        "idx": 221,
        "question": "How many grams of MgO must be added to 1kg of NiO to produce a ceramic that has a solidus temperature of 2200^{\\circ} C ?",
        "answer": "332g of mgo",
        "type": 1,
        "wrong_answers_1": "381g of mgo",
        "wrong_answers_2": "377g of mgo",
        "wrong_answers_3": "317g of mgo"
    },
    {
        "idx": 222,
        "question": "We would like to produce a MgO-\\mathrm{FeO} ceramic that is 30 wt% solid at 2000^{\\circ} C. Determine the original composition of the ceramic in wt%.",
        "answer": "56.9 wt% feo",
        "type": 1,
        "wrong_answers_1": "49.3 wt% feo",
        "wrong_answers_2": "49.5 wt% feo",
        "wrong_answers_3": "50.3 wt% feo"
    },
    {
        "idx": 223,
        "question": "A Nb-W alloy held at 2800°C is partly liquid and partly solid. If possible, determine the composition of each phase in the alloy.",
        "answer": "l: 49% w; α: 70% w",
        "type": 1,
        "wrong_answers_1": "l: 47% w; α: 61% w",
        "wrong_answers_2": "l: 51% w; α: 60% w",
        "wrong_answers_3": "l: 47% w; α: 66% w"
    },
    {
        "idx": 224,
        "question": "A Nb-W alloy held at 2800°C is partly liquid and partly solid. If possible, determine the amount of each phase in the alloy.",
        "answer": "not possible unless we know the original composition of the alloy.",
        "type": 1
    },
    {
        "idx": 225,
        "question": "A Nb-W alloy contains 55% α at 2600°C. Determine the composition of each phase.",
        "answer": "l: 22% w; α: 42% w",
        "type": 1,
        "wrong_answers_1": "l: 19% w; α: 48% w",
        "wrong_answers_2": "l: 24% w; α: 39% w",
        "wrong_answers_3": "l: 20% w; α: 45% w"
    },
    {
        "idx": 226,
        "question": "A Nb-W alloy contains 55% α at 2600°C. Determine the original composition of the alloy.",
        "answer": "33% w",
        "type": 1,
        "wrong_answers_1": "35% w",
        "wrong_answers_2": "31% w",
        "wrong_answers_3": "30% w"
    },
    {
        "idx": 227,
        "question": "Suppose a 1200-lb bath of a Nb-40 wt % W alloy is held at 2800 degrees C. How many pounds of tungsten can be added to the bath before any solid forms?",
        "answer": "212 additional pounds of W must be added before any solid forms.",
        "type": 1,
        "wrong_answers_1": "219 additional pounds of W must be added before any solid forms.",
        "wrong_answers_2": "222 additional pounds of W must be added before any solid forms.",
        "wrong_answers_3": "230 additional pounds of W must be added before any solid forms."
    },
    {
        "idx": 228,
        "question": "Suppose a 1200-lb bath of a Nb-40 wt % W alloy is held at 2800 degrees C. How many pounds of tungsten must be added to cause the entire bath to be solid?",
        "answer": "1200 additional pounds of W must be added to cause the entire bath to be solid.",
        "type": 1,
        "wrong_answers_1": "1320 additional pounds of W must be added to cause the entire bath to be solid.",
        "wrong_answers_2": "1320 additional pounds of W must be added to cause the entire bath to be solid.",
        "wrong_answers_3": "1090 additional pounds of W must be added to cause the entire bath to be solid."
    },
    {
        "idx": 229,
        "question": "Suppose a crucible made of pure nickel is used to contain 500g of liquid copper at 1150^{\\circ} C. Describe what happens to the system as it is held at this temperature for several hours. Explain.",
        "answer": "cu dissolves ni until the cu contains enough ni that it solidifies completely. when 10% ni is dissolved, freezing begins with x = 55.5 \\text{g ni}. when 18% ni is dissolved, the bath is completely solid with x = 109.8 \\text{g ni}.",
        "type": 4,
        "wrong_answers_1": "element  b  is arsenic (as), with an atomic weight of  74.92 \\, \\text{g/mol} . the two intermetallic compounds are  \\text{kas}  and  \\text{kas}_2 .",
        "wrong_answers_2": "the alloy is hypereutectoid since c_0 is greater than 0.76 \\, \\text{wt% c}.",
        "wrong_answers_3": "The electron configuration for a Cu+ ion is 1s2 2s2 2p6 3s2 3p6 3d10."
    },
    {
        "idx": 230,
        "question": "A NiO-60 mol% MgO ceramic is allowed to solidify. Determine the composition of the first solid to form under equilibrium conditions.",
        "answer": "80% mgo",
        "type": 1,
        "wrong_answers_1": "89% mgo",
        "wrong_answers_2": "69% mgo",
        "wrong_answers_3": "73% mgo"
    },
    {
        "idx": 231,
        "question": "A NiO-60 mol% MgO ceramic is allowed to solidify. Determine the composition of the last liquid to solidify under equilibrium conditions.",
        "answer": "35% mgo",
        "type": 1,
        "wrong_answers_1": "31% mgo",
        "wrong_answers_2": "33% mgo",
        "wrong_answers_3": "37% mgo"
    },
    {
        "idx": 232,
        "question": "A Nb-35% W alloy is allowed to solidify. Determine the composition of the first solid to form under equilibrium conditions.",
        "answer": "55% w",
        "type": 1,
        "wrong_answers_1": "59% w",
        "wrong_answers_2": "49% w",
        "wrong_answers_3": "53% w"
    },
    {
        "idx": 233,
        "question": "A Nb-35% W alloy is allowed to solidify. Determine the composition of the last liquid to solidify under equilibrium conditions.",
        "answer": "18% w",
        "type": 1,
        "wrong_answers_1": "20% w",
        "wrong_answers_2": "16% w",
        "wrong_answers_3": "20% w"
    },
    {
        "idx": 234,
        "question": "An intermetallic compound is found for 10 wt% \\mathrm{Si} in the Cu-Si phase diagram. Determine the formula for the compound.",
        "answer": "\\mathrm{sicu}_{4}",
        "type": 1,
        "wrong_answers_1": "\\mathrm{sicu}_{3}",
        "wrong_answers_2": "\\mathrm{sicu}_{5}",
        "wrong_answers_3": "\\mathrm{sicu}_{6}"
    },
    {
        "idx": 235,
        "question": "Consider a Pb-15% Sn alloy. During solidification, determine the composition of the first solid to form.",
        "answer": "8% Sn",
        "type": 1,
        "wrong_answers_1": "9% Sn",
        "wrong_answers_2": "7% Sn",
        "wrong_answers_3": "4% Sn"
    },
    {
        "idx": 236,
        "question": "Consider a Pb-15% Sn alloy. During solidification, determine the liquidus temperature, solidus temperature, solvus temperature, and freezing range of the alloy.",
        "answer": "liquidus temperature =290 c, solidus temperature =240 c, solvus temperature =170 c, freezing range =50 c",
        "type": 1,
        "wrong_answers_1": "liquidus temperature =301 c, solidus temperature =270 c, solvus temperature =183 c, freezing range =56 c",
        "wrong_answers_2": "liquidus temperature =265 c, solidus temperature =207 c, solvus temperature =149 c, freezing range =57 c",
        "wrong_answers_3": "liquidus temperature =325 c, solidus temperature =272 c, solvus temperature =180 c, freezing range =52 c"
    },
    {
        "idx": 237,
        "question": "Consider a Pb-15% Sn alloy. During solidification, determine the amounts and compositions of each phase at 260 c.",
        "answer": "l: 30% sn, α: 12% sn; % l=17%, % α=83%",
        "type": 1,
        "wrong_answers_1": "l: 26% sn, α: 11% sn; % l=18%, % α=87%",
        "wrong_answers_2": "l: 28% sn, α: 11% sn; % l=15%, % α=87%",
        "wrong_answers_3": "l: 32% sn, α: 11% sn; % l=16%, % α=91%"
    },
    {
        "idx": 238,
        "question": "Consider a Pb-15% Sn alloy. During solidification, determine the amounts and compositions of each phase at 183 c.",
        "answer": "α: 15% sn; 100% α",
        "type": 1,
        "wrong_answers_1": "α: 17% sn; 99% α",
        "wrong_answers_2": "α: 15% sn; 86% α",
        "wrong_answers_3": "α: 13% sn; 91% α"
    },
    {
        "idx": 239,
        "question": "Consider a Pb-15% Sn alloy. During solidification, determine the amounts and compositions of each phase at 25 c.",
        "answer": "α: 2% pb, β: 100% sn; % α=87%, % β=13%",
        "type": 1,
        "wrong_answers_1": "α: 2% pb, β: 91% sn; % α=92%, % β=12%",
        "wrong_answers_2": "α: 2% pb, β: 93% sn; % α=76%, % β=14%",
        "wrong_answers_3": "α: 2% pb, β: 91% sn; % α=79%, % β=14%"
    },
    {
        "idx": 240,
        "question": "Consider a Pb-35% Sn alloy. Determine if the alloy is hypoeutectic or hypereutectic.",
        "answer": "hypoeutectic",
        "type": 4,
        "wrong_answers_1": "In order for a material to have an index of refraction less than unity, the velocity of light in the material (v) would necessarily have to be greater than the velocity of light in a vacuum . This is not possible.",
        "wrong_answers_2": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_3": "The driving force for the formation of spheroidite is the net reduction in ferrite-cementite phase boundary area."
    },
    {
        "idx": 241,
        "question": "Consider a Pb-35% Sn alloy. Determine the composition of the first solid to form during solidification.",
        "answer": "14% sn",
        "type": 1,
        "wrong_answers_1": "13% sn",
        "wrong_answers_2": "12% sn",
        "wrong_answers_3": "11% sn"
    },
    {
        "idx": 242,
        "question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each phase at 184°C.",
        "answer": "α: 19% sn, l: 61.9% sn, % α = 63%, % l = 37%",
        "type": 1,
        "wrong_answers_1": "α: 21% sn, l: 55.7% sn, % α = 65%, % l = 34%",
        "wrong_answers_2": "α: 20% sn, l: 52.8% sn, % α = 69%, % l = 35%",
        "wrong_answers_3": "α: 18% sn, l: 56.2% sn, % α = 66%, % l = 41%"
    },
    {
        "idx": 243,
        "question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each base at 182°C.",
        "answer": "α: 19% sn, β: 97.5% sn, % α = 80%, % β = 20%",
        "type": 1,
        "wrong_answers_1": "α: 17% sn, β: 91.6% sn, % α = 86%, % β = 18%",
        "wrong_answers_2": "α: 18% sn, β: 86.1% sn, % α = 91%, % β = 21%",
        "wrong_answers_3": "α: 17% sn, β: 102.7% sn, % α = 71%, % β = 22%"
    },
    {
        "idx": 244,
        "question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each microconstituent at 182°C.",
        "answer": "primary α: 19% sn, % primary α = 63%, eutectic: 61.9% sn, % eutectic = 37%",
        "type": 1,
        "wrong_answers_1": "primary α: 18% sn, % primary α = 69%, eutectic: 68.8% sn, % eutectic = 33%",
        "wrong_answers_2": "primary α: 21% sn, % primary α = 57%, eutectic: 53.2% sn, % eutectic = 35%",
        "wrong_answers_3": "primary α: 16% sn, % primary α = 61%, eutectic: 69.8% sn, % eutectic = 40%"
    },
    {
        "idx": 245,
        "question": "Consider a Pb-35% Sn alloy. Determine the amounts and compositions of each phase at 25°C.",
        "answer": "α: 2% sn, β: 100% sn, % α = 66%, % β = 34%",
        "type": 1,
        "wrong_answers_1": "α: 2% sn, β: 100% sn, % α = 72%, % β = 36%",
        "wrong_answers_2": "α: 2% sn, β: 91% sn, % α = 63%, % β = 32%",
        "wrong_answers_3": "α: 2% sn, β: 100% sn, % α = 62%, % β = 32%"
    },
    {
        "idx": 246,
        "question": "Consider a Pb-70% Sn alloy. Determine if the alloy is hypoeutectic or hypereutectic.",
        "answer": "hypereutectic",
        "type": 3
    },
    {
        "idx": 247,
        "question": "Consider a Pb-70% Sn alloy. Determine the composition of the first solid to form during solidification.",
        "answer": "98% sn",
        "type": 1,
        "wrong_answers_1": "84% sn",
        "wrong_answers_2": "94% sn",
        "wrong_answers_3": "100% sn"
    },
    {
        "idx": 248,
        "question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 184 degrees C.",
        "answer": "beta: 97.5% sn, l: 61.9% sn, % beta = 22.8%, % l = 77.2%",
        "type": 1,
        "wrong_answers_1": "beta: 95.1% sn, l: 66.2% sn, % beta = 26.2%, % l = 82.4%",
        "wrong_answers_2": "beta: 90.9% sn, l: 56.4% sn, % beta = 23.7%, % l = 83.2%",
        "wrong_answers_3": "beta: 96.8% sn, l: 65.8% sn, % beta = 24.2%, % l = 80.7%"
    },
    {
        "idx": 249,
        "question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 182 degrees C.",
        "answer": "alpha: 19% sn, beta: 97.5% sn, % alpha = 35%, % beta = 65%",
        "type": 1,
        "wrong_answers_1": "alpha: 16% sn, beta: 88.7% sn, % alpha = 31%, % beta = 62%",
        "wrong_answers_2": "alpha: 18% sn, beta: 85.9% sn, % alpha = 39%, % beta = 71%",
        "wrong_answers_3": "alpha: 16% sn, beta: 90.4% sn, % alpha = 40%, % beta = 73%"
    },
    {
        "idx": 250,
        "question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each microconstituent at 182 degrees C.",
        "answer": "primary beta: 97.5% sn, % primary beta = 22.8%, eutectic: 61.9% sn, % eutectic = 77.2%",
        "type": 1,
        "wrong_answers_1": "primary beta: 99.5% sn, % primary beta = 26.2%, eutectic: 56.6% sn, % eutectic = 83.6%",
        "wrong_answers_2": "primary beta: 93.1% sn, % primary beta = 20.2%, eutectic: 57.9% sn, % eutectic = 86.7%",
        "wrong_answers_3": "primary beta: 91.7% sn, % primary beta = 20.5%, eutectic: 55.0% sn, % eutectic = 85.0%"
    },
    {
        "idx": 251,
        "question": "Consider a Pb-70% Sn alloy. Determine the amounts and compositions of each phase at 25 degrees C.",
        "answer": "alpha: 2% sn, beta: 100% sn, % alpha = 30%, % beta = 70%",
        "type": 1,
        "wrong_answers_1": "alpha: 2% sn, beta: 97% sn, % alpha = 32%, % beta = 79%",
        "wrong_answers_2": "alpha: 2% sn, beta: 100% sn, % alpha = 32%, % beta = 65%",
        "wrong_answers_3": "alpha: 2% sn, beta: 95% sn, % alpha = 34%, % beta = 60%"
    },
    {
        "idx": 252,
        "question": "Consider an Al-4% Si alloy. Determine if the alloy is hypo eutectic or hyper eutectic.",
        "answer": "hypo eutectic",
        "type": 3
    },
    {
        "idx": 253,
        "question": "Consider an Al-4% Si alloy. Determine the composition of the first solid to form during solidification.",
        "answer": "1% si",
        "type": 1,
        "wrong_answers_1": "2% si",
        "wrong_answers_2": "3% si",
        "wrong_answers_3": "4% si"
    },
    {
        "idx": 254,
        "question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 578 degrees C.",
        "answer": "alpha: 1.65% si, l: 12.6% si, % alpha=78.5%, % l=21.5%",
        "type": 1,
        "wrong_answers_1": "alpha: 1.82% si, l: 11.4% si, % alpha=89.9%, % l=23.5%",
        "wrong_answers_2": "alpha: 1.51% si, l: 12.2% si, % alpha=87.4%, % l=20.6%",
        "wrong_answers_3": "alpha: 1.45% si, l: 14.1% si, % alpha=86.2%, % l=18.8%"
    },
    {
        "idx": 255,
        "question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 576 degrees C.",
        "answer": "alpha: 1.65% si, beta: 99.83% si, % alpha=97.6%, % beta=2.4%",
        "type": 1,
        "wrong_answers_1": "alpha: 1.54% si, beta: 107.57% si, % alpha=86.1%, % beta=2.1%",
        "wrong_answers_2": "alpha: 1.51% si, beta: 88.44% si, % alpha=102.8%, % beta=2.7%",
        "wrong_answers_3": "alpha: 1.71% si, beta: 109.39% si, % alpha=107.9%, % beta=2.3%"
    },
    {
        "idx": 256,
        "question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each microconstituent at 576 degrees C.",
        "answer": "primary alpha: 1.65% si, % primary alpha=78.5%, eutectic: 12.6% si, % eutectic=21.5%",
        "type": 1,
        "wrong_answers_1": "primary alpha: 1.78% si, % primary alpha=72.5%, eutectic: 13.6% si, % eutectic=18.5%",
        "wrong_answers_2": "primary alpha: 1.77% si, % primary alpha=82.0%, eutectic: 11.1% si, % eutectic=18.6%",
        "wrong_answers_3": "primary alpha: 1.46% si, % primary alpha=82.2%, eutectic: 13.2% si, % eutectic=22.3%"
    },
    {
        "idx": 257,
        "question": "Consider an Al-4% Si alloy. Determine the amounts and compositions of each phase at 25 degrees C.",
        "answer": "alpha: 0% si, beta: 100% si, % alpha=96%, % beta=4%",
        "type": 1,
        "wrong_answers_1": "alpha: 0% si, beta: 100% si, % alpha=85%, % beta=4%",
        "wrong_answers_2": "alpha: 0% si, beta: 94% si, % alpha=83%, % beta=3%",
        "wrong_answers_3": "alpha: 0% si, beta: 87% si, % alpha=92%, % beta=4%"
    },
    {
        "idx": 258,
        "question": "Consider a Al-25% Si alloy. Determine if the alloy is hypo eutectic or hyper eutectic.",
        "answer": "hyper eutectic",
        "type": 3
    },
    {
        "idx": 259,
        "question": "Consider a Al-25% Si alloy. Determine the composition of the first solid to form during solidified.",
        "answer": "100% si",
        "type": 1,
        "wrong_answers_1": "112% si",
        "wrong_answers_2": "111% si",
        "wrong_answers_3": "114% si"
    },
    {
        "idx": 260,
        "question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 578 degrees C.",
        "answer": "β: 99.83% si, l: 12.6% si, %l: 85.8%, %β: 14.2%",
        "type": 1,
        "wrong_answers_1": "β: 91.87% si, l: 10.7% si, %l: 96.0%, %β: 13.6%",
        "wrong_answers_2": "β: 99.72% si, l: 13.3% si, %l: 94.8%, %β: 12.1%",
        "wrong_answers_3": "β: 93.49% si, l: 14.4% si, %l: 97.3%, %β: 16.1%"
    },
    {
        "idx": 261,
        "question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 576 degrees C.",
        "answer": "α: 1.65% si, β: 99.83% si, %α: 76.2%, %β: 23.8%",
        "type": 1,
        "wrong_answers_1": "α: 1.48% si, β: 107.84% si, %α: 73.4%, %β: 26.7%",
        "wrong_answers_2": "α: 1.70% si, β: 94.42% si, %α: 68.9%, %β: 27.2%",
        "wrong_answers_3": "α: 1.84% si, β: 88.85% si, %α: 67.4%, %β: 20.8%"
    },
    {
        "idx": 262,
        "question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each microconstituent at 576 degrees C.",
        "answer": "primary β: 99.83% si, % primary β: 14.2%; eutectic: 12.6% si, % eutectic: 85.8%",
        "type": 1,
        "wrong_answers_1": "primary β: 97.80% si, % primary β: 12.7%; eutectic: 11.0% si, % eutectic: 93.3%",
        "wrong_answers_2": "primary β: 96.71% si, % primary β: 12.8%; eutectic: 13.0% si, % eutectic: 96.2%",
        "wrong_answers_3": "primary β: 92.18% si, % primary β: 12.6%; eutectic: 13.7% si, % eutectic: 93.7%"
    },
    {
        "idx": 263,
        "question": "Consider a Al-25% Si alloy. Determine the amounts and compositions of each phase at 25 degrees C.",
        "answer": "α: 0% si, β: 100% si, %α: 75%, %β: 25%",
        "type": 1,
        "wrong_answers_1": "α: 0% si, β: 97% si, %α: 82%, %β: 23%",
        "wrong_answers_2": "α: 0% si, β: 92% si, %α: 67%, %β: 27%",
        "wrong_answers_3": "α: 0% si, β: 100% si, %α: 83%, %β: 28%"
    },
    {
        "idx": 264,
        "question": "A Pb-Sn alloy contains 45% alpha and 55% beta at 100C. Determine the composition of the alloy.",
        "answer": "56.15% sn",
        "type": 1,
        "wrong_answers_1": "60.21% sn",
        "wrong_answers_2": "61.19% sn",
        "wrong_answers_3": "48.53% sn"
    },
    {
        "idx": 265,
        "question": "Is the alloy hypoeutectic or hypereutectic?",
        "answer": "hypoeutectic",
        "type": 3
    },
    {
        "idx": 266,
        "question": "An Al-Si alloy contains 85 % α and 15 % β at 500°C. Determine the composition of the alloy.",
        "answer": "15.85% Si",
        "type": 1,
        "wrong_answers_1": "15.30% Si",
        "wrong_answers_2": "16.46% Si",
        "wrong_answers_3": "14.89% Si"
    },
    {
        "idx": 267,
        "question": "Is the alloy hypoeutectic or hypereutectic?",
        "answer": "hypereutectic",
        "type": 3
    },
    {
        "idx": 268,
        "question": "A Pb-Sn alloy contains 23% primary \\alpha and 77% eutectic microconstituent. Determine the composition of the alloy.",
        "answer": "x = 52% sn",
        "type": 1,
        "wrong_answers_1": "x = 47% sn",
        "wrong_answers_2": "x = 48% sn",
        "wrong_answers_3": "x = 59% sn"
    },
    {
        "idx": 269,
        "question": "An Al-Si alloy contains 15% primary \\beta and 85% eutectic microconstituent. Determine the composition of the alloy.",
        "answer": "25.71% si",
        "type": 1,
        "wrong_answers_1": "27.75% si",
        "wrong_answers_2": "29.14% si",
        "wrong_answers_3": "22.43% si"
    },
    {
        "idx": 270,
        "question": "Observation of a microstructure shows that there is 28% eutectic and 72% primary β in an Al-Li alloy. Determine the composition of the alloy and whether it is hypoeutectic or hypereutectic.",
        "answer": "17.46% li, hypereutectic",
        "type": 1,
        "wrong_answers_1": "16.33% li, hypereutectic",
        "wrong_answers_2": "15.73% li, hypereutectic",
        "wrong_answers_3": "15.07% li, hypereutectic"
    },
    {
        "idx": 271,
        "question": "Observation of a microstructure shows that there is 28% eutectic and 72% primary β in an Al-Li alloy. How much α and β are in the eutectic microconstituent?",
        "answer": "% α_est = 64% and % β_est = 36%",
        "type": 1,
        "wrong_answers_1": "% α_est = 68% and % β_est = 35%",
        "wrong_answers_2": "% α_est = 73% and % β_est = 32%",
        "wrong_answers_3": "% α_est = 56% and % β_est = 39%"
    },
    {
        "idx": 272,
        "question": "Calculate the total amount of alpha and beta in a Pb-50% Sn alloy at 182°C.",
        "answer": "total amount of alpha = 60.5%, beta = 39.5%.",
        "type": 1,
        "wrong_answers_1": "total amount of alpha = 52.0%, beta = 35.8%.",
        "wrong_answers_2": "total amount of alpha = 69.0%, beta = 41.1%.",
        "wrong_answers_3": "total amount of alpha = 52.1%, beta = 33.6%."
    },
    {
        "idx": 273,
        "question": "Calculate the amount of each microconstituent in a Pb-50% Sn alloy at 182°C.",
        "answer": "microconstituents: primary alpha = 27.7%, eutectic = 72.3%.",
        "type": 1,
        "wrong_answers_1": "microconstituents: primary alpha = 25.4%, eutectic = 68.3%.",
        "wrong_answers_2": "microconstituents: primary alpha = 23.6%, eutectic = 77.2%.",
        "wrong_answers_3": "microconstituents: primary alpha = 26.7%, eutectic = 62.3%."
    },
    {
        "idx": 274,
        "question": "What fraction of the total alpha in the alloy is contained in the eutectic microconstituent?",
        "answer": "fraction of total alpha in the eutectic microconstituent f = 0.54.",
        "type": 1,
        "wrong_answers_1": "fraction of total alpha in the eutectic microconstituent f = 0.59.",
        "wrong_answers_2": "fraction of total alpha in the eutectic microconstituent f = 0.52.",
        "wrong_answers_3": "fraction of total alpha in the eutectic microconstituent f = 0.59."
    },
    {
        "idx": 275,
        "question": "Recommend an artificial age-hardening heat treatment for a Cu-1.2% Be alloy (see Figure 12-34). Include appropriate temperatures.",
        "answer": "For the Cu-1.2% Be alloy, solution treat between 530°C and 870°C (typical at 780°C), quench, and age below 530°C (typical at 330°C).",
        "type": 4,
        "wrong_answers_1": "Quench to below the M_gamma (less than 130 degrees Celsius).",
        "wrong_answers_2": "Austenitize at approximately 750^{\\circ} C,\nQuench to below 130^{\\circ} C (the M_{\\gamma} temperature)\nTemper at 620^{\\circ} C or less.",
        "wrong_answers_3": "Oxidation occurs at the anode; reduction at the cathode."
    },
    {
        "idx": 276,
        "question": "Compare the amount of the γ2 precipitate that forms by artificial aging at 400°C with the amount of the precipitate that forms by natural aging for a Cu-1.2% Be alloy.",
        "answer": "The amount of γ2 precipitate at 400°C is 5.4%, and at room temperature, it is 8.5%.",
        "type": 4,
        "wrong_answers_1": "Oxidation occurs at the anode; reduction at the cathode.",
        "wrong_answers_2": "The melting temperature is, for a crystalline material and upon cooling, that temperature at which there is a sudden and discontinuous decrease in the specific-volume-versus-temperature curve.",
        "wrong_answers_3": "The glass transition temperature is, for a noncrystalline ceramic, that temperature at which there is a change of slope for the specific volume versus temperature curve."
    },
    {
        "idx": 277,
        "question": "For an Fe-0.35% C alloy, determine the temperature at which austenite first begins to transform on cooling.",
        "answer": "795°C",
        "type": 1,
        "wrong_answers_1": "709°C",
        "wrong_answers_2": "874°C",
        "wrong_answers_3": "753°C"
    },
    {
        "idx": 278,
        "question": "For an Fe-0.35% C alloy, determine the primary microconstituent that forms.",
        "answer": "primary α-ferrite",
        "type": 4,
        "wrong_answers_1": "primary Fe3C",
        "wrong_answers_2": "ferrite is the proeutectoid phase.",
        "wrong_answers_3": "martensite, proeutectoid ferrite, and bainite form."
    },
    {
        "idx": 279,
        "question": "For an Fe-0.35% C alloy, determine the composition and amount of each phase present at 728°C.",
        "answer": "α: 0.0218% C, 56.1% γ: 0.77% C, 43.9%",
        "type": 1,
        "wrong_answers_1": "α: 0.0202% C, 50.8% γ: 0.88% C, 49.2%",
        "wrong_answers_2": "α: 0.0186% C, 52.4% γ: 0.81% C, 38.3%",
        "wrong_answers_3": "α: 0.0238% C, 64.3% γ: 0.83% C, 38.4%"
    },
    {
        "idx": 280,
        "question": "For an Fe-0.35% C alloy, determine the composition and amount of each phase present at 726°C.",
        "answer": "α: 0.0218% C, 93.3% Fe3C: 6.67% C, 4.9%",
        "type": 1,
        "wrong_answers_1": "α: 0.0246% C, 88.0% Fe3C: 7.59% C, 5.1%",
        "wrong_answers_2": "α: 0.0227% C, 80.1% Fe3C: 7.21% C, 4.5%",
        "wrong_answers_3": "α: 0.0193% C, 85.3% Fe3C: 6.18% C, 4.6%"
    },
    {
        "idx": 281,
        "question": "For an Fe-0.35% C alloy, determine the composition and amount of each microconstituent present at 726°C.",
        "answer": "primary α: 0.0218% C, 56.1% pearlite: 0.77% C, 43.9%",
        "type": 1,
        "wrong_answers_1": "primary α: 0.0248% C, 52.9% pearlite: 0.70% C, 47.9%",
        "wrong_answers_2": "primary α: 0.0199% C, 64.3% pearlite: 0.70% C, 48.1%",
        "wrong_answers_3": "primary α: 0.0195% C, 52.9% pearlite: 0.67% C, 39.7%"
    },
    {
        "idx": 282,
        "question": "For an Fe-1.15% C alloy, determine the temperature at which austenite first begins to transform on cooling.",
        "answer": "880 degrees c",
        "type": 1,
        "wrong_answers_1": "973 degrees c",
        "wrong_answers_2": "773 degrees c",
        "wrong_answers_3": "760 degrees c"
    },
    {
        "idx": 283,
        "question": "For an Fe-1.15% C alloy, determine the primary microconstituent that forms.",
        "answer": "primary Fe3C",
        "type": 4,
        "wrong_answers_1": "primary α-ferrite",
        "wrong_answers_2": "the proeutectoid phase is fe3c.",
        "wrong_answers_3": "Metals are typically better thermal conductors than are ceramic materials because, for metals, most of the heat is transported by free electrons (of which there are relatively large numbers). In ceramic materials, the primary mode of thermal conduction is via phonons, and phonons are more easily scattered than are free electrons."
    },
    {
        "idx": 284,
        "question": "For an Fe-1.15% C alloy, determine the composition and amount of each phase present at 728 degrees C.",
        "answer": "Fe3C: 6.67% C, 6.4%; gamma: 0.77% C, 93.6%",
        "type": 1,
        "wrong_answers_1": "Fe3C: 6.21% C, 5.5%; gamma: 0.81% C, 90.5%",
        "wrong_answers_2": "Fe3C: 6.42% C, 6.1%; gamma: 0.74% C, 81.6%",
        "wrong_answers_3": "Fe3C: 5.87% C, 5.5%; gamma: 0.70% C, 97.5%"
    },
    {
        "idx": 285,
        "question": "For an Fe-1.15% C alloy, determine the composition and amount of each phase present at 720 degrees C.",
        "answer": "alpha: 0.0218% C, 83%; Fe3C: 6.67% C, 17%",
        "type": 1,
        "wrong_answers_1": "alpha: 0.0233% C, 88%; Fe3C: 6.02% C, 19%",
        "wrong_answers_2": "alpha: 0.0200% C, 95%; Fe3C: 7.27% C, 16%",
        "wrong_answers_3": "alpha: 0.0211% C, 71%; Fe3C: 7.22% C, 18%"
    },
    {
        "idx": 286,
        "question": "For an Fe-1.15% C alloy, determine the composition and amount of each microconstituent present at 726 degrees C.",
        "answer": "primary Fe3C: 6.67% C, 6.4%; pearlite: 0.77% C, 93.6%",
        "type": 1,
        "wrong_answers_1": "primary Fe3C: 5.69% C, 6.9%; pearlite: 0.83% C, 97.8%",
        "wrong_answers_2": "primary Fe3C: 5.77% C, 7.2%; pearlite: 0.71% C, 100.7%",
        "wrong_answers_3": "primary Fe3C: 6.10% C, 6.0%; pearlite: 0.75% C, 99.5%"
    },
    {
        "idx": 287,
        "question": "A steel contains 8 % cementite and 92 % ferrite at room temperature. Estimate the carbon content of the steel.",
        "answer": "the carbon content of the steel is 0.53 % c.",
        "type": 1,
        "wrong_answers_1": "the carbon content of the steel is 0.57 % c.",
        "wrong_answers_2": "the carbon content of the steel is 0.46 % c.",
        "wrong_answers_3": "the carbon content of the steel is 0.58 % c."
    },
    {
        "idx": 288,
        "question": "Is the steel hypoeutectoid or hypereutectoid?",
        "answer": "the steel is hypoeutectoid.",
        "type": 3
    },
    {
        "idx": 289,
        "question": "A steel contains 18 % cementite and 82 % ferrite at room temperature. Estimate the carbon content of the steel.",
        "answer": "the carbon content of the steel is 1.20 % c.",
        "type": 1,
        "wrong_answers_1": "the carbon content of the steel is 1.11 % c.",
        "wrong_answers_2": "the carbon content of the steel is 1.26 % c.",
        "wrong_answers_3": "the carbon content of the steel is 1.04 % c."
    },
    {
        "idx": 290,
        "question": "Is the steel hypoeutectoid or hypereutectoid?",
        "answer": "the steel is hypereutectoid.",
        "type": 3
    },
    {
        "idx": 291,
        "question": "A steel contains 18 % pearlite and 82 % primary ferrite at room temperature. Estimate the carbon content of the steel.",
        "answer": "the carbon content of the steel is 0.156 %c.",
        "type": 1,
        "wrong_answers_1": "the carbon content of the steel is 0.176 %c.",
        "wrong_answers_2": "the carbon content of the steel is 0.136 %c.",
        "wrong_answers_3": "the carbon content of the steel is 0.135 %c."
    },
    {
        "idx": 292,
        "question": "Is the steel hypoeutectoid or hypereutectoid?",
        "answer": "the steel is hypoeutectoid.",
        "type": 3
    },
    {
        "idx": 293,
        "question": "A steel contains 94 % pearlite and 6 % primary cementite at room temperature. Estimate the carbon content of the steel.",
        "answer": "the carbon content of the steel is 1.124 % c.",
        "type": 1,
        "wrong_answers_1": "the carbon content of the steel is 1.078 % c.",
        "wrong_answers_2": "the carbon content of the steel is 1.223 % c.",
        "wrong_answers_3": "the carbon content of the steel is 0.968 % c."
    },
    {
        "idx": 294,
        "question": "Is the steel hypoeutectoid or hypereutectoid?",
        "answer": "the steel is hypereutectoid.",
        "type": 3
    },
    {
        "idx": 295,
        "question": "A steel contains 55% \\alpha and 45% \\gamma at 750^{\\circ} C. Estimate the carbon content of the steel.",
        "answer": "the carbon content of the steel is 0.281% C.",
        "type": 1,
        "wrong_answers_1": "the carbon content of the steel is 0.293% C.",
        "wrong_answers_2": "the carbon content of the steel is 0.312% C.",
        "wrong_answers_3": "the carbon content of the steel is 0.266% C."
    },
    {
        "idx": 296,
        "question": "A steel contains 96% \\gamma and 4% \\mathrm{Fe}_{3} C at 800^{\\circ} C. Estimate the carbon content of the steel.\n\\[\n\\begin{array}{l}\n\\text {",
        "answer": "the carbon content of the steel is 1.15% C.",
        "type": 1,
        "wrong_answers_1": "the carbon content of the steel is 1.31% C.",
        "wrong_answers_2": "the carbon content of the steel is 0.98% C.",
        "wrong_answers_3": "the carbon content of the steel is 1.10% C."
    },
    {
        "idx": 297,
        "question": "A steel is heated until 40% austenite, with a carbon content of 0.5%, forms. Estimate the temperature.",
        "answer": "the temperature is about 760 degrees c.",
        "type": 1,
        "wrong_answers_1": "the temperature is about 828 degrees c.",
        "wrong_answers_2": "the temperature is about 688 degrees c.",
        "wrong_answers_3": "the temperature is about 824 degrees c."
    },
    {
        "idx": 298,
        "question": "A steel is heated until 40% austenite, with a carbon content of 0.5%, forms. Estimate the overall carbon content of the steel.",
        "answer": "the overall carbon content of the steel is 0.212% c.",
        "type": 1,
        "wrong_answers_1": "the overall carbon content of the steel is 0.189% c.",
        "wrong_answers_2": "the overall carbon content of the steel is 0.230% c.",
        "wrong_answers_3": "the overall carbon content of the steel is 0.186% c."
    },
    {
        "idx": 299,
        "question": "An isothermally transformed eutectoid steel is found to have a yield strength of 410 MPa. Estimate the transformation temperature.",
        "answer": "transformation temperature =615 C",
        "type": 1,
        "wrong_answers_1": "transformation temperature =580 C",
        "wrong_answers_2": "transformation temperature =634 C",
        "wrong_answers_3": "transformation temperature =682 C"
    },
    {
        "idx": 300,
        "question": "An isothermally transformed eutectoid steel is found to have a yield strength of 410 MPa. Estimate the interlamellar spacing in the pearlite.",
        "answer": "1 / lambda=60,000 or lambda=1.67 × 10^-5 cm",
        "type": 1,
        "wrong_answers_1": "1 / lambda=67718 or lambda=1.88 × 10^-5 cm",
        "wrong_answers_2": "1 / lambda=67940 or lambda=1.50 × 10^-5 cm",
        "wrong_answers_3": "1 / lambda=55458 or lambda=1.80 × 10^-5 cm"
    },
    {
        "idx": 301,
        "question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have the hardness value HRC 38",
        "answer": "600 degrees C",
        "type": 1,
        "wrong_answers_1": "577 degrees C",
        "wrong_answers_2": "666 degrees C",
        "wrong_answers_3": "304 degrees C"
    },
    {
        "idx": 302,
        "question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have the hardness value HRC 42",
        "answer": "400 degrees C",
        "type": 1,
        "wrong_answers_1": "384 degrees C",
        "wrong_answers_2": "401 degrees C",
        "wrong_answers_3": "405 degrees C"
    },
    {
        "idx": 303,
        "question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have the hardness value HRC 48",
        "answer": "340 degrees C",
        "type": 1,
        "wrong_answers_1": "370 degrees C",
        "wrong_answers_2": "359 degrees C",
        "wrong_answers_3": "309 degrees C"
    },
    {
        "idx": 304,
        "question": "Determine the required transformation temperature and microconstituent if an eutectoid steel is to have the hardness value HRC 52",
        "answer": "300 degrees C",
        "type": 1,
        "wrong_answers_1": "264 degrees C",
        "wrong_answers_2": "342 degrees C",
        "wrong_answers_3": "323 degrees C"
    },
    {
        "idx": 305,
        "question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to 800^{\\circ} C for 1h, quenched to 350^{\\circ} C and held for 750s, and finally quenched to room temperature.",
        "answer": "HRC =47 and the microstructure is all baninite.",
        "type": 4,
        "wrong_answers_1": "HRC =25 and the microstructure is all pearlite.",
        "wrong_answers_2": "HRC=66 and the microstructure is all martensite.",
        "wrong_answers_3": "HRC=42 and the microstructure is all tempered martensite."
    },
    {
        "idx": 306,
        "question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to 700^{\\circ} C, quenched to 650^{\\circ} C and held for 500s, and finally quenched to room temperature.",
        "answer": "HRC =25 and the microstructure is all pearlite.",
        "type": 4,
        "wrong_answers_1": "HRC=66 and the microstructure is all martensite.",
        "wrong_answers_2": "HRC =47 and the microstructure is all baninite.",
        "wrong_answers_3": "HRC=42 and the microstructure is all tempered martensite."
    },
    {
        "idx": 307,
        "question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to 300^{\\circ} C, quenched to 300^{\\circ} C and held for 10s, and finally quenched to room temperature.",
        "answer": "HRC=66 and the microstructure is all martensite.",
        "type": 4,
        "wrong_answers_1": "HRC=42 and the microstructure is all tempered martensite.",
        "wrong_answers_2": "HRC =25 and the microstructure is all pearlite.",
        "wrong_answers_3": "HRC =47 and the microstructure is all baninite."
    },
    {
        "idx": 308,
        "question": "Describe the hardness and microstructure in an eutectoid steel that has been heated to 1000^{\\circ} C, quenched to 3010^{\\circ} C and held for 10 s, quenched to room temperature, and then reheated to 400^{\\circ} C before finally cooling to room temperature again.",
        "answer": "HRC=42 and the microstructure is all tempered martensite.",
        "type": 4,
        "wrong_answers_1": "HRC=66 and the microstructure is all martensite.",
        "wrong_answers_2": "HRC =25 and the microstructure is all pearlite.",
        "wrong_answers_3": "HRC =47 and the microstructure is all baninite."
    },
    {
        "idx": 309,
        "question": "Describe the complete heat treatment required to produce a quenched and tempered eutectoid steel having a tensile strength of at least 125,000 psi. Include appropriate temperatures.",
        "answer": "Austenitize at approximately 750^{\\circ} C,\nQuench to below 130^{\\circ} C (the M_{\\gamma} temperature)\nTemper at 620^{\\circ} C or less.",
        "type": 4,
        "wrong_answers_1": "The softening point of a glass is that temperature at which the viscosity is 4 × 10^{6} \\mathrm{~Pa}·s; from Figure 13.7, these temperatures for the 96% silica, borosilicate, and soda-lime glasses are 1540^{\\circ} C\\left(2800^{\\circ} F\\right), 830^{\\circ} C\\left(1525^{\\circ} F\\right), and 700^{\\circ} C\\left(1290^{\\circ} F\\right), respectively.",
        "wrong_answers_2": "Quench to below the M_gamma (less than 130 degrees Celsius).",
        "wrong_answers_3": "This question asks us to name which, of several polymers, would be suitable for the fabrication of cups to contain hot coffee. At its glass transition temperature, an amorphous polymer begins to soften. The maximum temperature of hot coffee is probably slightly below 100^{\\circ} C\\left(212^{\\circ} F\\right). Of the polymers listed, only polystyrene and polycarbonate have glass transition temperatures of 100^{\\circ} C or above, and would be suitable for this application."
    },
    {
        "idx": 310,
        "question": "What is the austenitizing temperature required to produce a quenched and tempered eutectoid steel with a HRC hardness of less than 50?",
        "answer": "Austenitize at approximately 75 degrees Celsius.",
        "type": 4,
        "wrong_answers_1": "Temper at a temperature higher than 330 degrees Celsius, but less than 727 degrees Celsius.",
        "wrong_answers_2": "Quench to below the M_gamma (less than 130 degrees Celsius).",
        "wrong_answers_3": "Austenitize at approximately 750^{\\circ} C,\nQuench to below 130^{\\circ} C (the M_{\\gamma} temperature)\nTemper at 620^{\\circ} C or less."
    },
    {
        "idx": 311,
        "question": "What is the quenching temperature required to produce a quenched and tempered eutectoid steel with a HRC hardness of less than 50?",
        "answer": "Quench to below the M_gamma (less than 130 degrees Celsius).",
        "type": 4,
        "wrong_answers_1": "Temper at a temperature higher than 330 degrees Celsius, but less than 727 degrees Celsius.",
        "wrong_answers_2": "Austenitize at approximately 750^{\\circ} C,\nQuench to below 130^{\\circ} C (the M_{\\gamma} temperature)\nTemper at 620^{\\circ} C or less.",
        "wrong_answers_3": "Austenitize at approximately 75 degrees Celsius."
    },
    {
        "idx": 312,
        "question": "What is the tempering temperature range required to produce a quenched and tempered eutectoid steel with a HRC hardness of less than 50?",
        "answer": "Temper at a temperature higher than 330 degrees Celsius, but less than 727 degrees Celsius.",
        "type": 4,
        "wrong_answers_1": "Quench to below the M_gamma (less than 130 degrees Celsius).",
        "wrong_answers_2": "Austenitize at approximately 75 degrees Celsius.",
        "wrong_answers_3": "for w: 1200 degrees c (2190 degrees f)"
    },
    {
        "idx": 313,
        "question": "Materials Science and Engineering is the study of material behavior & performance and how this is simultaneously related to structure, properties, and processing. Which of the following is the best example of a material property? (a) Density (b) Annealing (c) Forging (d) Single-crystal (e) Crystalline",
        "answer": "(a) Density",
        "type": 2
    },
    {
        "idx": 314,
        "question": "Materials Science and Engineering is the study of material behavior & performance and how this is simultaneously related to structure, properties, and processing. Which of the following is the best example of material processing?(a) Extrusion(b) Crystalline(c) Amorphous(d) Glassy (e) Elastic Modulus",
        "answer": "(a)Extrusion",
        "type": 2
    },
    {
        "idx": 315,
        "question": "Materials Science and Engineering is the study of material behavior & performance and how this is simultaneously related to structure, properties, and processing. Which of the following is the best example of material structure? (a) Single-phase (b) Elastic Modulus (c) Sintering (d) Magnetic Permeability (e) Brittle",
        "answer": "(a)Single-phase",
        "type": 2
    },
    {
        "idx": 316,
        "question": "Which class of material is generally associated with the highest density values at room temperature? (a) Composites (b) Ceramics (c) Metals (d) Polymers",
        "answer": "(c) Metals",
        "type": 2
    },
    {
        "idx": 317,
        "question": "By how many orders of magnitude (powers of ten, approximately) does density vary for metals? (a) 0.13 (b) 1.3 (c) 13 (d) 130",
        "answer": "(b)1.3",
        "type": 2
    },
    {
        "idx": 318,
        "question": "The atomic mass of an atom may be expressed as the sum of the masses of (a) Electrons (b) Neutrons (c) Protons (d) Choose all that apply.",
        "answer": "(b) Neutrons and (c) Protons",
        "type": 2
    },
    {
        "idx": 319,
        "question": "The nucleus of an atom contains electrons?",
        "answer": "No, the nucleus of an atom does not contain electrons.",
        "type": 3
    },
    {
        "idx": 320,
        "question": "The nucleus of an atom contains neutrons?",
        "answer": "Yes, the nucleus of an atom contains neutrons.",
        "type": 3
    },
    {
        "idx": 321,
        "question": "The nucleus of an atom contains protons?",
        "answer": "Yes, the nucleus of an atom contains protons.",
        "type": 3
    },
    {
        "idx": 322,
        "question": "The atomic number of an electrically neutral atom is equal to the number of:(a) protons (b) electrons (c) neutrons (d)Choose all that apply.",
        "answer": "(b)electrons.",
        "type": 2
    },
    {
        "idx": 323,
        "question": "Hafnium has six naturally occurring isotopes: 0.16% of { }^{174} Hf, with an atomic weight of 173.940 amu; 5.26% of { }^{176} Hf, with an atomic weight of 175.941 amu ; 18.60% of { }^{177} Hf, with an atomic weight of 176.943 amu ; 27.28% of { }^{178} Hf, with an atomic weight of 177.944 amu ; 13.62% of { }^{179} Hf, with an atomic weight of 178.946 amu;. and 35.08% of { }^{180} Hf, with an atomic weight of 179.947 amu. Calculate the average atomic weight of Hf. Give your answer to three decimal places.",
        "answer": "the average atomic weight of hf is 178.485 amu.",
        "type": 1,
        "wrong_answers_1": "the average atomic weight of hf is 193.448 amu.",
        "wrong_answers_2": "the average atomic weight of hf is 197.354 amu.",
        "wrong_answers_3": "the average atomic weight of hf is 184.530 amu."
    },
    {
        "idx": 324,
        "question": "Bromium has two naturally occurring isotopes: { }^{79} Br, with an atomic weight of 78.918 amu, and { }^{81} Br, with an atomic weight of 80.916 amu. If the average atomic weight for Br is 79.903 amu, calculate the fraction-of-occurrences of these two isotopes.Give your answer to three decimal places.",
        "answer": "the fraction-of-occurrence of { }^{79}\\mathrm{br} is 0.507, and the fraction-of-occurrence of { }^{81}\\mathrm{br} is 0.493.",
        "type": 1,
        "wrong_answers_1": "the fraction-of-occurrence of { }^{89}\\mathrm{br} is 0.531, and the fraction-of-occurrence of { }^{75}\\mathrm{br} is 0.514.",
        "wrong_answers_2": "the fraction-of-occurrence of { }^{86}\\mathrm{br} is 0.481, and the fraction-of-occurrence of { }^{89}\\mathrm{br} is 0.452.",
        "wrong_answers_3": "the fraction-of-occurrence of { }^{75}\\mathrm{br} is 0.558, and the fraction-of-occurrence of { }^{86}\\mathrm{br} is 0.546."
    },
    {
        "idx": 325,
        "question": "An element that has the electron configuration 1 s^{2} 2 s^{2} 2 p^{6} has how many electrons? Enter numeric values only.",
        "answer": "9",
        "type": 1,
        "wrong_answers_1": "10",
        "wrong_answers_2": "8",
        "wrong_answers_3": "6"
    },
    {
        "idx": 326,
        "question": "The electrons that occupy the outermost filled shell are called electrons.",
        "answer": "The electrons that occupy the outermost filled shell are called valence electrons.",
        "type": 4,
        "wrong_answers_1": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities.",
        "wrong_answers_2": "The n quantum number designates the electron shell.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) electron configuration is that of an alkaline earth metal because of two (s) electrons."
    },
    {
        "idx": 327,
        "question": "When all the electrons in an atom occupy the lowest possible energy states, the atom is said to be in its: (a)ground state(b)ionized state (c) cold state (d) regular state",
        "answer": "(a)ground state",
        "type": 2
    },
    {
        "idx": 328,
        "question": "How many p electrons at the outermost orbital do the Group VIIA elements have?",
        "answer": "5 p electrons.",
        "type": 1,
        "wrong_answers_1": "6 p electrons.",
        "wrong_answers_2": "7 p electrons.",
        "wrong_answers_3": "8 p electrons."
    },
    {
        "idx": 329,
        "question": "To what group in the periodic table would an element with atomic number 119 belong?(a) Group 0 (or 18) (b) Group IA (or 1) (c) Group IIA (or 2) (d) Group VIIA (or 17)",
        "answer": "(b) Group IA (or 1)",
        "type": 2
    },
    {
        "idx": 330,
        "question": "Calculate the energy of attraction between a cation with a valence of +2 and an anion with a valence of -2 , the centers of which are separated by a distance of 3.7nm.",
        "answer": "-2.5 × 10^{-19} J",
        "type": 1,
        "wrong_answers_1": "-2.2 × 10^{-19} J",
        "wrong_answers_2": "-2.8 × 10^{-19} J",
        "wrong_answers_3": "-2.7 × 10^{-19} J"
    },
    {
        "idx": 331,
        "question": "Materials whose constituent particles are bound by which type of bond are generally expected to have the lowest melting temperatures?(a) Covalent(b) Metallic(c) Ionic(d) Van der Waals(e) Hydrogen",
        "answer": "(d) Van der Waals",
        "type": 2
    },
    {
        "idx": 332,
        "question": "Calculate % IC of the interatomic bonds for the intermetallic compound TiAl3.",
        "answer": "0%",
        "type": 1,
        "wrong_answers_1": "1.0%",
        "wrong_answers_2": "0.4%",
        "wrong_answers_3": "0.3%"
    },
    {
        "idx": 333,
        "question": "On the basis of this result what type of interatomic bonding would you expect to be found in TiAl3 ? (a) van der Waals (b) ionic (c) metallic (d) covalent",
        "answer": " (c)metallic.",
        "type": 2
    },
    {
        "idx": 334,
        "question": "Which of the following microstructures is expected to be most similar to a single crystal in terms of structure and properties? Assume all of the options offer the same volumes and only consider grain boundaries as a crystalline defect for this question. (a) Textured polycrystal with about 10,000 grains (b) Random polycrystal with about 1,000,000 grains (c) Random polycrystal with about 1,000,000,000 grains (d) Amorphous",
        "answer": "(a) Textured polycrystal with about 10,000 grains",
        "type": 2
    },
    {
        "idx": 335,
        "question": "For a metal that has the simple cubic crystal structure, calculate the atomic radius if the metal has a density of 2.05g / {cm}^{3} and an atomic weight of 77.84g / mol.",
        "answer": "the atomic radius is 0.122nm.",
        "type": 1,
        "wrong_answers_1": "the atomic radius is 0.130nm.",
        "wrong_answers_2": "the atomic radius is 0.135nm.",
        "wrong_answers_3": "the atomic radius is 0.114nm."
    },
    {
        "idx": 336,
        "question": "Some metal is known to have a cubic unit cell with an edge length of 0.437 nm. In addition, it has a density of 4.37 g/cm^3 and an atomic weight of 54.85 g/mol. Determine the crystal structure of the metal.",
        "answer": "The metal has an FCC crystal structure.",
        "type": 1
    },
    {
        "idx": 337,
        "question": "Some metal is known to have a cubic unit cell with an edge length of 0.437 nm. In addition, it has a density of 4.37 g/cm^3 and an atomic weight of 54.85 g/mol. Determine the atomic radius of the metal.",
        "answer": "The metal has an atomic radius of 0.155 nm.",
        "type": 1,
        "wrong_answers_1": "The metal has an atomic radius of 0.168 nm.",
        "wrong_answers_2": "The metal has an atomic radius of 0.145 nm.",
        "wrong_answers_3": "The metal has an atomic radius of 0.175 nm."
    },
    {
        "idx": 338,
        "question": "Some metal is known to have a cubic unit cell with an edge length of 0.437 nm. In addition, it has a density of 4.37 g/cm^3 and an atomic weight of 54.85 g/mol. Indicate the letter of the metal listed in the following table that has these characteristics.",
        "answer": "The metal with these characteristics is metal C, which has an atomic radius of 0.155 nm and an FCC crystal structure.",
        "type": 2
    },
    {
        "idx": 339,
        "question": "If the atomic radius of a metal that has the body-centered cubic crystal structure is 0.181nm, calculate the volume of its unit cell.",
        "answer": "the volume of the unit cell is 0.0271 nm^3.",
        "type": 1,
        "wrong_answers_1": "the volume of the unit cell is 0.0254 nm^3.",
        "wrong_answers_2": "the volume of the unit cell is 0.0310 nm^3.",
        "wrong_answers_3": "the volume of the unit cell is 0.0257 nm^3."
    },
    {
        "idx": 340,
        "question": "If the atomic radius of a metal that has the face-centered cubic crystal structure is 0.123nm, calculate the volume of its unit cell.",
        "answer": "\\[\nv_{c}=0.258nm^{3}\n\\]",
        "type": 1,
        "wrong_answers_1": "\\[\nv_{c}=0.236nm^{3}\n\\]",
        "wrong_answers_2": "\\[\nv_{c}=0.233nm^{3}\n\\]",
        "wrong_answers_3": "\\[\nv_{c}=0.293nm^{3}\n\\]"
    },
    {
        "idx": 341,
        "question": "Rhodium has an atomic radius of 0.1345nm, a density of (12.41g / {cm}^{3}) and an atomic weight of 102.91 g / mol. What is rhodium's crystal structure?(a) Simple cubic(b) \\mathrm{BCC}(c) fcc",
        "answer": "the crystal structure for rhodium is fcc.",
        "type": 2
    },
    {
        "idx": 342,
        "question": "Niobium (Nb) has a BCC crystal structure, an atomic radius of 0.143nm and an atomic weight of 92.91g / mol. Calculate the theoretical density for nb.",
        "answer": "the theoretical density for nb is 8.48g / {cm}^{3}. the experimental density for nb is 8.57g / {cm}^{3}.",
        "type": 1,
        "wrong_answers_1": "the theoretical density for nb is 9.64g / {cm}^{3}. the experimental density for nb is 9.12g / {cm}^{3}.",
        "wrong_answers_2": "the theoretical density for nb is 9.14g / {cm}^{3}. the experimental density for nb is 9.46g / {cm}^{3}.",
        "wrong_answers_3": "the theoretical density for nb is 7.62g / {cm}^{3}. the experimental density for nb is 8.06g / {cm}^{3}."
    },
    {
        "idx": 343,
        "question": "Consider the ideal barium titanate \\left(\\mathrm{BaTiO}_{3}\\right) structure. What is the coordination number of the \\mathrm{Ti}^{4+} ion in terms of surrounding \\mathrm{O}^{2-} ions?\n[a] 1\n[b] 2\n[c] 3\n[d] 4\n[e] 5\n[f] 6\n[g] 7\n[h] 8",
        "answer": "According to Fig. 3.10, the central \\mathrm{Ti}^{4+} ion is surrounded by six \\mathrm{O}^{2-} ions, one residing at each of the six cube faces.",
        "type": 2
    },
    {
        "idx": 344,
        "question": "Consider the ideal barium titanate \\left(\\mathrm{BaTiO}_{3}\\right) structure. What is the coordination number of the \\mathrm{Ba}^{2+} ion in terms of surrounding \\mathrm{Ti}^{4+} ions?\n[a] 4\n[b] 6\n[c] 8\n[d] 10\n[e] 12",
        "answer": "Careful consideration of Fig. 3.10, and consideration of the perovskite compound stoichiometry reveals that 12 \\mathrm{Ti}^{4+} ions surround each \\mathrm{Ba}^{2+} ion. A change in perspective is recommended to help visualize this, where the unit cell cube corners feature \\mathrm{O}^{2-} ions in the corners, \\mathrm{Ti}^{4+} ions centered along each cube edge, and \\mathrm{Ba}^{2+} ions centered in the middle of the cube.",
        "type": 2
    },
    {
        "idx": 345,
        "question": "Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?\n\\[\na=b=c\n\\](a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic(g) Triclinic",
        "answer": "For both cubic and rhombohedral crystal systems all the unit cell edge lengths are equal.",
        "type": 2
    },
    {
        "idx": 346,
        "question": "Which crystal system(s) listed below has (have) the following relationship for the unit cell edge lengths?\n\\[\na=b \\neq c\n\\](a) Cubic(b) Hexagonal(c) Tetragonal\n\\mathrm{d}) Rhombohedral(e) Orthorhombic(f) Monoclinic (g) Triclinic",
        "answer": "For both hexagonal and tetragonal crystal systems two of the unit cell edge lengths are equal to one another, but unequal to the third length.",
        "type": 2
    },
    {
        "idx": 347,
        "question": "Which crystal system(s) listed below has (have) the following relationship for the unit cell e edge lengths?a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic(g) Triclinic",
        "answer": "For orthorhombic, monoclinic and triclinic crystal systems, all three of the unit cell edge lengths are unequal to one another.",
        "type": 2
    },
    {
        "idx": 348,
        "question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?\n\\[\n\\alpha \\neq \\beta \\neq \\gamma \\neq 90^{\\circ}\n\\](a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic(g) Triclinic",
        "answer": "Triclinic is the only crystal system for which none of the interaxial angles are equal to one another and also not equal to 90^{\\circ}.",
        "type": 2
    },
    {
        "idx": 349,
        "question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?(a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic\ng) Triclinic",
        "answer": "Cubic, tetragonal and orthorhombic crystal systems all have the three interaxial angles equal to 90^{\\circ}.",
        "type": 2
    },
    {
        "idx": 350,
        "question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?\n\\[\n\\alpha=\\beta=90^{\\circ}, \\gamma=120^{\\circ}\n\\](a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic(g) Triclinic",
        "answer": "Only the hexagonal crystal system has two of the interaxial angles equal to 90^{\\circ}, while the third angle is equal to 120^{\\circ}.",
        "type": 2
    },
    {
        "idx": 351,
        "question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship? \\[\n\\alpha=\\beta=\\gamma \\neq 90^{\\circ}\n\\](a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f)\nMonoclinic(g) Triclinic",
        "answer": "Only the rhombohedral crystal system has all three interaxial angles equal to one other, but not equal to 90^{\\circ}.",
        "type": 2
    },
    {
        "idx": 352,
        "question": "Which crystal system(s) listed below has (have) the following interaxial angle relationship?a) Cubic(b) Hexagonal(c) Tetragonal(d) Rhombohedral(e) Orthorhombic(f) Monoclinic(g) Triclinic",
        "answer": "Only the monoclinic crystal system has two of the interaxial angles equal to one another and to 90^{\\circ}, which is different than the third interaxial angle.",
        "type": 2
    },
    {
        "idx": 353,
        "question": "Polyethylene may be fluorinated by inducing the random substitution of fluorine atoms for hydrogen. For this polymer, determine the concentration of F (in wt %) that must be added if this substitution occurs for 18.6 % of all of the original hydrogen atoms. Atomic weights for several elements are included in the following table: Carbon 12.01 g/mol, Chlorine 35.45 g/mol, Fluorine 19.00 g/mol, Hydrogen 1.008 g/mol, Oxygen 16.00 g/mol.",
        "answer": "the concentration of F that must be added is 18.0 wt %.",
        "type": 1,
        "wrong_answers_1": "the concentration of F that must be added is 15.5 wt %.",
        "wrong_answers_2": "the concentration of F that must be added is 18.8 wt %.",
        "wrong_answers_3": "the concentration of F that must be added is 20.3 wt %."
    },
    {
        "idx": 354,
        "question": "Polyethylene may be fluorinated by inducing the random substitution of fluorine atoms for hydrogen. For this polymer, determine the concentration of F (in wt %) that must be added to completely fluorinate the material, i.e. to produce polytetrafluoroethylene (PTFE). Atomic weights for several elements are included in the following table: Carbon 12.01 g/mol, Chlorine 35.45 g/mol, Fluorine 19.00 g/mol, Hydrogen 1.008 g/mol, Oxygen 16.00 g/mol.",
        "answer": "the concentration of F required to completely fluorinate the material is 76.0 wt %.",
        "type": 1,
        "wrong_answers_1": "the concentration of F required to completely fluorinate the material is 83.1 wt %.",
        "wrong_answers_2": "the concentration of F required to completely fluorinate the material is 84.4 wt %.",
        "wrong_answers_3": "the concentration of F required to completely fluorinate the material is 65.3 wt %."
    },
    {
        "idx": 355,
        "question": "Which of the following may form linear polymers?(a) Rubber(b) Epoxy(c) Polyethylene(d) Phenol-formaldehyde(e) Polystyrene(f) Nylon",
        "answer": "Polyethylene, polystyrene and nylon may form linear polymers.",
        "type": 2
    },
    {
        "idx": 356,
        "question": "Which of the following form network polymers?(a) Rubber(b) Epoxy(c) Polyethylene\nd) Phenol-formaldehyde(e) Polystyrene(f) Nylon\n\\",
        "answer": "Epoxy and phenol-formaldehyde form network polymers.",
        "type": 2
    },
    {
        "idx": 357,
        "question": "For most polymers, which configuration predominates?(a) Head-to-head(b) Head-to-tail",
        "answer": "For most polymers, the head-to-tail configuration predominates.",
        "type": 2
    },
    {
        "idx": 358,
        "question": "Is it possible to produce a polymer that is 100% crystalline? (a) True (b) False",
        "answer": "False. It is not possible to produce a polymer that is 100% crystalline. The maximum crystallinity that can be obtained is about 95%, with the remaining material being amorphous.",
        "type": 3
    },
    {
        "idx": 359,
        "question": "The number of vacancies in some hypothetical metal increases by a factor of 5 when the temperature is increased from 1040 K to 1150 K. Calculate the energy (in kj/mol ) for vacancy formation assuming that the density of the metal remains the same over this temperature range.",
        "answer": "the energy for vacancy formation is 115 kj/mol.",
        "type": 1,
        "wrong_answers_1": "the energy for vacancy formation is 123 kj/mol.",
        "wrong_answers_2": "the energy for vacancy formation is 100 kj/mol.",
        "wrong_answers_3": "the energy for vacancy formation is 105 kj/mol."
    },
    {
        "idx": 360,
        "question": "The number of vacancies present in some metal at 864^{\\circ} C is 1.1 × 10^{24}{m}^{-3}. Calculate the number of vacancies at 463^{\\circ} C given that the energy for vacancy formation is 1.25 \\mathrm{eV} / atom; assume that the density at both temperatures is the same",
        "answer": "the number of vacancies at 463^{\\circ} C is 5.45 × 10^{22}{m}^{-3}.",
        "type": 1,
        "wrong_answers_1": "the number of vacancies at 517^{\\circ} C is 5.99 × 10^{22}{m}^{-3}.",
        "wrong_answers_2": "the number of vacancies at 437^{\\circ} C is 6.02 × 10^{22}{m}^{-3}.",
        "wrong_answers_3": "the number of vacancies at 508^{\\circ} C is 5.82 × 10^{22}{m}^{-3}."
    },
    {
        "idx": 361,
        "question": "In metals, there are significantly more vacancies than self-interstitials.(a) True(b) False",
        "answer": "True. In metals, there are significantly more vacancies than self-interstitials; the reason for this is that the atom is significantly larger than the interstitial position in which it is situated, and, consequently significant lattice strains result.",
        "type": 3
    },
    {
        "idx": 362,
        "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answers for MgO.",
        "answer": "For MgO, the ionic radii of the Mn2+ and Mg2+ are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.072 nm - 0.067 nm) / 0.072 nm × 100 = 6.9%, which value is within the acceptable range for a high degree of solubility.",
        "type": 4,
        "wrong_answers_1": "For NiO, the ionic radii of the Mn2+ and Ni2+ are 0.067 nm and 0.069 nm, respectively. Therefore, the percentage difference in ionic radii Δr% is determined as follows: Δr% = (0.069 nm - 0.067 nm) / 0.069 nm × 100 = 3%, which value is, of course, within the acceptable range for a high degree of solubility.",
        "wrong_answers_2": "For CaO, the ionic radii of the Mn2+ and Ca2+ are 0.067 nm and 0.100 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.100 nm - 0.067 nm) / 0.100 nm × 100 = 33%. This Δr% value is much larger than the ±15% range, and, therefore, CaO is not expected to experience any appreciable solubility in MnO.",
        "wrong_answers_3": "For BeO, the ionic radii of the Mn2+ and Be2+ are 0.067 nm and 0.035 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.067 nm - 0.035 nm) / 0.067 nm × 100 = 48%. This Δr% value is much larger than the ±15% range, and, therefore, BeO is not expected to experience any appreciable solubility in MnO."
    },
    {
        "idx": 363,
        "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answers for CaO.",
        "answer": "For CaO, the ionic radii of the Mn2+ and Ca2+ are 0.067 nm and 0.100 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.100 nm - 0.067 nm) / 0.100 nm × 100 = 33%. This Δr% value is much larger than the ±15% range, and, therefore, CaO is not expected to experience any appreciable solubility in MnO.",
        "type": 4,
        "wrong_answers_1": "For BeO, the ionic radii of the Mn2+ and Be2+ are 0.067 nm and 0.035 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.067 nm - 0.035 nm) / 0.067 nm × 100 = 48%. This Δr% value is much larger than the ±15% range, and, therefore, BeO is not expected to experience any appreciable solubility in MnO.",
        "wrong_answers_2": "For MgO, the ionic radii of the Mn2+ and Mg2+ are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.072 nm - 0.067 nm) / 0.072 nm × 100 = 6.9%, which value is within the acceptable range for a high degree of solubility.",
        "wrong_answers_3": "For NiO, the ionic radii of the Mn2+ and Ni2+ are 0.067 nm and 0.069 nm, respectively. Therefore, the percentage difference in ionic radii Δr% is determined as follows: Δr% = (0.069 nm - 0.067 nm) / 0.069 nm × 100 = 3%, which value is, of course, within the acceptable range for a high degree of solubility."
    },
    {
        "idx": 364,
        "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answers for BeO.",
        "answer": "For BeO, the ionic radii of the Mn2+ and Be2+ are 0.067 nm and 0.035 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.067 nm - 0.035 nm) / 0.067 nm × 100 = 48%. This Δr% value is much larger than the ±15% range, and, therefore, BeO is not expected to experience any appreciable solubility in MnO.",
        "type": 4,
        "wrong_answers_1": "For CaO, the ionic radii of the Mn2+ and Ca2+ are 0.067 nm and 0.100 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.100 nm - 0.067 nm) / 0.100 nm × 100 = 33%. This Δr% value is much larger than the ±15% range, and, therefore, CaO is not expected to experience any appreciable solubility in MnO.",
        "wrong_answers_2": "For MgO, the ionic radii of the Mn2+ and Mg2+ are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.072 nm - 0.067 nm) / 0.072 nm × 100 = 6.9%, which value is within the acceptable range for a high degree of solubility.",
        "wrong_answers_3": "For NiO, the ionic radii of the Mn2+ and Ni2+ are 0.067 nm and 0.069 nm, respectively. Therefore, the percentage difference in ionic radii Δr% is determined as follows: Δr% = (0.069 nm - 0.067 nm) / 0.069 nm × 100 = 3%, which value is, of course, within the acceptable range for a high degree of solubility."
    },
    {
        "idx": 365,
        "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MnO? Explain your answers for NiO.",
        "answer": "For NiO, the ionic radii of the Mn2+ and Ni2+ are 0.067 nm and 0.069 nm, respectively. Therefore, the percentage difference in ionic radii Δr% is determined as follows: Δr% = (0.069 nm - 0.067 nm) / 0.069 nm × 100 = 3%, which value is, of course, within the acceptable range for a high degree of solubility.",
        "type": 4,
        "wrong_answers_1": "For MgO, the ionic radii of the Mn2+ and Mg2+ are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.072 nm - 0.067 nm) / 0.072 nm × 100 = 6.9%, which value is within the acceptable range for a high degree of solubility.",
        "wrong_answers_2": "For CaO, the ionic radii of the Mn2+ and Ca2+ are 0.067 nm and 0.100 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.100 nm - 0.067 nm) / 0.100 nm × 100 = 33%. This Δr% value is much larger than the ±15% range, and, therefore, CaO is not expected to experience any appreciable solubility in MnO.",
        "wrong_answers_3": "For BeO, the ionic radii of the Mn2+ and Be2+ are 0.067 nm and 0.035 nm, respectively. Therefore, the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.067 nm - 0.035 nm) / 0.067 nm × 100 = 48%. This Δr% value is much larger than the ±15% range, and, therefore, BeO is not expected to experience any appreciable solubility in MnO."
    },
    {
        "idx": 366,
        "question": "Which of these elements would you expect to form a substitutional solid solution having complete solubility with copper? The elements are: Ni, O, H, Pt, Pd, Co, C, Zn, Ag, Al, Cr, Fe. The criteria for complete solubility are: 1) the difference in atomic radii between Cu and the other element must be less than ±15%, 2) the crystal structures must be the same, 3) the electronegativities must be similar, and 4) the valences should be the same, or nearly the same.",
        "answer": "Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures.",
        "type": 2
    },
    {
        "idx": 367,
        "question": "Which of these elements would you expect to form a substitutional solid solution of incomplete solubility with copper? The elements are: Pd, Al, Cr, Fe, H, Pt, Ni, Zn, C, Ag, O, Co. The criteria for incomplete solubility are: these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Cu are greater than ±15%, and/or have a valence different than 2+.",
        "answer": "Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Cu are greater than ±15%, and/or have a valence different than 2+.",
        "type": 2
    },
    {
        "idx": 368,
        "question": "Which of these elements would you expect to form an interstitial solid solution with copper? The elements are: C, Pd, Pt, H, Cr, O, Zn, Ag, Al, Co, Fe, Ni. The criteria for interstitial solid solutions are: these elements have atomic radii that are significantly smaller than the atomic radius of Cu.",
        "answer": "C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Cu.",
        "type": 2
    },
    {
        "idx": 369,
        "question": "The concentration of carbon in an iron-carbon alloy is 0.57 wt%. What is the concentration in kilograms of carbon per cubic meter of alloy? The densities of iron and carbon are 7.87 and 2.25 g / {cm}^{3}, respectively.",
        "answer": "the concentration of carbon is 11.8 kg / m^{3}.",
        "type": 1,
        "wrong_answers_1": "the concentration of carbon is 13.1 kg / m^{3}.",
        "wrong_answers_2": "the concentration of carbon is 13.2 kg / m^{3}.",
        "wrong_answers_3": "the concentration of carbon is 13.3 kg / m^{3}."
    },
    {
        "idx": 370,
        "question": "13}\nSome hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and \\mathrm{B} are 4.27 and 6.35g / {cm}^{3}, respectively, whereas their respective atomic weights are 61.4 and 125.7g / mol, determine whether the crystal structure for this alloy is simple cubic, facercentered cubic, or body-centered cubic. Assume a unit cell edge length of 0.395nm.",
        "answer": "the crystal structure is body-centered cubic.",
        "type": 1
    },
    {
        "idx": 371,
        "question": "17}\nIron and vanadium both have the BCC crystal structure andV forms a substitutional solid solution in \\mathrm{Fe} for concentrations up to approximately 20 wt%V at room temperature. Determine the concentration in weight percent ofV that must be added to iron to yield a unit cell edge length of 0.289nm.",
        "answer": "the concentration of \\mathrm{v} that must be added to iron to yield a unit cell edge length of 0.289nm is 12.9 wt%.",
        "type": 1,
        "wrong_answers_1": "the concentration of \\mathrm{v} that must be added to iron to yield a unit cell edge length of 0.299nm is 11.7 wt%.",
        "wrong_answers_2": "the concentration of \\mathrm{v} that must be added to iron to yield a unit cell edge length of 0.324nm is 12.3 wt%.",
        "wrong_answers_3": "the concentration of \\mathrm{v} that must be added to iron to yield a unit cell edge length of 0.265nm is 14.3 wt%."
    },
    {
        "idx": 372,
        "question": "Which of the following is a (are) linear defect(s)?(a) An edge dislocation(b) A Frenkel defect(c) A Schottky defect",
        "answer": "Edge and screw dislocations are linear defects.",
        "type": 2
    },
    {
        "idx": 373,
        "question": "A photomicrograph was taken of a specimen at a magnification of 100 ×, and it was determined that the average number of grains per square inch was 200 . What is this specimen's ASTM grain size number?",
        "answer": "the specimen's astm grain size number is 4.9.",
        "type": 1,
        "wrong_answers_1": "the specimen's astm grain size number is 5.2.",
        "wrong_answers_2": "the specimen's astm grain size number is 5.5.",
        "wrong_answers_3": "the specimen's astm grain size number is 4.4."
    },
    {
        "idx": 374,
        "question": "Diffusion by which mechanism occurs more rapidly in metal alloys?(a) Vacancy diffusion(b) Interstitial diffusion",
        "answer": "In metal alloys, interstitial diffusion takes place more rapidly than vacancy diffusion because the interstitial atoms are smaller and are more mobile. Also, there are more vacant adjacent interstitial sites than there are vacancies.",
        "type": 2
    },
    {
        "idx": 375,
        "question": "As temperature decreases, the fraction of total number of atoms that are capable of diffusive motion(a) increases.(b) decreases.",
        "answer": "As temperature decreases, the fraction of the total number of atoms that are capable of diffusive motion decreases.",
        "type": 2
    },
    {
        "idx": 376,
        "question": "If [m] atoms of helium pass through a [a] square meter plate area every [t] hours, and if this flux is constant with time, compute the flux of helium in units of atoms per square meter per second.",
        "answer": "\\[\nj = \\frac{[m]}{[a] × [t] × 3600} \\text{ atoms per square meter per second}\n\\]",
        "type": 1,
        "wrong_answers_1": "\\[\nj = \\frac{[m]}{[a] × [t] × 3290} \\text{ atoms per square meter per second}\n\\]",
        "wrong_answers_2": "\\[\nj = \\frac{[m]}{[a] × [t] × 3180} \\text{ atoms per square meter per second}\n\\]",
        "wrong_answers_3": "\\[\nj = \\frac{[m]}{[a] × [t] × 3370} \\text{ atoms per square meter per second}\n\\]"
    },
    {
        "idx": 377,
        "question": "If water molecules pass through a membrane with a steady state flux of [j] mole /\\left(m^{2}\\right. day ), how long will it take, in hours, for [m] kg of water to pass through a [\\mathrm{a}] square centimeter of the membrane?",
        "answer": "\\[\nt = \\frac{[m] × 1000 × 24}{[a] × [j] × 18 × 10000} \\text{ hours}\n\\]",
        "type": 1,
        "wrong_answers_1": "\\[\nt = \\frac{[m] × 1100 × 26}{[a] × [j] × 17 × 10000} \\text{ hours}\n\\]",
        "wrong_answers_2": "\\[\nt = \\frac{[m] × 1150 × 25}{[a] × [j] × 17 × 10000} \\text{ hours}\n\\]",
        "wrong_answers_3": "\\[\nt = \\frac{[m] × 880 × 27}{[a] × [j] × 17 × 10000} \\text{ hours}\n\\]"
    },
    {
        "idx": 378,
        "question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the contact lens thickness likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.",
        "answer": "If the contact lens thickness is increased and all other factors are preserved, the concentration gradient decrease and the oxygen diffusion flux should decrease.",
        "type": 4,
        "wrong_answers_1": "Increasing the concentration of oxygen in the ambient environment will increase the concentration gradient, thereby increasing the diffusion flux of oxygen.",
        "wrong_answers_2": "The driving force for steady-state diffusion is the concentration gradient.",
        "wrong_answers_3": "If the lens features more voids, we expect the oxygen to permeate the membrane faster since oxygen should be able to transit voids faster than the bulk lens material."
    },
    {
        "idx": 379,
        "question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the diffusivity of oxygen gas by decreasing the contact lens porosity likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.",
        "answer": "If the lens features more voids, we expect the oxygen to permeate the membrane faster since oxygen should be able to transit voids faster than the bulk lens material.",
        "type": 4,
        "wrong_answers_1": "If the contact lens thickness is increased and all other factors are preserved, the concentration gradient decrease and the oxygen diffusion flux should decrease.",
        "wrong_answers_2": "diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.",
        "wrong_answers_3": "oxygen vacancies. Two Li+ ions added, a single oxygen vacancy is formed."
    },
    {
        "idx": 380,
        "question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the ambient temperature likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.",
        "answer": "Increasing the temperature should increase the diffusivity and therefore increase the diffusion flux and flow rate.",
        "type": 4,
        "wrong_answers_1": "Increasing the concentration of oxygen in the ambient environment will increase the concentration gradient, thereby increasing the diffusion flux of oxygen.",
        "wrong_answers_2": "Three factors that affect the rate of drying are temperature, humidity, and rate of air flow. The rate of drying is enhanced by increasing both the temperature and rate of air flow, and by decreasing the humidity of the air.",
        "wrong_answers_3": "Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species--i.e., the diffusion flux is independent of time."
    },
    {
        "idx": 381,
        "question": "If an increased steady-state flow rate of O2 (oxygen molecules per second) to the cornea is desired, is increasing the ambient partial pressure of oxygen gas likely to be useful? Note: the flow rate is equal to product of the diffusion flux and an area of interest through which diffusion occurs.",
        "answer": "Increasing the concentration of oxygen in the ambient environment will increase the concentration gradient, thereby increasing the diffusion flux of oxygen.",
        "type": 4,
        "wrong_answers_1": "diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.",
        "wrong_answers_2": "Increasing the temperature should increase the diffusivity and therefore increase the diffusion flux and flow rate.",
        "wrong_answers_3": "If the contact lens thickness is increased and all other factors are preserved, the concentration gradient decrease and the oxygen diffusion flux should decrease."
    },
    {
        "idx": 382,
        "question": "For a steel alloy it has been determined that a carburizing heat treatment of 16h duration at 757^{\\circ} C will raise the carbon concentration to 0.5 wt% at a point 2.3mm from the surface. Estimate the time necessary to achieve the same concentration at a 8mm position for an identical steel and at a carburizing temperature of 1130^{\\circ} C. Assume that D_{0} is 4.6 × 10^{-5}{m}^{2} / s and Q_{d} is 104kJ / mol.",
        "answer": "the time necessary to achieve the same concentration at a 8mm position for an identical steel and at a carburizing temperature of 1130^{\\circ} C is 17.9h.",
        "type": 1,
        "wrong_answers_1": "the time necessary to achieve the same concentration at a 7mm position for an identical steel and at a carburizing temperature of 1290^{\\circ} C is 19.4h.",
        "wrong_answers_2": "the time necessary to achieve the same concentration at a 9mm position for an identical steel and at a carburizing temperature of 990^{\\circ} C is 15.6h.",
        "wrong_answers_3": "the time necessary to achieve the same concentration at a 9mm position for an identical steel and at a carburizing temperature of 1180^{\\circ} C is 16.8h."
    },
    {
        "idx": 383,
        "question": "The diffusion coefficient for aluminum in silicon is D_{\\mathrm{Al}} in_{\\mathrm{Si}}=3 × 10^{-16} cm^{2} / s at 300 K (note that 300 K is about room temperature).\nWhat is a reasonable value for D_{\\mathrm{Al} \\text { in } \\mathrm{Si}} at 600 K ?\nNote: Rather than performing a specific calculation, you should be able to justify your answer from the options below based on the mathematical temperature dependence of the diffusion coefficient. (a) D<3 × 10^{16} cm^{2} /s (b) D=3 × 10^{16} cm^{2} /s (c) D=6 × 10^{16} cm^{2}/s (d) D=1.5 × 10^{16} cm^{2}/s (e) D>6 × 10^{16} cm^{2}/s (f) D=6 × 10^{-17} cm^{2}/s",
        "answer": "(e) D>6 × 10^{16} cm^{2}",
        "type": 2
    },
    {
        "idx": 384,
        "question": "A specimen of some metal having a rectangular cross section 11.2mm × 12.4mm is pulled in tension with a force of 31200N, which produces only elastic deformation. Given that the elastic modulus of this metal is 63 \\mathrm{GPa}, calculate the resulting strain.",
        "answer": "the resulting strain is 2.92 × 10^{-3}.",
        "type": 1,
        "wrong_answers_1": "the resulting strain is 2.81 × 10^{-3}.",
        "wrong_answers_2": "the resulting strain is 2.79 × 10^{-3}.",
        "wrong_answers_3": "the resulting strain is 3.04 × 10^{-3}."
    },
    {
        "idx": 385,
        "question": "For a bronze alloy, the stress at which plastic deformation begins is 277 MPa and the modulus of elasticity is 117 GPa. What is the maximum load that may be applied to a specimen having a cross-sectional area of 327 mm^2 without plastic deformation?",
        "answer": "the maximum load that may be applied without plastic deformation is 86,700 N.",
        "type": 1,
        "wrong_answers_1": "the maximum load that may be applied without plastic deformation is 97,795 N.",
        "wrong_answers_2": "the maximum load that may be applied without plastic deformation is 98,291 N.",
        "wrong_answers_3": "the maximum load that may be applied without plastic deformation is 90,914 N."
    },
    {
        "idx": 386,
        "question": "For a bronze alloy, the stress at which plastic deformation begins is 277 MPa and the modulus of elasticity is 117 GPa. If the original specimen length is 148 mm, what is the maximum length to which it may be stretched without causing plastic deformation?",
        "answer": "the maximum length to which the specimen may be stretched without causing plastic deformation is 131.30 mm.",
        "type": 1,
        "wrong_answers_1": "the maximum length to which the specimen may be stretched without causing plastic deformation is 135.58 mm.",
        "wrong_answers_2": "the maximum length to which the specimen may be stretched without causing plastic deformation is 140.97 mm.",
        "wrong_answers_3": "the maximum length to which the specimen may be stretched without causing plastic deformation is 144.47 mm."
    },
    {
        "idx": 387,
        "question": "A steel bar 100mm (4.0 in.) long and having a square cross section 20mm (0.8 in.) on an edge is pulled in tension with a load of 89,000N\\left(20,000 lb_{0}\\right), and experiences an elongation of 0.10mm (4.0 × 10^{-3} in.). Assuming that the deformation is entirely elastic, calculate the elastic modulus of the steel.",
        "answer": "the elastic modulus of the steel is 223 gpa (31.3 × 10^6 psi).",
        "type": 1,
        "wrong_answers_1": "the elastic modulus of the steel is 216 gpa (29.2 × 10^6 psi).",
        "wrong_answers_2": "the elastic modulus of the steel is 236 gpa (30.0 × 10^6 psi).",
        "wrong_answers_3": "the elastic modulus of the steel is 238 gpa (29.5 × 10^6 psi)."
    },
    {
        "idx": 388,
        "question": "13: Typical relationship between E and G}\nFor most metals, the relationship between elastic and shear moduli is approximately which of the following?(a) G=0.1 E(b) G=0.2 E(c) G=0.3 E(d) G=0.4 E(e) G=0.5 E",
        "answer": "(d) G=0.4 E",
        "type": 2
    },
    {
        "idx": 389,
        "question": "A cylindrical rod 380mm (15.0 in.) long and having a diameter of 10.0mm (0.40 in.), is to be subjected to a tensile load of 24,500 N (5500 lbf). Which of the four metals or alloys listed below will not experience plastic deformation under this load? The materials are: (a)Aluminum alloy (Yield Strength: 255 MPa) (b)Brass alloy (Yield Strength: 345 MPa)(c)Copper (Yield Strength: 250 MPa) (d)Steel alloy (Yield Strength: 450 MPa).",
        "answer": " (b)Brass alloy (Yield Strength: 345 MPa) and (d)Steel alloy (Yield Strength: 450 MPa)",
        "type": 2
    },
    {
        "idx": 390,
        "question": "A cylindrical rod 380mm (15.0 in.) long and having a diameter of 10.0mm (0.40 in.), is to be subjected to a tensile load of 24,500 N (5500 lbf). Which of the four metals or alloys listed below will experience an elongation of no more than 0.9mm (0.035 in.) under this load? The materials are: (a) Aluminum alloy (Modulus of Elasticity: 70 GPa) (b)Brass alloy (Modulus of Elasticity: 100 GPa) (c)Copper (Modulus of Elasticity: 110 GPa) (d)Steel alloy (Modulus of Elasticity: 207 GPa).",
        "answer": "(b)Brass alloy (c)Copper(d)Steel alloy",
        "type": 2
    },
    {
        "idx": 392,
        "question": "A tensile test is performed on a specimen of some metal alloy, and it is found that a true plastic strain of 0.12 is produced when a true stress of 280 MPa is applied. For this alloy, the value of the strain hardening exponent is 0.3 . On the basis of these data what true plastic strain would be expected for a total true plastic stress of 330 MPa ?",
        "answer": "the true plastic strain expected for a total true plastic stress of 330 mpa is 0.383.",
        "type": 1,
        "wrong_answers_1": "the true plastic strain expected for a total true plastic stress of 350 mpa is 0.350.",
        "wrong_answers_2": "the true plastic strain expected for a total true plastic stress of 352 mpa is 0.422.",
        "wrong_answers_3": "the true plastic strain expected for a total true plastic stress of 355 mpa is 0.409."
    },
    {
        "idx": 393,
        "question": "A cylindrical specimen of a metal alloy 48.8mm long and 9.09mm in diameter is stressed in tension. A true stress of 327 MPa causes the specimen to plastically elongate to a length of 55 mm. If it is known that the strain-hardening exponent for this alloy is 0.3 , calculate the true stress (in MPa )necessary to plastically elongate a specimen of this same material from a length of 48.8mm to a length of 57.6 mm.",
        "answer": "390 MPa",
        "type": 1,
        "wrong_answers_1": "372 MPa",
        "wrong_answers_2": "430 MPa",
        "wrong_answers_3": "351 MPa"
    },
    {
        "idx": 394,
        "question": "A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 390 MPa (56,600 psi). If the specimen radius is 2.5mm (0.10 in.) and the support point separation distance is 30mm (1.2 in.), predict whether or not you would expect the specimen to fracture when a load of 620 N (140 lbf) is applied.",
        "answer": "Since the flexural strength for this test is 379 MPa (53,500 psi), which is less than the reported flexural strength of 390 MPa, fracture is not predicted. However, there is some chance of fracture due to variability in material strength.",
        "type": 1,
        "wrong_answers_1": "Since the flexural strength for this test is 356 MPa (60,750 psi), which is less than the reported flexural strength of 347 MPa, fracture is not predicted. However, there is some chance of fracture due to variability in material strength.",
        "wrong_answers_2": "Since the flexural strength for this test is 392 MPa (56,067 psi), which is less than the reported flexural strength of 430 MPa, fracture is not predicted. However, there is some chance of fracture due to variability in material strength.",
        "wrong_answers_3": "Since the flexural strength for this test is 393 MPa (50,863 psi), which is less than the reported flexural strength of 444 MPa, fracture is not predicted. However, there is some chance of fracture due to variability in material strength."
    },
    {
        "idx": 395,
        "question": "Calculate the value of flexural strength for this test given a cylindrical specimen of aluminum oxide with a radius of 2.5mm (0.10 in.) and a support point separation distance of 30mm (1.2 in.) when a load of 620 N (140 lbf) is applied.",
        "answer": "The flexural strength for this test is 379 MPa (53,500 psi).",
        "type": 1,
        "wrong_answers_1": "The flexural strength for this test is 323 MPa (51177 psi).",
        "wrong_answers_2": "The flexural strength for this test is 421 MPa (57461 psi).",
        "wrong_answers_3": "The flexural strength for this test is 335 MPa (47098 psi)."
    },
    {
        "question": "The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows: \begin{tabular}{cc} \\hline \\sigma_{\\mathrm{fs}} (MPa) & \\mathbf{P} \\ \\hline 100 & 0.05 \\ 50 & 0.20 \\ \\hline \\end{tabular} (a) Compute the flexural strength for a completely nonporous specimen of this material.",
        "answer": "the flexural strength for a completely nonporous specimen of this material is 126 mpa.",
        "idx": 396,
        "type": 1,
        "wrong_answers_1": "the flexural strength for a completely nonporous specimen of this material is 120 mpa.",
        "wrong_answers_2": "the flexural strength for a completely nonporous specimen of this material is 112 mpa.",
        "wrong_answers_3": "the flexural strength for a completely nonporous specimen of this material is 109 mpa."
    },
    {
        "question": "The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows: \begin{tabular}{cc} \\hline \\sigma_{\\mathrm{fs}} (MPa) & \\mathbf{P} \\ \\hline 100 & 0.05 \\ 50 & 0.20 \\ \\hline \\end{tabular} (b) Compute the flexural strength for a 0.10 volume fraction porosity.",
        "answer": "the flexural strength for a 0.10 volume fraction porosity is 79.4 mpa.",
        "idx": 397,
        "type": 1,
        "wrong_answers_1": "the flexural strength for a 0.09 volume fraction porosity is 74.2 mpa.",
        "wrong_answers_2": "the flexural strength for a 0.10 volume fraction porosity is 71.0 mpa.",
        "wrong_answers_3": "the flexural strength for a 0.09 volume fraction porosity is 69.3 mpa."
    },
    {
        "idx": 398,
        "question": "Mechanical twinning occurs in metals having which type(s) of crystal structure(s)? (a) bcc (b) fcc (c) hcp",
        "answer": " (a) bcc (c) hcp",
        "type": 2
    },
    {
        "idx": 399,
        "question": "How does grain size influence strength of a polycrystalline material?(a) Strengthfine-grained < strength_course-grained(b) Strength_fine-grained = strength_course-grained (c) Strength fine-grained > strength_course-grained",
        "answer": " (c) Strength fine-grained > strength_course-grained",
        "type": 2
    },
    {
        "idx": 400,
        "question": "Reducing the grain size of metal improves toughness.(a) True(b) False",
        "answer": "True.",
        "type": 3
    },
    {
        "idx": 401,
        "question": "As dislocation density increases, the resistance to dislocation movement(a) increases.(b) decreases.",
        "answer": "(a) increases",
        "type": 2
    },
    {
        "idx": 402,
        "question": "On the average, dislocation-dislocation strain interactions are(a) repulsive.(b) attractive.",
        "answer": "(a) repulsive",
        "type": 2
    },
    {
        "idx": 403,
        "question": "As a metal is strain hardened, its ductility(a) increases(b) decreases",
        "answer": "(b) decreases",
        "type": 2
    },
    {
        "idx": 404,
        "question": "Most metals strain harden at room temperature.(a) True(b) False",
        "answer": "True",
        "type": 3
    },
    {
        "idx": 405,
        "question": "During the recovery of a cold-worked material, is some of the internal strain energy relieved?",
        "answer": "Some of the internal strain energy is relieved.",
        "type": 3
    },
    {
        "idx": 406,
        "question": "During the recovery of a cold-worked material, is all of the internal strain energy relieved?",
        "answer": "All of the internal strain energy is not relieved.",
        "type": 3
    },
    {
        "idx": 407,
        "question": "During the recovery of a cold-worked material, is there some reduction in the number of dislocations?",
        "answer": "There is some reduction in the number of dislocations.",
        "type": 3
    },
    {
        "idx": 408,
        "question": "During the recovery of a cold-worked material, is there a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state?",
        "answer": "There is not a significant reduction in the number of dislocations, to approximately the number found in the precold-worked state.",
        "type": 3
    },
    {
        "idx": 409,
        "question": "During the recovery of a cold-worked material, is the electrical conductivity recovered to its precold-worked state?",
        "answer": "The electrical conductivity is recovered to its precold-worked state.",
        "type": 3
    },
    {
        "idx": 410,
        "question": "During the recovery of a cold-worked material, is the thermal conductivity recovered to its precold-worked state?",
        "answer": "The thermal conductivity is recovered to its precold-worked state.",
        "type": 3
    },
    {
        "idx": 411,
        "question": "During the recovery of a cold-worked material, does the metal become more ductile, as in its precold-worked state?",
        "answer": "The metal does not become more ductile, as in its precold-worked state.",
        "type": 3
    },
    {
        "idx": 412,
        "question": "During the recovery of a cold-worked material, are grains with high strains replaced with new, unstrained grains?",
        "answer": "Grains with high strains are not replaced with new, unstrained grains.",
        "type": 3
    },
    {
        "idx": 413,
        "question": "During the recrystallization of a cold-worked material, is some of the internal strain energy relieved?",
        "answer": "All of the internal strain energy is relieved.",
        "type": 3
    },
    {
        "idx": 414,
        "question": "During the recrystallization of a cold-worked material, is there some reduction in the number of dislocations?",
        "answer": "There is significant reduction in the number of dislocations.",
        "type": 3
    },
    {
        "idx": 415,
        "question": "During the recrystallization of a cold-worked material, does the metal become more ductile, as in its precold-worked state?",
        "answer": "The metal becomes more ductile, as in its precold-worked state.",
        "type": 3
    },
    {
        "idx": 416,
        "question": "During the recrystallization of a cold-worked material, are grains with high strains replaced with new, unstrained grains?",
        "answer": "Grains with high strains are replaced with new, unstrained grains.",
        "type": 3
    },
    {
        "idx": 417,
        "question": "Grain growth requirements\n}\nGrain growth must always be preceded by recovery and recrystallization.(a) True(b) False",
        "answer": "False. Grain growth does not always need to be preceded by recovery and recrystallization; it may occur in materials that have not been cold worked.",
        "type": 3
    },
    {
        "idx": 418,
        "question": "A hypothetical metal alloy has a grain diameter of 2.4 × 10^{-2} mm. After a heat treatment at 575^{\\circ} C for 500min, the grain diameter has increased to 7.3 × 10^{-2} mm. Compute the time required for a specimen of this same material (i.e., \\mathrm{d}_{0}=2.4 × 10^{-2}mm ) to achieve a grain diameter of 5.5 × 10^{-2}mm while being heated at 575^{\\circ} C. Assume the N grain diameter exponent has a value of 2.2.",
        "answer": "the time required for the specimen to achieve a grain diameter of 5.5 × 10^{-2} mm is 246min.",
        "type": 1,
        "wrong_answers_1": "the time required for the specimen to achieve a grain diameter of 6.0 × 10^{-2} mm is 211min.",
        "wrong_answers_2": "the time required for the specimen to achieve a grain diameter of 5.2 × 10^{-2} mm is 255min.",
        "wrong_answers_3": "the time required for the specimen to achieve a grain diameter of 5.3 × 10^{-2} mm is 267min."
    },
    {
        "idx": 419,
        "question": "Tensile strengths and number-average molecular weights for two polymers are as follows:\nTensile strength Number average molecular weight\n(MPa)\n(g / mol)\n37.7\n36800\n131\n62400\nEstimate the tensile strength (in MPa) for a number-average molecular weight of 51500g / mol.",
        "answer": "the tensile strength for a number-average molecular weight of 51500g/mol is 95.2 MPa.",
        "type": 1,
        "wrong_answers_1": "the tensile strength for a number-average molecular weight of 51500g/mol is 106.0 MPa.",
        "wrong_answers_2": "the tensile strength for a number-average molecular weight of 51500g/mol is 84.5 MPa.",
        "wrong_answers_3": "the tensile strength for a number-average molecular weight of 51500g/mol is 104.2 MPa."
    },
    {
        "idx": 420,
        "question": "Tensile strengths and number-average molecular weights for two polymers are as follows\n\\[\n\\begin{array}{l}\n\\text { Tensile strength Number average molecular weight } \\\\\n(MPa) \\\\\n138 \\\\\n184 \\\\\n\\text { (g/mol) } \\\\\n12600 \\\\\n28100\n\\end{array}\n\\]\nEstimate number average molecular weight (in g / mol ) at a tensile strength of 141 MPa.",
        "answer": "the estimated number average molecular weight at a tensile strength of 141 \\, MPa is 14600 \\, \\mathrm{g/mol}.",
        "type": 1,
        "wrong_answers_1": "the estimated number average molecular weight at a tensile strength of 141 \\, MPa is 1260 \\, \\mathrm{g/mol}.",
        "wrong_answers_2": "the estimated number average molecular weight at a tensile strength of 141 \\, MPa is 1630 \\, \\mathrm{g/mol}.",
        "wrong_answers_3": "the estimated number average molecular weight at a tensile strength of 141 \\, MPa is 1320 \\, \\mathrm{g/mol}."
    },
    {
        "idx": 421,
        "question": "The bonding forces between adhesive and adherend surfaces are thought to be(a) Electrostatic(b) Covalent(c) Chemical",
        "answer": "(a) Electrostatic",
        "type": 2
    },
    {
        "idx": 422,
        "question": "Deformation of a semicrystalline polymer by drawing produces which of the following in the direction of drawing?",
        "answer": "Increase in strength in the direction of drawing.",
        "type": 4,
        "wrong_answers_1": "Decrease in strength perpendicular to the direction of drawing.",
        "wrong_answers_2": "Increasing the temperature should increase the diffusivity and therefore increase the diffusion flux and flow rate.",
        "wrong_answers_3": "Laminar composites are a series of sheets or panels, each of which has a preferred high-strength direction. These sheets are stacked and then cemented together such that the orientation of the high-strength direction varies from layer to layer."
    },
    {
        "idx": 423,
        "question": "Deformation of a semicrystalline polymer by drawing produces which of the following perpendicular to the direction of drawing?",
        "answer": "Decrease in strength perpendicular to the direction of drawing.",
        "type": 4,
        "wrong_answers_1": "Increase in strength in the direction of drawing.",
        "wrong_answers_2": "screw dislocation--perpendicular",
        "wrong_answers_3": "Laminar composites are a series of sheets or panels, each of which has a preferred high-strength direction. These sheets are stacked and then cemented together such that the orientation of the high-strength direction varies from layer to layer."
    },
    {
        "idx": 424,
        "question": "How does deformation by drawing of a semicrystalline polymer affect its tensile strength?(a) Increases(b) Decreases",
        "answer": "Deformation by drawing increases the tensile strength of a semicrystalline polymer. This effect is due to the highly oriented chain structure that is produced by drawing, which gives rise to higher interchain secondary bonding forces.",
        "type": 2
    },
    {
        "idx": 425,
        "question": "How does increasing the degree of crystallinity of a semicrystalline polymer affect its tensile strength?(a) Increases(b) Decreases",
        "answer": "(a) Increases",
        "type": 2
    },
    {
        "idx": 426,
        "question": "How does increasing the molecular weight of a semicrysatlline polymer affect its tensile strength?(a) Increases(b) Decreases",
        "answer": "(a) Increases",
        "type": 2
    },
    {
        "idx": 427,
        "question": "Which kind of fracture (ductile or brittle) is associated with intergranular crack propagation?",
        "answer": "Intergranular fracture is brittle.",
        "type": 3
    },
    {
        "idx": 428,
        "question": "Which kind of fracture (ductile or brittle) is associated with transgranular crack propagation?",
        "answer": "Transgranular fracture is brittle.",
        "type": 3
    },
    {
        "idx": 429,
        "question": "The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter the surface crack geometry (i.e. reduce crack length and increase tip radius). Calculate the ratio of the etched and original crack tip radii if the fracture strength is increased by a factor of 7 when 28% of the crack length is removed.",
        "answer": "the ratio of the etched and original crack tip radii is: \\[\n\\frac{\\rho_{t}^{\\prime}}{\\rho_{t}} = 32\n\\]",
        "type": 1,
        "wrong_answers_1": "the ratio of the etched and original crack tip radii is: \\[\n\\frac{\\rho_{t}^{\\prime}}{\\rho_{t}} = 29\n\\]",
        "wrong_answers_2": "the ratio of the etched and original crack tip radii is: \\[\n\\frac{\\rho_{t}^{\\prime}}{\\rho_{t}} = 33\n\\]",
        "wrong_answers_3": "the ratio of the etched and original crack tip radii is: \\[\n\\frac{\\rho_{t}^{\\prime}}{\\rho_{t}} = 28\n\\]"
    },
    {
        "idx": 430,
        "question": "A structural component in the shape of a flat plate 27.3mm thick is to be fabricated from a metal alloy for which the yield strength and plane strain fracture toughness values are 535 MPa and 31.3 MPa·m^{-1 / 2}, respectively. For this particular geometry, the value of Y is 1.8 . Assuming a design stress of 0.5 times the yield strength, calculate the critical length of a surface flaw.",
        "answer": "the critical length of a surface flaw is 2.0 mm.",
        "type": 1,
        "wrong_answers_1": "the critical length of a surface flaw is 1.9 mm.",
        "wrong_answers_2": "the critical length of a surface flaw is 1.8 mm.",
        "wrong_answers_3": "the critical length of a surface flaw is 1.7 mm."
    },
    {
        "idx": 431,
        "question": "How would the plane strain fracture toughness of a metal be expected to change with rising temperature?(a) Increase(b) Decrease(c) Remain constant",
        "answer": "(a) Increase",
        "type": 2
    },
    {
        "idx": 432,
        "question": "Does increasing temperature favor brittle fracture in polymers?",
        "answer": "No, increasing temperature does not favor brittle fracture in polymers.",
        "type": 3
    },
    {
        "idx": 433,
        "question": "Does increasing strain rate favor brittle fracture in polymers?",
        "answer": "Yes, increasing strain rate favors brittle fracture in polymers.",
        "type": 3
    },
    {
        "idx": 434,
        "question": "Does the presence of a sharp notch favor brittle fracture in polymers?",
        "answer": "Yes, the presence of a sharp notch favors brittle fracture in polymers.",
        "type": 3
    },
    {
        "idx": 435,
        "question": "Does decreasing specimen thickness favor brittle fracture in polymers?",
        "answer": "No, decreasing specimen thickness does not favor brittle fracture in polymers.",
        "type": 3
    },
    {
        "idx": 436,
        "question": "A plate of an alloy steel has a plane-strain fracture toughness of 50 MPa·{m}^{1 / 2}. If it is known that the largest surface crack is 0.5mm long, and that the value of Y is 1.1 , which of the following can be said about this plate when a tensile stress of 1200 MPa is applied?(a) The plate will definitely fracture.(b) The plate will definitely not fracture. (c) It is not possible to determine whether or not the plate will fracture.",
        "answer": "(a)the plate will definitely fracture.",
        "type": 2
    },
    {
        "idx": 437,
        "question": "The effect of a stress raiser is more significant for which of the following types of materials?(a) Brittle materials(b) Ductile materials",
        "answer": "(a) Brittle materials",
        "type": 2
    },
    {
        "idx": 438,
        "question": "A structural component is fabricated from an alloy that has a plane strain fracture toughness of 45 MPa. It has been determined that this component fails at a stress of 300 MPa when the maximum length of a surface crack is 0.95 mm. What is the maximum allowable surface crack length (in mm ) without fracture for this same component exposed to a stress of 300 MPa and made from another alloy with a plane strain fracture toughness of 100.0 MPa ? The geometry factor Y is the same in both cases.",
        "answer": "the maximum allowable surface crack length without fracture is 1.55 mm.",
        "type": 1,
        "wrong_answers_1": "the maximum allowable surface crack length without fracture is 1.76 mm.",
        "wrong_answers_2": "the maximum allowable surface crack length without fracture is 1.48 mm.",
        "wrong_answers_3": "the maximum allowable surface crack length without fracture is 1.64 mm."
    },
    {
        "idx": 439,
        "question": "A cylindrical specimen 7.5mm in diameter of an S-590 alloy is to be exposed to a tensile load of 9000 N. At approximately what temperature will the steady-state creep be 10^{-2}h^{-1} ?",
        "answer": "approximately 815^\\circ C.",
        "type": 1,
        "wrong_answers_1": "approximately 907^\\circ C.",
        "wrong_answers_2": "approximately 785^\\circ C.",
        "wrong_answers_3": "approximately 847^\\circ C."
    },
    {
        "idx": 440,
        "question": "The mineral olivine is a solid solution of the silicate compounds forsterite \\left(Mg_{2} \\mathrm{SiO}_{4}\\right) and fayalite \\left(\\mathrm{Fe}_{2} \\mathrm{SiO}_{4}\\right). How many chemical components are there in a sample of olivine? (a) 1 (b) 2 (c) 3 (d) 4",
        "answer": "(b)2",
        "type": 2
    },
    {
        "idx": 441,
        "question": "Seawater, which covers the majority of the earth, is composed primarily of molecules of H_{2} \\mathrm{O} and equal numbers of \\mathrm{Na}^{+}ions and \\mathrm{Cl}^{-}ions. Suppose we have a thoroughly mixed solution (containing these species only) at 25^{\\circ} C. How many components and how many phases are in such a system?\n(a) 1 component, 1 phase\n(b) 1 component, 2 phase\n(c) 1 component, 3 phase\n(d) 1 component, 4 phase\n(e) 2 component, 1 phase\n(f) 2 component, 2 phase\n(g) 2 component, 3 phase\n(h) 2 component, 4 phase\n(i) 3 component, 1 phase\n(j) 3 component, 2 phase\n(k) 3 component, 3 phase(l) 3 component, 4 phase\n(m) 4 component, 1 phase\n(n) 4 component, 2 phase\n(o) 4 component, 3 phase\n(p) 4 component, 4 phase",
        "answer": "(e) 2 component, 1 phase",
        "type": 2
    },
    {
        "idx": 442,
        "question": "Is solid ductile cast iron (ferrite solid solution + embedded graphite spheres) a two-phase material system?",
        "answer": "Yes, it is a two-phase system because there is a physical boundary beyond the particle level that separates chemically and structurally distinct volumes.",
        "type": 3
    },
    {
        "idx": 443,
        "question": "Is solid sodium chloride (salt, NaCl) a two-phase material system?",
        "answer": "No, it is a single-phase compound, not a solution. The ratio of the two ions is in a fixed stoichiometry and they adopt an orderly arrangement in the crystal.",
        "type": 3
    },
    {
        "idx": 444,
        "question": "Is liquid bronze (Cu + Sn liquid solution) a two-phase material system?",
        "answer": "No, it is a liquid solution, and solutions are single-phase by definition.",
        "type": 3
    },
    {
        "idx": 445,
        "question": "Is solid gray cast iron (ferrite solid solution + embedded graphite flakes) a two-phase material system?",
        "answer": "Yes, it is a two-phase system because there is a physical boundary beyond the particle level that separates chemically and structurally distinct volumes.",
        "type": 3
    },
    {
        "idx": 446,
        "question": "Is solid aluminum featuring dissolved silicon a two-phase material system?",
        "answer": "No, it is a solid solution since silicon is implied to substitute for aluminum. Solutions are single-phase by definition.",
        "type": 3
    },
    {
        "idx": 447,
        "question": "Is solid lead-tin solder (a mixture of Pb-rich and Sn-rich solid solutions) a two-phase material system?",
        "answer": "Yes, it is a two-phase system because there is a physical boundary beyond the particle level that separates chemically and structurally distinct volumes.",
        "type": 3
    },
    {
        "idx": 448,
        "question": "Is partially melted aluminum a two-phase material system?",
        "answer": "No, the physical boundary merely separates volumes that are structurally distinct (crystalline and liquid).",
        "type": 3
    },
    {
        "idx": 449,
        "question": "Is frozen water with trapped air bubbles a two-phase material system?",
        "answer": "Yes, it is a two-phase system because there is a physical boundary beyond the particle level that separates chemically and structurally distinct volumes.",
        "type": 3
    },
    {
        "idx": 450,
        "question": "Is epoxy embedded with carbon fibers a two-phase material system?",
        "answer": "Yes, it is a two-phase system because there is a physical boundary beyond the particle level that separates chemically and structurally distinct volumes.",
        "type": 3
    },
    {
        "idx": 451,
        "question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility. The solute and host species must have a very [w] sizes. (w = similar, different)",
        "answer": "The solute and host species must have a very similar sizes.",
        "type": 4,
        "wrong_answers_1": "The solute and host species must feature a similar (or the same) valence electron configuration.",
        "wrong_answers_2": "The solute and host species must attempt to pack with a similar (or the same) crystal structure.",
        "wrong_answers_3": "The solute and host species must feature a similar ability to attract electrons (electronegativity)."
    },
    {
        "idx": 452,
        "question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility. The solute and host species must attempt to pack with [x] crystal structure. (x = a similar (or the same), a different)",
        "answer": "The solute and host species must attempt to pack with a similar (or the same) crystal structure.",
        "type": 4,
        "wrong_answers_1": "The solute and host species must feature a similar (or the same) valence electron configuration.",
        "wrong_answers_2": "The solute and host species must have a very similar sizes.",
        "wrong_answers_3": "The solute and host species must feature a similar ability to attract electrons (electronegativity)."
    },
    {
        "idx": 453,
        "question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility. The solute and host species must feature [y] valence electron configuration. (y = a similar (or the same), a different)",
        "answer": "The solute and host species must feature a similar (or the same) valence electron configuration.",
        "type": 4,
        "wrong_answers_1": "The solute and host species must feature a similar ability to attract electrons (electronegativity).",
        "wrong_answers_2": "The solute and host species must have a very similar sizes.",
        "wrong_answers_3": "The solute and host species must attempt to pack with a similar (or the same) crystal structure."
    },
    {
        "idx": 454,
        "question": "Complete the following statement regarding conditions that must be satisfied in order for a solid solution to exhibit extensive solubility. The solute and host species must feature [z] ability to attract electrons (electronegativity). (z = a similar, a different)",
        "answer": "The solute and host species must feature a similar ability to attract electrons (electronegativity).",
        "type": 4,
        "wrong_answers_1": "The solute and host species must feature a similar (or the same) valence electron configuration.",
        "wrong_answers_2": "The solute and host species must have a very similar sizes.",
        "wrong_answers_3": "The solute and host species must attempt to pack with a similar (or the same) crystal structure."
    },
    {
        "idx": 455,
        "question": "For a solution, which of the following is present in the higher concentration?(a) Solvent(b) Solute",
        "answer": "Solvent. By definition, solvent is the element/compound that is present in a solution in the greatest amount.",
        "type": 2
    },
    {
        "idx": 456,
        "question": "A liquidus line separates which of the following combinations of phase fields?(a) Liquid and Liquid +\\alpha(b) \\alpha and Liquid +\\alpha(c) \\alpha and \\alpha+\\beta(d) Liquid +\\alpha and \\alpha+\\beta",
        "answer": "A liquidus line separates Liquid and Liquid +\\alpha phase fields.",
        "type": 2
    },
    {
        "idx": 457,
        "question": "A solvus line separates which of the following pairs of phase fields?(a) Liquid and Liquid +\\alpha(b) \\alpha \\operatorname{and} Liquid +\\alpha(c) \\alpha and \\alpha+\\beta d) Liquid +\\alpha and \\alpha+\\beta",
        "answer": "A solvus line separates \\alpha and \\alpha+\\beta phase fields.",
        "type": 2
    },
    {
        "idx": 458,
        "question": "Which of the following kinds of information may be determined with the aid of a phase diagram? The phase(s) present at a specified temperature and composition.",
        "answer": "With the aid of a phase diagram the following may be determined: The phase(s) present at a specified temperature and composition.",
        "type": 2
    },
    {
        "idx": 459,
        "question": "Which of the following kinds of information may be determined with the aid of a phase diagram? The composition(s) of phase(s) present at a specified temperature and composition.",
        "answer": "With the aid of a phase diagram the following may be determined: The composition(s) of phase(s) present at a specified temperature and composition.",
        "type": 2
    },
    {
        "idx": 460,
        "question": "Which of the following kinds of information may be determined with the aid of a phase diagram? The fraction(s) of phase(s) present at specified temperature and composition.",
        "answer": "With the aid of a phase diagram the following may be determined: The fraction(s) of phase(s) present at a specified temperature and composition.",
        "type": 2
    },
    {
        "idx": 461,
        "question": "At a eutectic point on a binary temperature-composition phase diagrams, how many phases are present when the system is at equilibrium?(a) 0 (b) 1 (c) 2 (d) 3",
        "answer": "(d) 3",
        "type": 2
    },
    {
        "idx": 462,
        "question": "An intermetallic compound is found in the magnesium-gallium system that has a composition of 41.1 wt% Mg - 58.9 wt% Ga. Specify the formula for this compound.(a) MgGa (b) Mg_{2} \\mathrm{Ga} (c) \\mathrm{MgGa}_{2} (d) Mg_{3} \\mathrm{Ga}_{2}",
        "answer": "(b) Mg_{2}Ga",
        "type": 2
    },
    {
        "idx": 463,
        "question": "A eutectoid reaction involves which of the following phases? (a) One liquid and one solid (b) One liquid and two solid (c) Two liquids and one solid (d) Three solid",
        "answer": "(d) Three solid",
        "type": 2
    },
    {
        "idx": 464,
        "question": "A peritectic reaction involves which of the following combinations of phase fields? (a) One liquid and one solid (b) One liquid and two solid (c) two liquids and one solid (d) Three solid",
        "answer": "(b) One liquid and two solid",
        "type": 2
    },
    {
        "idx": 465,
        "question": "For a congruent phase transformations there are (a) no composition alterations.(b) compositional alterations.",
        "answer": "(a) no composition alterations",
        "type": 2
    },
    {
        "idx": 466,
        "question": "Which of the following have a significant influence on a material's electrical resistivity?(a) impurity concentration(b) temperature(c) grain size(d) cold work(e) vacancy concentration",
        "answer": "(a) impurity concentration(b) temperature (d) cold work(e) vacancy concentration",
        "type": 2
    },
    {
        "idx": 467,
        "question": "Conductivity in a metal is almost always reduced by the introduction of defects into the lattice.\nThe factor primarily affected by defects is: (a) free electron concentration (b) electron charge (c) electron mobility (d) electron spin",
        "answer": "(c) electron mobility",
        "type": 2
    },
    {
        "idx": 468,
        "question": "Compute the number of electrons that each aluminum atom donates, on average, to a bulk piece of aluminum metal. Room temperature data for aluminum:\nThe resistivity of aluminum is 2.63 × 10^{-8} \\Omega·m The electron mobility of aluminum is 0.0012{m}^{2} /(V·s)\nThe mass density of aluminum is 2.7g / {cm}^{3} The atomic weight of aluminum is 27g / mol",
        "answer": "3.29 electron/atom",
        "type": 1,
        "wrong_answers_1": "3.47 electron/atom",
        "wrong_answers_2": "3.18 electron/atom",
        "wrong_answers_3": "3.62 electron/atom"
    },
    {
        "idx": 469,
        "question": "Compute the number of electrons that each atom donates, on average, to a bulk piece of hypothetical metal. Room temperature data for the metal:\nThe resistivity of the metal is [\\mathrm{r}] \\Omega·{cm}\nThe electron mobility of the metal is [m] {cm}^{2} /(V·s)\nThe mass density of the metal is [\\mathrm{d}] g / {cm}^{3}\nThe atomic weight of the metal is [\\mathrm{w}] g / mol",
        "answer": "the number of free electrons donated by each atom, on average, is: \\[\n\\frac{[w]}{\\left([r]\\left(1.6 × 10^{-19}\\right)[m]\\right)\\left([d]\\left(6.022 × 10^{23}\\right)\\right)}\n\\]",
        "type": 1,
        "wrong_answers_1": "the number of free electrons donated by each atom, on average, is: \\[\n\\frac{[w]}{\\left([r]\\left(1.7 × 10^{-19}\\right)[m]\\right)\\left([d]\\left(6.591 × 10^{23}\\right)\\right)}\n\\]",
        "wrong_answers_2": "the number of free electrons donated by each atom, on average, is: \\[\n\\frac{[w]}{\\left([r]\\left(1.6 × 10^{-19}\\right)[m]\\right)\\left([d]\\left(6.839 × 10^{23}\\right)\\right)}\n\\]",
        "wrong_answers_3": "the number of free electrons donated by each atom, on average, is: \\[\n\\frac{[w]}{\\left([r]\\left(1.7 × 10^{-19}\\right)[m]\\right)\\left([d]\\left(5.308 × 10^{23}\\right)\\right)}\n\\]"
    },
    {
        "idx": 470,
        "question": "When an electric field is applied, in which direction are the free electrons accelerated?(a) Opposite to the direction of the electric field.(b) In the same direction as the electric field.",
        "answer": "(a) Opposite to the direction of the electric field",
        "type": 2
    },
    {
        "idx": 471,
        "question": "The electrical conductivity of an intrinsic semiconductor is(a) characteristic of the high-purity metal.(b) due to the presence of impurities.",
        "answer": "(a) characteristic of the high-purity material",
        "type": 2
    },
    {
        "idx": 472,
        "question": "How do the electrical conductivities of metals compare with those of semiconductors?(a) \\sigma_{\\text {metals }}>\\sigma_{\\text {semiconductors }}(b) \\sigma_{\\text {metals }}=\\sigma_{\\text {semiconductors }}(c) \\sigma_{\\text {metals }}<\\sigma_{\\text {semiconductors }}",
        "answer": "(a) \\sigma_{\\text {metals }}>\\sigma_{\\text {semiconductors }}",
        "type": 2
    },
    {
        "idx": 473,
        "question": "How does increasing temperature affect the concentration of both electrons and holes in an intrinsic semiconductor?(a) Increases the concentration.(b) Decreases the concentration.(c) May increase and/or decrease the concentration, depending on the temperature range.",
        "answer": "(a) Increases the concentration",
        "type": 2
    },
    {
        "idx": 474,
        "question": "Which type of charge carrier will be introduced into a semiconductor by the presence of an acceptor impurity?(a) Electron(b) Hole",
        "answer": "(b) Hole",
        "type": 2
    },
    {
        "idx": 475,
        "question": "Which type of charge carrier will be introduced into a semiconductor by the presence of a donor impurity?(a) Impurity(b) Hole",
        "answer": "(a) Electron",
        "type": 2
    },
    {
        "idx": 476,
        "question": "The electrical conductivity of an extrinsic semiconductor is(a) Characteristic of the high-purity material.(b) due to the presence of impurities.",
        "answer": "(b) due to the presence of impurities",
        "type": 2
    },
    {
        "idx": 477,
        "question": "For an n-type semiconductor, which type of charge carrier is present in the greater concentration?(a) Hole(b) Electron",
        "answer": "(b) Electron",
        "type": 2
    },
    {
        "idx": 478,
        "question": "For an n-type semiconductor(a) Concentration _{electrons< concentration _holes (b) Concentration _{electrons= concentration _holes (c) Concentration _{electrons< concentration _{\\text { holes }}",
        "answer": "(c) Concentration _{electrons< concentration _{\\text { holes }}",
        "type": 2
    },
    {
        "idx": 479,
        "question": "For a p-type semiconductor, which type of charge carrier is present in the greater concentration?(a) Holes(b) Electrons",
        "answer": "(a) Holes",
        "type": 2
    },
    {
        "idx": 480,
        "question": "For a p-type semiconductor(a) Concentration _{electrons< concentration _holes (b) Concentration _{electrons= concentration _holes (c) Concentration _electrons < concentration _holes",
        "answer": "(a) Concentration _ electrons< concentration _holes ",
        "type": 2
    },
    {
        "idx": 481,
        "question": "In order for a semiconductor to exhibit extrinsic electrical characteristics, relatively high impurity concentrations are required.(a) True(b) False",
        "answer": "False. Even when minute impurity concentrations (e.g., 1 atom in 10^{12} ) are present in a semiconductor, its electrical characteristics will be extrinsic.",
        "type": 3
    },
    {
        "idx": 482,
        "question": "The following electrical characteristics have been determined for both intrinsic and n-type extrinsic indium phosphide (InP) at room temperature:\n\\[\n\\begin{array}{llll}\n\\sigma(\\Omega·\\mathbf{m})^{-1} & \\boldsymbol{n}\\left(\\mathbf{m}^{-3}\\right) & \\boldsymbol{p}\\left(\\mathbf{m}^{-3}\\right) \\\\\n\\text { Intrinsic } & 2.5 × 10^{-6} & 3.0 × 10^{13} & 3.0 × 10^{13} \\\\\n\\text { Extrinsic } & 3.6 × 10^{-5} & 4.5 × 10^{14} & 2.0 × 10^{12}\n\\end{array}\n\\]\nCalculate electron and hole mobilities.\n\\mu_{\\mathrm{e}}=\n\\mu_{h}=",
        "answer": "\\mu_{\\mathrm{e}} = 0.50 \\, m^{2} / \\mathrm{v}·s   \\mu_{h} = 0.02 \\, m^{2} / \\mathrm{v}·s",
        "type": 1,
        "wrong_answers_1": "\\mu_{\\mathrm{e}} = 0.57 \\, m^{2} / \\mathrm{v}·s   \\mu_{h} = 0.02 \\, m^{2} / \\mathrm{v}·s",
        "wrong_answers_2": "\\mu_{\\mathrm{e}} = 0.43 \\, m^{2} / \\mathrm{v}·s   \\mu_{h} = 0.02 \\, m^{2} / \\mathrm{v}·s",
        "wrong_answers_3": "\\mu_{\\mathrm{e}} = 0.53 \\, m^{2} / \\mathrm{v}·s   \\mu_{h} = 0.02 \\, m^{2} / \\mathrm{v}·s"
    },
    {
        "idx": 483,
        "question": "Which of the following are preferred for semiconducting devices?(a) Single crystals(b) Polycrystalline materials",
        "answer": "(a) Single crystals",
        "type": 2
    },
    {
        "idx": 484,
        "question": "As temperature increases, the electrical conductivities of polymers and ionic ceramics(a) Increase(b) Decrease",
        "answer": "(a) Increase",
        "type": 2
    },
    {
        "idx": 485,
        "question": "Most polymers and ionic ceramics have energy band gap structures that are most similar to those of(a) Insulators(b) Semiconductors(c) Metals",
        "answer": "(a) Insulators",
        "type": 2
    },
    {
        "idx": 486,
        "question": "The charge carriers in ionic ceramics and polymers can be(a) Electrons(b) Holes(c) Anion(d) Cations",
        "answer": "(a) Electrons(b) Holes(c) Anion(d) Cations",
        "type": 2
    },
    {
        "idx": 487,
        "question": "Select T/F for the following statement regarding aluminum / aluminum alloys: Aluminum alloys are generally not viable as lightweight structural materials in humid environments because they are highly susceptible to corrosion by water vapor.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 488,
        "question": "Select T/F for the following statement regarding aluminum / aluminum alloys: Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 489,
        "question": "Select T/F for the following statement regarding aluminum / aluminum alloys: Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 490,
        "question": "Select T/F for the following statement regarding aluminum / aluminum alloys: Compared to most other metals, like steel, pure aluminum is very resistant to creep deformation.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 491,
        "question": "Select T/F for the following statement regarding aluminum / aluminum alloys: The relatively low melting point of aluminum is often considered a significant limitation for structural applications.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 492,
        "question": "[a] Aluminum alloys are generally viable as lightweight structural materials in humid environments because they are not very susceptible to corrosion by water vapor.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 493,
        "question": "[b] Aluminum is not very workable at high temperatures in air, in terms of extrusion and rolling, because a non-protective oxide grows and consumes the metal, converting it to a hard and brittle ceramic.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 494,
        "question": "[c] Aluminum alloys are generally superior to pure aluminum, in terms of yield strength, because their microstructures often contain precipitate phases that strain the lattice, thereby hardening the alloy relative to pure aluminum.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 495,
        "question": "[d] Compared to other metals, like steel, pure aluminum is very resistant to failure via fatigue.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 496,
        "question": "[e] Aluminum exhibits one of the highest melting points of all metals, which makes it difficult and expensive to cast.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 497,
        "question": "[a] Copper has a higher elastic modulus than aluminum.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 498,
        "question": "[b] The density of copper is closer to that of aluminum than it is to iron.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 499,
        "question": "[c] Bronze is an alloy of copper and zinc.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 500,
        "question": "[d] Copper and its alloys form a green tarnish over time, consisting of sulfides and carbonates.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 501,
        "question": "[e] Copper is relatively resistant to corrosion by neutral and even mildly basic water, making it useful for freshwater plumbing applications.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 502,
        "question": "Select T / F for the following statement regarding copper & copper alloys: Copper is much more abundant in the earth's crust compared to iron or aluminum.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 503,
        "question": "Select T / F for the following statement regarding copper & copper alloys: Copper is one of just a few metals that can be found in metallic form in nature.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 504,
        "question": "Select T / F for the following statement regarding copper & copper alloys: Pure and/or annealed copper is more difficult to machine compared to its work-hardened form or its alloys.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 505,
        "question": "Select T / F for the following statement regarding copper & copper alloys: Copper is a minor component (by weight) of most brass & bronze alloys.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 506,
        "question": "Select T / F for the following statement regarding copper & copper alloys: Amongst metals and alloys copper is one of the best conductors of heat.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 507,
        "question": "[a] Nickel is majority component (by mass) in certain superalloys such as Waspaloy TM.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 508,
        "question": "[b] Tungsten is the lowest density metal that has structural use.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 509,
        "question": "[c] Tantalum offers extremely good corrosion resistance, especially at low temperatures.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 510,
        "question": "[d] Magnesium metal is very similar to aluminum, in terms of its physical and mechanical properties.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 511,
        "question": "[e] Beryllium metal is commonly used as an alloying agent in copper metal.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 512,
        "question": "Increasing the alumina \\left(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\right) content of fireclays results in(a) an increase in maximum service temperature.(b) a decrease in maximum service temperature.",
        "answer": "(a) an increase in maximum service temperature",
        "type": 2
    },
    {
        "idx": 513,
        "question": "The high-temperature performance of silica refractories is compromised by the presence of even small concentrations of alumina \\left(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\end{array}(a) True(b) False",
        "answer": "True",
        "type": 3
    },
    {
        "idx": 514,
        "question": "As the porosity of refractory ceramic bricks increases, what happens to the strength?",
        "answer": "Strength decreases.",
        "type": 4,
        "wrong_answers_1": "Chemical resistance decreases.",
        "wrong_answers_2": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probability of the existence of a flaw that is capable of initiating a crack diminishes.",
        "wrong_answers_3": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probably of the existence of a flaw of that is capable of initiating a crack diminishes."
    },
    {
        "idx": 515,
        "question": "As the porosity of refractory ceramic bricks increases, what happens to the chemical resistance?",
        "answer": "Chemical resistance decreases.",
        "type": 4,
        "wrong_answers_1": "Strength decreases.",
        "wrong_answers_2": "Five important characteristics for polymers that are to be used in thin-film applications are: (1) low density; (2) high flexibility; (3) high tensile and tear strengths; (4) resistance to moisture/chemical attack; and (5) low gas permeability.\n}",
        "wrong_answers_3": "Distinctive features: relatively low density, high melting temperatures, and high strengths are possible. Limitation: because of chemical reactivity with other materials at elevated temperatures, these alloys are expensive to refine. Applications: aircraft structures, space vehicles, and in chemical and petroleum industries."
    },
    {
        "idx": 516,
        "question": "As the porosity of refractory ceramic bricks increases, what happens to the thermal insulation?",
        "answer": "Thermal insulation increases.",
        "type": 4,
        "wrong_answers_1": "For refractory ceramic materials, three characteristics that improve with increasing porosity are (1) decreased thermal expansion and contraction upon thermal cycling, (2) improved thermal insulation, and (3) improved resistance to thermal shock.",
        "wrong_answers_2": "For some ceramic materials, the thermal conductivity first decreases with rising temperature because the scattering of lattice vibrations increases with temperature. At higher temperatures, the thermal conductivity will increase for some ceramics that are porous because radiant heat transfer across pores may become important, which process increases with rising temperature.",
        "wrong_answers_3": "The two thermal properties of a liquid medium that influence its quenching effectiveness are thermal conductivity and heat capacity."
    },
    {
        "idx": 517,
        "question": "The presence of silica \\left(\\mathrm{SiO}_{2}\\right) in basic refractory ceramics is beneficial to their high-temperature performance.(a) True(b) False",
        "answer": "False",
        "type": 3
    },
    {
        "idx": 518,
        "question": "Basic refractory ceramics are often used for the containment of slags that are rich in(a) silica (b) CaO (c) MgO",
        "answer": "(b) CaO (c) MgO",
        "type": 2
    },
    {
        "idx": 519,
        "question": "Silica refractory ceramics are often used for the containment of slags that are rich in(a) silica (b) CaO (c) MgO",
        "answer": "(a) silica",
        "type": 2
    },
    {
        "idx": 520,
        "question": "Such that the bond strength across the fiber-epoxy interface is [s] MPa, and the shear yield strength of the epoxy is [y] MPa, compute the minimum fiber length, in millimeters, to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite. The tensile strength of these carbon fibers is [f] MPa.",
        "answer": "the minimum fiber length to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite is \\frac{[f][d]}{2[s] × 1000} millimeters.",
        "type": 1,
        "wrong_answers_1": "the minimum fiber length to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite is \\frac{[f][d]}{2[s] × 1060} millimeters.",
        "wrong_answers_2": "the minimum fiber length to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite is \\frac{[f][d]}{2[s] × 1100} millimeters.",
        "wrong_answers_3": "the minimum fiber length to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite is \\frac{[f][d]}{2[s] × 1060} millimeters."
    },
    {
        "idx": 521,
        "question": "For an aligned fibrous composite, when a stress is applied in a direction that is parallel to the fibers, what is the reinforcement efficiency?(a) 0 (b) \\frac{1}{5}(c) \\frac{3}{8}(d) \\frac{3}{4}(e) 1",
        "answer": "(e) 1",
        "type": 2
    },
    {
        "idx": 522,
        "question": "For an aligned fibrous composite, when a stress is applied perpendicular to the fibers, what is the reinforcement efficiency?(a) 0(b)  \\frac{1}{5}(c) \\frac{3}{8}(d)\n\\frac{3}{4}(e) 1",
        "answer": "(a) 0",
        "type": 2
    },
    {
        "idx": 523,
        "question": "For a fibrous composite with fibers that are randomly and uniformly oriented within a specific plane, when a stress is applied in any direction within the plane of the fibers, what is the reinforcement efficiency?(a) 0(b) \\frac{1}{5}(c) \\frac{3}{8}(d) \\frac{3}{4}(e) 1",
        "answer": "(c) \\frac{3}{8}",
        "type": 2
    },
    {
        "idx": 524,
        "question": "For a fibrous composite with fibers that are uniformly distributed and randomly oriented in all directions, when a stress is applied in any direction, what is the reinforcement efficiency?(a) 0(b) \\frac{1}{{ }_{5}}(c) \\frac{3}{8}(d) \\frac{3}{{ }_{4}}(e) 1",
        "answer": "(b) \\frac{1}{5}",
        "type": 2
    },
    {
        "idx": 525,
        "question": "A stress-strain test is performed on an aligned fibrous composite such that the force is applied in the longitudinal direction. During the initial stage of the test, which phase bears most of the load?(a) Fibers(b) Matrix",
        "answer": "(a) Fibers",
        "type": 2
    },
    {
        "idx": 526,
        "question": "Once the fibers fail in a fibrous composite, catastrophic failure of the piece takes place.(a) True(b) False",
        "answer": "False. Once the fibers fail in a composite, catastrophic failure of the piece does not take place. Since the broken fibers are still embedded within the matrix, they are still capable of sustaining a diminished load.",
        "type": 3
    },
    {
        "idx": 527,
        "question": "How are continuous fibers typically oriented in fibrous composites?(a) Aligned(b) Partially oriented(c) Randomly oriented",
        "answer": "(a) Aligned",
        "type": 2
    },
    {
        "idx": 528,
        "question": "How are discontinuous fibers typically oriented in fibrous composites?(a) Aligned(b) Partially oriented",
        "answer": "(b) Partially oriented",
        "type": 2
    },
    {
        "idx": 529,
        "question": "If the fiber orientation is random, which type of fibers is normally used?(a) Discontinuous(b) Continuous",
        "answer": "(a) Discontinuous",
        "type": 2
    },
    {
        "idx": 530,
        "question": "Composites are single-phase materials by definition.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 531,
        "question": "The term \"composite\" applies to materials that feature polymeric materials only.",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 532,
        "question": "Structural composites are, in general, highly regarded for their specific strengths.",
        "answer": "T",
        "type": 3
    },
    {
        "idx": 533,
        "question": "Composites featuring continuous and aligned fibers for reinforcement generally offer properties that are highly isotropic compared to most metals (random polycrystals).",
        "answer": "F",
        "type": 3
    },
    {
        "idx": 534,
        "question": "Which of the following materials are typically used as fibers?(a) Graphite/carbon(b) Silicon carbide(c) Silicon nitride(d) Aluminum oxide(e) Glass(f) Boron(g) Steel (h) Tungsten (i) Molybdenum",
        "answer": "(a) (b) (d) (e) (f) ",
        "type": 2
    },
    {
        "idx": 535,
        "question": "Match the fiber type 'Whiskers' with its description.",
        "answer": "Whiskers are single crystals with extremely large length-to-diameter ratios.",
        "type": 4,
        "wrong_answers_1": "Wires are large-diameter metals.",
        "wrong_answers_2": "Single crystal electrical insulators are transparent.",
        "wrong_answers_3": "Yes, it is possible to reduce the average grain diameter of an undeformed alloy specimen from 0.050mm to 0.020 mm. In order to do this, plastically deform the material at room temperature (i.e., cold work it), and then anneal at an elevated temperature in order to allow recrystallization and some grain growth to occur until the average grain diameter is 0.020 mm."
    },
    {
        "idx": 536,
        "question": "Match the fiber type 'Fibers' with its description.",
        "answer": "Fibers are polycrystalline or amorphous materials with small diameters.",
        "type": 4,
        "wrong_answers_1": "For polymer-matrix fiber-reinforced composites, three functions of the polymer-matrix phase are: (1) to bind the fibers together so that the applied stress is distributed among the fibers; (2) to protect the surface of the fibers from being damaged; and (3) to separate the fibers and inhibit crack propagation.",
        "wrong_answers_2": "Polycrystalline and nonporous electrical insulators are translucent.",
        "wrong_answers_3": "Amorphous polymers are normally transparent because there is no scattering of a light beam within the material. However, for semicrystalline polymers, visible light will be scattered at boundaries between amorphous and crystalline regions since they have different indices of refraction. This leads to translucency or, for extensive scattering, opacity, except for semicrystalline polymers having very small crystallites."
    },
    {
        "idx": 537,
        "question": "Match the fiber type 'Wires' with its description.",
        "answer": "Wires are large-diameter metals.",
        "type": 4,
        "wrong_answers_1": "Whiskers are single crystals with extremely large length-to-diameter ratios.",
        "wrong_answers_2": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP.",
        "wrong_answers_3": "Metals are typically better thermal conductors than are ceramic materials because, for metals, most of the heat is transported by free electrons (of which there are relatively large numbers). In ceramic materials, the primary mode of thermal conduction is via phonons, and phonons are more easily scattered than are free electrons."
    },
    {
        "idx": 538,
        "question": "For a composite material, which phase normally has the higher elastic modulus?(a) Fiber phase(b) Matrix phase",
        "answer": "(a) Fiber phase",
        "type": 2
    },
    {
        "idx": 539,
        "question": "For a composite material, how does the ductility of the matrix phase normally compare with the ductility of the dispersed phase?(a) more ductile(b) less ductile",
        "answer": "(a) more ductile",
        "type": 2
    },
    {
        "idx": 540,
        "question": "Aramid fiber-reinforced composites have very high tensile strengths and relatively low compressive strengths.(a) True(b) False",
        "answer": "True. Aramid fiber-reinforced composites have very high tensile strengths and relatively low compressive strength.",
        "type": 3
    },
    {
        "idx": 541,
        "question": "Which of aramid and metal fibers have higher strength-to-weight ratios?(a) Aramid fibers(b) Metal fibers",
        "answer": "(a) Aramid fibers",
        "type": 2
    },
    {
        "idx": 542,
        "question": "Carbon fiber-reinforced composites have relatively high strengths?",
        "answer": "Carbon fiber-reinforced composites have relatively high strengths.",
        "type": 3
    },
    {
        "idx": 543,
        "question": "Carbon fiber-reinforced composites have relatively high stiffnesses?",
        "answer": "Carbon fiber-reinforced composites have relatively high stiffnesses.",
        "type": 3
    },
    {
        "idx": 544,
        "question": "Carbon fiber-reinforced composites have high service temperatures (>200 degrees C)?",
        "answer": "Carbon fiber-reinforced composites have high service temperatures (>200 degrees C).",
        "type": 3
    },
    {
        "idx": 545,
        "question": "Which of the following materials are typically used as whiskers?(a) graphite/carbon(b) silicon carbide(c) silicon nitride(d) aluminum oxide(e) glass(f) boron(g) steel (h) tungsten (i) molybdenum",
        "answer": "(a) graphite/carbon(b) silicon carbide(c) silicon nitride(d) aluminum ",
        "type": 2
    },
    {
        "idx": 546,
        "question": "Which of the following materials are typically used as wires in composites?(a) graphite/carbon(b) silicon carbide(c) silicon nitride (d) aluminum oxide (e) glass (f) boron (g) steel (h) tungsten (i) molybdenum",
        "answer": "(g) steel (h) tungsten (i) molybdenum",
        "type": 2
    },
    {
        "idx": 547,
        "question": "Compared to other ceramic materials, do ceramic-matrix composites have better fracture toughnesses?",
        "answer": "Ceramic-matrix composites have higher fracture toughnesses than other ceramic materials.",
        "type": 3
    },
    {
        "idx": 548,
        "question": "Compared to other ceramic materials, do ceramic-matrix composites have better oxidation resistance?",
        "answer": "The answer is not provided in the given information.",
        "type": 3
    },
    {
        "idx": 549,
        "question": "Compared to other ceramic materials, do ceramic-matrix composites have better stability at elevated temperatures?",
        "answer": "The answer is not provided in the given information.",
        "type": 3
    },
    {
        "idx": 550,
        "question": "Which of the following materials are typically used as stabilizers in transformation-toughened ceramic-matrix composites?(a) CaO(b) MgO(c) \\mathrm{Y}_{2} \\mathrm{O}_{3}(d) \\mathrm{CeO} (e) \\mathrm{Al}_{2} \\mathrm{O}_{3}(f) \\mathrm{SiC}",
        "answer": "(a) CaO, (b) MgO (c) \\mathrm{Y}_{2} \\mathrm{O}_{3} {3} (d) CeO",
        "type": 2
    },
    {
        "idx": 551,
        "question": "Do carbon-carbon composites exhibit high tensile moduli at elevated temperatures?",
        "answer": "Carbon-carbon composites have high tensile moduli at elevated temperatures.",
        "type": 3
    },
    {
        "idx": 552,
        "question": "Do carbon-carbon composites exhibit high tensile strengths at elevated temperatures?",
        "answer": "Carbon-carbon composites have high tensile strengths at elevated temperatures.",
        "type": 3
    },
    {
        "idx": 553,
        "question": "Do carbon-carbon composites exhibit resistance to creep?",
        "answer": "Carbon-carbon composites are highly resistant to creep.",
        "type": 3
    },
    {
        "idx": 554,
        "question": "Do carbon-carbon composites exhibit large fracture toughness values?",
        "answer": "Carbon-carbon composites have large fracture toughness values.",
        "type": 3
    },
    {
        "idx": 555,
        "question": "Do carbon-carbon composites exhibit high thermal conductivities?",
        "answer": "Carbon-carbon composites have high thermal conductivities.",
        "type": 3
    },
    {
        "idx": 556,
        "question": "Do carbon-carbon composites exhibit low coefficients of thermal expansion?",
        "answer": "Carbon-carbon composites have low coefficients of thermal expansion.",
        "type": 3
    },
    {
        "idx": 557,
        "question": "Do carbon-carbon composites exhibit resistance to oxidation at elevated temperatures?",
        "answer": "The answer does not specify resistance to oxidation at elevated temperatures.",
        "type": 3
    },
    {
        "idx": 558,
        "question": "Are carbon-carbon composites low cost?",
        "answer": "The answer does not specify low cost.",
        "type": 3
    },
    {
        "idx": 559,
        "question": "Laminar composites have high strengths in all directions (in three dimensions).(a) True(b) False",
        "answer": "False. Laminar composites have high strengths in all directions only in their two-dimensional planes.",
        "type": 3
    },
    {
        "idx": 560,
        "question": "The strong outer sheets of sandwich panels are separated by a layer of material that is(a) less dense than the outer sheet material(b) more dense than the outer sheet material",
        "answer": "The strong outer sheets of sandwich panels are separated by a layer of material that is less dense than the outer sheet material.",
        "type": 2
    },
    {
        "idx": 561,
        "question": "To what temperature would 23.0kg of some material at 100^{\\circ} C be raised if 255kJ of heat is supplied? Assume a C_{p} value of 423 J/ kg-K for this material.\n(A) 26.2^{\\circ} C\n(B) 73.8^{\\circ} C\n(C) 126^{\\circ} C\n(D) 152^{\\circ} C",
        "answer": "the final temperature is 126^{\\circ} C, which corresponds to answer C.",
        "type": 2
    },
    {
        "idx": 562,
        "question": "A rod of some material 0.50{m} long elongates 0.40mm on heating from 50^{\\circ} C to 151^{\\circ} C. What is the value of the linear coefficient of thermal expansion for this material?\n(A) 5.30 × 10^{-6}\\left({ }^{\\circ} C\\right)^{-1}\n(B) 7.92 × 10^{-6}\\left({ }^{\\circ} C C\\right)^{-1}\n(C) 1.60 × 10^{-5}\\left({ }^{\\circ} C\\right)^{-1}\n(D) 1.24 × 10^{-6}\\left({ }^{\\circ} C_{-}^{1}\\right.",
        "answer": "the linear coefficient of thermal expansion for this material is 7.92 × 10^{-6} \\left(^{\\circ}c\\right)^{-1}, which is answer b.",
        "type": 2
    },
    {
        "idx": 563,
        "question": "A 10 meter long square bar of 316 stainless steel (edge length of 5 cm, with a modulus of 193 \\mathrm{GPa} and a yield point of 290 MPa ) is bolted securely in place when its installation temperature was around [1]^{\\circ} C. What is the expected thermal stress in the bar when its service temperature reaches [F]^{o} C ? Enter a negative indicate compressive stress, if necessary. The thermal expansion coefficient of 316 stainless steel is 16.0 × 10^{-6} \\mathrm{I} /{ }^{\\circ} C.",
        "answer": "sigma_{th} = (193,000)(16 × 10^{-6})([i] - [f]) MPa",
        "type": 1,
        "wrong_answers_1": "sigma_{th} = (212,168)(14 × 10^{-6})([i] - [f]) MPa",
        "wrong_answers_2": "sigma_{th} = (186,908)(14 × 10^{-6})([i] - [f]) MPa",
        "wrong_answers_3": "sigma_{th} = (204,921)(14 × 10^{-6})([i] - [f]) MPa"
    },
    {
        "idx": 564,
        "question": "Which of the following 1kg samples is expected to change temperature the least if 100kJ of heat is perfectly transferred to each of them at a constant pressure of 1 atmosphere. The initial temperature of each specimen is 25^{\\circ} C. (a) Aluminum (b) Copper (c) Gold (d) Borosilicate Glass (e) Polystyrene",
        "answer": "polystyrene will experience the smallest temperature change.",
        "type": 2
    },
    {
        "idx": 565,
        "question": "If you were to locally heat identical geometry plates of the materials listed below with the same heat source, which would increase in temperature the fastest? (a) Polystyrene (b) Aluminum (c) Copper (d) Gold (e) Borosilicate Glass",
        "answer": "In this scenario, the material with the highest thermal conductivity will conduct heat the fastest. Therefore this material will absorb heat and increase in temperature the fastest. In this case, copper has the highest thermal conductivity.",
        "type": 2
    },
    {
        "idx": 566,
        "question": "This ceramic outlier has the highest room-temperature thermal conductivity of about 2,000 watts per meter per kelvin, a value that is five time higher than the best thermally conductive metals.\n(a) \\mathrm{Al}_{2} \\mathrm{O}_{3}\n(b) \\mathrm{CaF}_{2}\n(c) \\mathrm{TiO}_{2}\n(d) \\mathrm{SrTiO}_{3}\n(e) C (diamond)",
        "answer": "Diamond has the highest thermal conductivity of these options.",
        "type": 2
    },
    {
        "idx": 567,
        "question": "Which two of the following are ferromagnetic materials? \\\\(a) Aluminum oxide \\\\(b) Copper \\\\ c) Aluminum \\\\ d) Titanium \\\\ e) Iron ( \\alpha ferrite) \\\\ f) Nickel \\\\ g) MnO \\\\ h) \\mathrm{Fe}_{3} \\mathrm{O}_{4} \\\\ i) \\mathrm{NiFe}_{2} \\mathrm{O}_{4}\n}",
        "answer": "Iron ( \\alpha ferrite) and nickel are ferromagnetic materials.",
        "type": 2
    },
    {
        "idx": 568,
        "question": "Which of the following materials display(s) antiferromagnetic behavior?(a) Aluminum oxide(b) Copper(c) Aluminum(d) Titanium(e) Iron ( \\alpha ferrite)(f) Nickel(g) MnO (h) \\mathrm{Fe}_{3} \\mathrm{O}_{4} (i) \\mathrm{NiFe}_{2} \\mathrm{O}_{4}",
        "answer": "(\\mathrm{MnO} is an antiferromagnetic material.",
        "type": 2
    },
    {
        "idx": 569,
        "question": "Which of the following are ferrimagnetic materials?(a) Aluminum oxide(b) Copper(c) Aluminum(d) Titanium(e) Iron ( \\alpha ferrite)(f) Nickel(g) MnO (h) \\mathrm{Fe}_{3} \\mathrm{O}_{4} (i) \\mathrm{NiFe}_{2} \\mathrm{O}_{4}",
        "answer": "(\\mathrm{Fe}_{3} \\mathrm{O}_{4} and \\mathrm{NiFe}_{2} \\mathrm{O}_{4} are ferrimagnetic materials.",
        "type": 2
    },
    {
        "idx": 570,
        "question": "The formula for yttrium iron garnet \\left(\\mathrm{Y}_{3} \\mathrm{Fe}_{3} \\mathrm{O}_{12}\\right) may be written in the form \\mathrm{Y}_{3}^{*} \\mathrm{Fe}_{2}^{*} \\mathrm{Fe}_{3}^{d} \\mathrm{O}_{12}^{2}, where the superscripts a, c, and d represent different sites on which the \\mathrm{Y}^{3+} and \\mathrm{Fe}^{3+} ions are located. The spin magnetic moments for the \\mathrm{Y}^{3+} and \\mathrm{Fe}^{\\text {out }} ions positioned in the \\mathrm{a} and C sites are oriented parallel to one another and antiparallel to the \\mathrm{Fe}^{3+} ions in \\mathrm{d} sites. Compute the number of Bohr magnetons associated with each \\mathrm{Y}^{3+} ion, given the following information: (1) each unit cell consists of eight formula \\left(\\mathrm{Y}_{3} \\mathrm{Fe}_{5} \\mathrm{O}_{12}\\right) units; (2) the unit cell is cubic with an edge length of 1.2376nm; (3) the saturation magnetization for this material is 1.0 × 10^{4} \\mathrm{~A} / m; and (4) assume that there are 5 Bohr magnetons associated with each \\mathrm{Fe}^{3+} ion.",
        "answer": "the number of bohr magnetons associated with each \\mathrm{y}^{3+} ion is 1.75 bm.",
        "type": 1,
        "wrong_answers_1": "the number of bohr magnetons associated with each \\mathrm{y}^{3+} ion is 1.63 bm.",
        "wrong_answers_2": "the number of bohr magnetons associated with each \\mathrm{y}^{3+} ion is 1.51 bm.",
        "wrong_answers_3": "the number of bohr magnetons associated with each \\mathrm{y}^{3+} ion is 1.95 bm."
    },
    {
        "idx": 571,
        "question": "At the Curie temperature, the saturation magnetization abruptly diminishes. Which of the following magnetic material types will have Curie temperatures?(a) Diamagnetics(b) Paramagnetics(c) Ferromagnetics(d) Antiferromagnetics(e) Ferrimagnetics",
        "answer": "Ferromagnetic and ferrimagnetic materials will have Curie temperatures.",
        "type": 2
    },
    {
        "idx": 572,
        "question": "In a polycrystalline material, each grain will always consist of just a single domain.(a) True(b) False",
        "answer": "False. In a polycrystalline material, each grain may consist of more than one domain.",
        "type": 3
    },
    {
        "idx": 573,
        "question": "With increasing temperature antiferromagnetic materials eventually become which of the following?(a) Diamagnetics(b) Paramagnetics(c) Ferromagnetics(d) Antiferromagnetics(e) Ferrimagnetics",
        "answer": "With increasing temperature antiferromagnetic materials eventually become paramagnetic.",
        "type": 2
    },
    {
        "idx": 574,
        "question": "Which type(s) of magnetic materials may be classified as either soft or hard?(a) Diamagnetic(b) Paramagnetic(c) Ferromagnetic(d) Antiferromagnetic(e) Ferrimagnetic",
        "answer": "Ferromagnetic and ferrimagnetic materials may be classified as either soft or hard.",
        "type": 2
    },
    {
        "idx": 575,
        "question": "Which of the following characteristics are displayed by soft magnetic materials in terms of hysteresis loop size?",
        "answer": "Soft materials have relatively small relative hysteresis loops.",
        "type": 2
    },
    {
        "idx": 576,
        "question": "Which of the following characteristics are displayed by soft magnetic materials in terms of magnetization and demagnetization fields?",
        "answer": "Magnetization and demagnetization may be achieved using relatively low applied fields.",
        "type": 2
    },
    {
        "idx": 577,
        "question": "Which of the following characteristics are displayed by hard magnetic materials in terms of hysteresis loop size?",
        "answer": "Hard materials have relatively large hysteresis loops.",
        "type": 2
    },
    {
        "idx": 578,
        "question": "Which of the following characteristics are displayed by hard magnetic materials in terms of magnetization and demagnetization field requirements?",
        "answer": "Magnetization and demagnetization require relatively high applied fields.",
        "type": 2
    },
    {
        "idx": 579,
        "question": "What is the order of magnitude wavelength for visible light?(a) 0.5 Angstroms(b) 0.5 nanometers(c) 0.5 micrometers(d) 0.5 millimeters(e) 0.5 meters(f) 0.5 kilometers",
        "answer": "(c) 0.5 micrometers",
        "type": 2
    },
    {
        "idx": 580,
        "question": "In the visible spectrum, a thick metal specimen will be(a) Transparent(b) Translucent(c) Opaque",
        "answer": "In the visible spectrum, a thick metal specimen is opaque.",
        "type": 2
    },
    {
        "idx": 581,
        "question": "In the visible spectrum, an electrical insulator that is a single crystal and without porosity is normally(a) Transparent(b) Opaque(c) Translucent",
        "answer": "In the visible spectrum, an electrical insulator that is a single crystal and without porosity\nnormally transparent.",
        "type": 2
    },
    {
        "idx": 582,
        "question": "Match the type of light transmission with its description: Transmits light with relative little absorption.",
        "answer": "A transparent material transmits light with relatively little absorption.",
        "type": 4,
        "wrong_answers_1": "A translucent material transmits light diffusely.",
        "wrong_answers_2": "For a transparent material that appears colorless, any absorption within its interior is the same for all visible wavelengths. On the other hand, if there is any selective absorption of visible light (usually by electron excitations), the material will appear colored, its color being dependent on the frequency distribution of the transmitted light beam.",
        "wrong_answers_3": "The characteristic color of a transparent nonmetal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is transmitted through the material."
    },
    {
        "idx": 583,
        "question": "Match the type of light transmission with its description: Transmits light diffusely.",
        "answer": "A translucent material transmits light diffusely.",
        "type": 4,
        "wrong_answers_1": "Light is transmitted diffusely through translucent materials (there is some internal light scattering). Objects are not clearly distinguishable when viewed through a translucent material.",
        "wrong_answers_2": "Light is transmitted diffusely through translucent materials (there is some internal light scattering). Objects are not clearly distinguishable when viewed through a translucent material.",
        "wrong_answers_3": "A transparent material transmits light with relatively little absorption."
    },
    {
        "idx": 584,
        "question": "Match the type of light transmission with its description: Is impervious to light transmission.",
        "answer": "An opaque material is impervious to light transmission.",
        "type": 4,
        "wrong_answers_1": "Opaque materials are impervious to light transmission; it is not possible to see through them.",
        "wrong_answers_2": "Opaque materials are impervious to light transmission; it is not possible to see through them.",
        "wrong_answers_3": "Porous electrical insulators are opaque to visible light."
    },
    {
        "idx": 585,
        "question": "In the visible spectrum, a semiconductor that is a single crystal and nonporous may be(a) Transparent(b) Translucent(c) Opaque",
        "answer": "In the visible spectrum, a semiconductor that is a single crystal and nonporous may be transparent or opaque.",
        "type": 2
    },
    {
        "idx": 586,
        "question": "To which of the following electromagnetic radiation types are bulk metals opaque?(a) radio waves(b) microwaves(c) infrared radiation(d) ultraviolet radiation(e) X-rays",
        "answer": "Bulk metals are opaque to the following radiation types: radio waves, microwaves, infrared radiation, and ultraviolet radiation.",
        "type": 2
    },
    {
        "idx": 587,
        "question": "Every nonmetallic material becomes opaque to electromagnetic radiation having some wavelength.(a) True(b) False",
        "answer": "True. Every nonmetallic material becomes opaque to electromagnetic radiation having some wavelength. For all nonmetallic materials there is some maximum wavelength below which electronexcitations across the band gap will occur. This results in the adsorption of radiation, and, consequently, the material becomes opaque.",
        "type": 3
    },
    {
        "idx": 588,
        "question": "Match the following material types with their light transmission characteristics: Single crystal electrical insulators - Opaque, Translucent, Transparent",
        "answer": "Single crystal electrical insulators are transparent.",
        "type": 4,
        "wrong_answers_1": "Polycrystalline and nonporous electrical insulators are translucent.",
        "wrong_answers_2": "Porous electrical insulators are opaque to visible light.",
        "wrong_answers_3": "The two characteristics of component ions that determine the crystal structure of a ceramic compound are: 1) the magnitude of the electrical charge on each ion, and 2) the relative sizes of the cations and anions."
    },
    {
        "idx": 589,
        "question": "Match the following material types with their light transmission characteristics: Polycrystalline and nonporous electrical insulators - Translucent, Transparent, Opaque",
        "answer": "Polycrystalline and nonporous electrical insulators are translucent.",
        "type": 4,
        "wrong_answers_1": "Single crystal electrical insulators are transparent.",
        "wrong_answers_2": "Porous electrical insulators are opaque to visible light.",
        "wrong_answers_3": "A translucent material transmits light diffusely."
    },
    {
        "idx": 590,
        "question": "Match the following material types with their light transmission characteristics: Porous electrical insulators - Transparent, Opaque, Translucent",
        "answer": "Porous electrical insulators are opaque to visible light.",
        "type": 4,
        "wrong_answers_1": "Single crystal electrical insulators are transparent.",
        "wrong_answers_2": "Polycrystalline and nonporous electrical insulators are translucent.",
        "wrong_answers_3": "An opaque material is impervious to light transmission."
    },
    {
        "idx": 591,
        "question": "For noncubic crystals, the index of refraction is lowest in the crystallographic direction that has the(a) Highest atomic packing density(b) Lowest atomic packing density",
        "answer": "(b)",
        "type": 2
    },
    {
        "idx": 592,
        "question": "A completely amorphous and nonporous polymer will be(a) Transparent(b) Translucent(c) Opaque",
        "answer": "A completely amorphous and nonporous polymer will be transparent.",
        "type": 2
    },
    {
        "idx": 593,
        "question": "A beam of light is shined on a thin (sub-millimeter thick) single crystal wafer of material. The light source is special since it can be tuned to provide any wavelength of visible light on demand. The specimen is illuminated such that the frequency of light is decreased over time while the transmitted intensity of the light is measured. If the sample becomes transparent when the frequency is less than [F] THz, what is the band gap of the material, in eV? Assume that an intrinsic excitation of electrons is responsible for the absorption.",
        "answer": "the band gap of the material is 4.135 × 10^{-3} [f] ev.",
        "type": 1,
        "wrong_answers_1": "the band gap of the material is 4.303 × 10^{-3} [f] ev.",
        "wrong_answers_2": "the band gap of the material is 3.531 × 10^{-3} [f] ev.",
        "wrong_answers_3": "the band gap of the material is 3.839 × 10^{-3} [f] ev."
    },
    {
        "idx": 594,
        "question": "Select the word combination that best completes this statement.\nWhen a semiconductor is exposed to a light source, its intrinsic carrier concentration will increase if the [a] of the light is [b] than band gap of the semiconductor.\n[a]: intensity, energy, wavelength, frequency, voltage, current, resistance\n[b]: greater, less",
        "answer": "([a] energy\n[b] greater",
        "type": 2
    },
    {
        "idx": 595,
        "question": "Match the luminescence characteristics with their descriptions.\nReemission of photons occurs in much less than one second after excitation.\n- Phosphorescence\n- Fluorescence\nReemission of photons occurs in more than one second after excitation.\n- Fluorescence\n- Phosphorescence",
        "answer": "Fluorescence involves reemission of photons in much less than one second after excitation; while phosphorescence involves reemission of photons in more than one second after excitation.",
        "type": 4,
        "wrong_answers_1": "The second criterion is previous experience.",
        "wrong_answers_2": "No hole is generated by an electron excitation involving a donor impurity atom because the excitation comes from a level within the band gap, and thus, no missing electron is created within the normally filled valence band.",
        "wrong_answers_3": "The second stage is growth. The growth stage is simply the increase in size of the new phase particles."
    },
    {
        "idx": 596,
        "question": "Only pure materials luminesce.(a) True(b) False",
        "answer": "False. Luminescent materials contain impurities.",
        "type": 3
    },
    {
        "idx": 597,
        "question": "Cite the difference between atomic mass and atomic weight.",
        "answer": "Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the atomic masses of an atom's naturally occurring isotopes.",
        "type": 4,
        "wrong_answers_1": "Oxidation is the process by which an atom gives up an electron (or electrons) to become a cation. Reduction is the process by which an atom acquires an extra electron (or electrons) and becomes an anion.",
        "wrong_answers_2": "Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material.",
        "wrong_answers_3": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom."
    },
    {
        "idx": 598,
        "question": "Chromium has four naturally-occurring isotopes: 4.34% of { }^{50} \\mathrm{Cr}, with an atomic weight of 49.9460 amu, 83.79% of { }^{52} \\mathrm{Cr}, with an atomic weight of 51.9405 amu, 9.50% of { }^{53} \\mathrm{Cr}, with an atomic weight of 52.9407 amu, and 2.37% of { }^{54} \\mathrm{Cr}, with an atomic weight of 53.9389 amu. On the basis of these data, confirm that the average atomic weight of \\mathrm{Cr} is 51.9963 amu.",
        "answer": "the average atomic weight of \\mathrm{cr} is 51.9963 \\, amu.",
        "type": 1,
        "wrong_answers_1": "the average atomic weight of \\mathrm{cr} is 45.1310 \\, amu.",
        "wrong_answers_2": "the average atomic weight of \\mathrm{cr} is 55.0804 \\, amu.",
        "wrong_answers_3": "the average atomic weight of \\mathrm{cr} is 56.0126 \\, amu."
    },
    {
        "idx": 599,
        "question": "How many grams are there in one amu of a material?",
        "answer": "1.66 x 10^-24 g/amu",
        "type": 1,
        "wrong_answers_1": "1.83 x 10^-24 g/amu",
        "wrong_answers_2": "1.59 x 10^-24 g/amu",
        "wrong_answers_3": "1.90 x 10^-24 g/amu"
    },
    {
        "idx": 600,
        "question": "Mole, in the context of this book, is taken in units of gram-mole. On this basis, how many atoms are there in a pound-mole of a substance?",
        "answer": "2.73 x 10^26 atoms/lb-mol",
        "type": 1,
        "wrong_answers_1": "2.94 x 10^26 atoms/lb-mol",
        "wrong_answers_2": "2.59 x 10^26 atoms/lb-mol",
        "wrong_answers_3": "3.10 x 10^26 atoms/lb-mol"
    },
    {
        "idx": 601,
        "question": "(a) Cite two important quantum-mechanical concepts associated with the Bohr model of the atom.",
        "answer": "Two important quantum-mechanical concepts associated with the Bohr model of the atom are (1) that electrons are particles moving in discrete orbitals, and (2) electron energy is quantized into shells.",
        "type": 4,
        "wrong_answers_1": "Two important refinements resulting from the wave-mechanical atomic model are (1) that electron position is described in terms of a probability distribution, and (2) electron energy is quantized into both shells and subshells--each electron is characterized by four quantum numbers.",
        "wrong_answers_2": "For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid.",
        "wrong_answers_3": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities."
    },
    {
        "idx": 602,
        "question": "(b) Cite two important additional refinements that resulted from the wave-mechanical atomic model.",
        "answer": "Two important refinements resulting from the wave-mechanical atomic model are (1) that electron position is described in terms of a probability distribution, and (2) electron energy is quantized into both shells and subshells--each electron is characterized by four quantum numbers.",
        "type": 4,
        "wrong_answers_1": "Two important quantum-mechanical concepts associated with the Bohr model of the atom are (1) that electrons are particles moving in discrete orbitals, and (2) electron energy is quantized into shells.",
        "wrong_answers_2": "For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid.",
        "wrong_answers_3": "The mj quantum number designates the number of electron states in each electron subshell."
    },
    {
        "idx": 603,
        "question": "Relative to electrons and electron states, what does the n quantum number specify?",
        "answer": "The n quantum number designates the electron shell.",
        "type": 4,
        "wrong_answers_1": "The l quantum number designates the electron subshell.",
        "wrong_answers_2": "The mj quantum number designates the number of electron states in each electron subshell.",
        "wrong_answers_3": "The ms quantum number designates the spin moment on each electron."
    },
    {
        "idx": 604,
        "question": "Relative to electrons and electron states, what does the l quantum number specify?",
        "answer": "The l quantum number designates the electron subshell.",
        "type": 4,
        "wrong_answers_1": "The mj quantum number designates the number of electron states in each electron subshell.",
        "wrong_answers_2": "The n quantum number designates the electron shell.",
        "wrong_answers_3": "The ms quantum number designates the spin moment on each electron."
    },
    {
        "idx": 605,
        "question": "Relative to electrons and electron states, what does the mj quantum number specify?",
        "answer": "The mj quantum number designates the number of electron states in each electron subshell.",
        "type": 4,
        "wrong_answers_1": "The l quantum number designates the electron subshell.",
        "wrong_answers_2": "The n quantum number designates the electron shell.",
        "wrong_answers_3": "The ms quantum number designates the spin moment on each electron."
    },
    {
        "idx": 606,
        "question": "Relative to electrons and electron states, what does the ms quantum number specify?",
        "answer": "The ms quantum number designates the spin moment on each electron.",
        "type": 4,
        "wrong_answers_1": "The l quantum number designates the electron subshell.",
        "wrong_answers_2": "The n quantum number designates the electron shell.",
        "wrong_answers_3": "The mj quantum number designates the number of electron states in each electron subshell."
    },
    {
        "idx": 607,
        "question": "For the L shell (n=2), write the four quantum numbers for all of the electrons and note which correspond to the s and p subshells.",
        "answer": "For the L state, n=2, and eight electron states are possible. Possible l values are 0 and 1, while possible ml values are 0 and ±1; and possible ms values are ±1/2. Therefore, for the s states, the quantum numbers are 200(1/2) and 200(-1/2). For the p states, the quantum numbers are 210(1/2), 210(-1/2), 211(1/2), 211(-1/2), 21(-1)(1/2), and 21(-1)(-1/2).",
        "type": 4,
        "wrong_answers_1": "For the M state, n=3, and 18 states are possible. Possible l values are 0, 1, and 2; possible ml values are 0, ±1, and ±2; and possible ms values are ±1/2. Therefore, for the s states, the quantum numbers are 300(1/2), 300(-1/2), for the p states they are 310(1/2), 310(-1/2), 311(1/2), 311(-1/2), 31(-1)(1/2), and 31(-1)(-1/2); for the d states they are 320(1/2), 320(-1/2), 321(1/2), 321(-1/2), 32(-1)(1/2), 32(-1)(-1/2), 322(1/2), 322(-1/2), 32(-2)(1/2), and 32(-2)(-1/2).",
        "wrong_answers_2": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities.",
        "wrong_answers_3": "The mj quantum number designates the number of electron states in each electron subshell."
    },
    {
        "idx": 608,
        "question": "For the M shell (n=3), write the four quantum numbers for all of the electrons and note which correspond to the s, p, and d subshells.",
        "answer": "For the M state, n=3, and 18 states are possible. Possible l values are 0, 1, and 2; possible ml values are 0, ±1, and ±2; and possible ms values are ±1/2. Therefore, for the s states, the quantum numbers are 300(1/2), 300(-1/2), for the p states they are 310(1/2), 310(-1/2), 311(1/2), 311(-1/2), 31(-1)(1/2), and 31(-1)(-1/2); for the d states they are 320(1/2), 320(-1/2), 321(1/2), 321(-1/2), 32(-1)(1/2), 32(-1)(-1/2), 322(1/2), 322(-1/2), 32(-2)(1/2), and 32(-2)(-1/2).",
        "type": 4,
        "wrong_answers_1": "For the L state, n=2, and eight electron states are possible. Possible l values are 0 and 1, while possible ml values are 0 and ±1; and possible ms values are ±1/2. Therefore, for the s states, the quantum numbers are 200(1/2) and 200(-1/2). For the p states, the quantum numbers are 210(1/2), 210(-1/2), 211(1/2), 211(-1/2), 21(-1)(1/2), and 21(-1)(-1/2).",
        "wrong_answers_2": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities.",
        "wrong_answers_3": "For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid."
    },
    {
        "idx": 609,
        "question": "Give the electron configuration for the ion Fe2+.",
        "answer": "The electron configuration for an Fe2+ ion is 1s2 2s2 2p6 3s2 3p6 3d6.",
        "type": 4,
        "wrong_answers_1": "The electron configuration for a Cu+ ion is 1s2 2s2 2p6 3s2 3p6 3d10.",
        "wrong_answers_2": "The electron configuration for an O2- ion is 1s2 2s2 2p6.",
        "wrong_answers_3": "The electron configuration for an Al3+ ion is 1s2 2s2 2p6."
    },
    {
        "idx": 610,
        "question": "Give the electron configuration for the ion Al3+.",
        "answer": "The electron configuration for an Al3+ ion is 1s2 2s2 2p6.",
        "type": 4,
        "wrong_answers_1": "The electron configuration for an O2- ion is 1s2 2s2 2p6.",
        "wrong_answers_2": "The electron configuration for an Fe2+ ion is 1s2 2s2 2p6 3s2 3p6 3d6.",
        "wrong_answers_3": "The electron configuration for a Cu+ ion is 1s2 2s2 2p6 3s2 3p6 3d10."
    },
    {
        "idx": 611,
        "question": "Give the electron configuration for the ion Cu+.",
        "answer": "The electron configuration for a Cu+ ion is 1s2 2s2 2p6 3s2 3p6 3d10.",
        "type": 4,
        "wrong_answers_1": "The electron configuration for a Br- ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6.",
        "wrong_answers_2": "The electron configuration for an Fe2+ ion is 1s2 2s2 2p6 3s2 3p6 3d6.",
        "wrong_answers_3": "The electron configuration for a Ba2+ ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6."
    },
    {
        "idx": 612,
        "question": "Give the electron configuration for the ion Ba2+.",
        "answer": "The electron configuration for a Ba2+ ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6.",
        "type": 4,
        "wrong_answers_1": "The electron configuration for a Br- ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6.",
        "wrong_answers_2": "The electron configuration for a Cu+ ion is 1s2 2s2 2p6 3s2 3p6 3d10.",
        "wrong_answers_3": "The electron configuration for an Fe2+ ion is 1s2 2s2 2p6 3s2 3p6 3d6."
    },
    {
        "idx": 613,
        "question": "Give the electron configuration for the ion Br-.",
        "answer": "The electron configuration for a Br- ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6.",
        "type": 4,
        "wrong_answers_1": "The electron configuration for a Cu+ ion is 1s2 2s2 2p6 3s2 3p6 3d10.",
        "wrong_answers_2": "The electron configuration for a Ba2+ ion is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6.",
        "wrong_answers_3": "The electron configuration for an Fe2+ ion is 1s2 2s2 2p6 3s2 3p6 3d6."
    },
    {
        "idx": 614,
        "question": "Give the electron configuration for the ion O2-.",
        "answer": "The electron configuration for an O2- ion is 1s2 2s2 2p6.",
        "type": 4,
        "wrong_answers_1": "The electron configuration for an Al3+ ion is 1s2 2s2 2p6.",
        "wrong_answers_2": "The electron configuration for an Fe2+ ion is 1s2 2s2 2p6 3s2 3p6 3d6.",
        "wrong_answers_3": "The electron configuration for a Cu+ ion is 1s2 2s2 2p6 3s2 3p6 3d10."
    },
    {
        "idx": 615,
        "question": "The Na+ ion has an electron structure that is identical to which inert gas?",
        "answer": "The Na+ ion has an electron configuration the same as neon.",
        "type": 4,
        "wrong_answers_1": "The Cl- ion has an electron configuration the same as argon.",
        "wrong_answers_2": "The I- ion is an iodine atom that has acquired one extra electron; therefore, it has an electron configuration the same as xenon.",
        "wrong_answers_3": "The K+ ion is just a potassium atom that has lost one electron; therefore, it has an electron configuration the same as argon."
    },
    {
        "idx": 616,
        "question": "The Cl- ion has an electron structure that is identical to which inert gas?",
        "answer": "The Cl- ion has an electron configuration the same as argon.",
        "type": 4,
        "wrong_answers_1": "The K+ ion is just a potassium atom that has lost one electron; therefore, it has an electron configuration the same as argon.",
        "wrong_answers_2": "The Na+ ion has an electron configuration the same as neon.",
        "wrong_answers_3": "The I- ion is an iodine atom that has acquired one extra electron; therefore, it has an electron configuration the same as xenon."
    },
    {
        "idx": 617,
        "question": "With regard to electron configuration, what do all the elements in Group VIIA of the periodic table have in common?",
        "answer": "Each of the elements in Group VIIA has five p electrons.",
        "type": 4,
        "wrong_answers_1": "Each of the elements in Group IIA has two s electrons.",
        "wrong_answers_2": "Three variables that determine the microstructure of an alloy are (1) the alloying elements present, (2) the concentrations of these alloying elements, and (3) the heat treatment of the alloy.",
        "wrong_answers_3": "Yes, it is possible to determine which polymer has the higher melting temperature. The polypropylene will have the higher Tm because it has a bulky phenyl side group in its repeat unit structure, which is absent in the polyethylene. Furthermore, the polypropylene has a higher degree of polymerization."
    },
    {
        "idx": 618,
        "question": "To what group in the periodic table would an element with atomic number 114 belong?",
        "answer": "From the periodic table the element having atomic number 114 would belong to group IVA. Ds, having an atomic number of 110 lies below Pt in the periodic table and in the rightmost column of group VIII. Moving four columns to the right puts element 114 under \\mathrm{Pb} and in group IVA.",
        "type": 4,
        "wrong_answers_1": "From the periodic table the element having atomic number 112 would belong to group IIB. According to the periodic table, Ds, having an atomic number of 110 lies below Pt in the periodic table and in the right-most column of group VIII. Moving two columns to the right puts element 112 under \\mathrm{Hg} and in group IIB.\nThis element has been artificially created and given the name Copernicium with the symbol Cn. It was named after Nicolaus Copernicus, the Polish scientist who proposed that the earth moves around the sun (and not vice versa).",
        "wrong_answers_2": "For tungsten, the bonding is metallic since it is a metallic element from the periodic table.",
        "wrong_answers_3": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom."
    },
    {
        "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "idx": 619,
        "type": 4,
        "wrong_answers_1": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{5} 5 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron."
    },
    {
        "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}) electron configuration is that of an inert gas because of filled (3 s) and (3 p) subshells.",
        "idx": 620,
        "type": 4,
        "wrong_answers_1": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}) electron configuration is that of an inert gas because of filled (4 s) and (4 p) subshells.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron."
    },
    {
        "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{5}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{5}) electron configuration is that of a halogen because it is one electron deficient from having a filled (L) shell.",
        "idx": 621,
        "type": 2
    },
    {
        "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) electron configuration is that of an alkaline earth metal because of two (s) electrons.",
        "idx": 622,
        "type": 2
    },
    {
        "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "idx": 623,
        "type": 2
    },
    {
        "question": "Determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron.",
        "idx": 624,
        "type": 4,
        "wrong_answers_1": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell."
    },
    {
        "idx": 625,
        "question": "What electron subshell is being filled for the rare earth series of elements on the periodic table?",
        "answer": "The 4 f subshell is being filled for the rare earth series of elements.",
        "type": 4,
        "wrong_answers_1": "The 5 f subshell is being filled for the actinide series of elements.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}) electron configuration is that of a halogen because it is one electron deficient from having a filled (p) subshell.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) electron configuration is that of an alkaline earth metal because of two (s) electrons."
    },
    {
        "idx": 626,
        "question": "What electron subshell is being filled for the actinide series?",
        "answer": "The 5 f subshell is being filled for the actinide series of elements.",
        "type": 4,
        "wrong_answers_1": "The 4 f subshell is being filled for the rare earth series of elements.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}) electron configuration is that of a halogen because it is one electron deficient from having a filled (p) subshell.",
        "wrong_answers_3": "Three variables that determine the microstructure of an alloy are (1) the alloying elements present, (2) the concentrations of these alloying elements, and (3) the heat treatment of the alloy."
    },
    {
        "idx": 627,
        "question": "Calculate the force of attraction between a K^{+}and an \\mathrm{O}^{2-} ion the centers of which are separated by a distance of 1.5nm.",
        "answer": "the force of attraction between the k^{+} and \\mathrm{o}^{2-} ions is 2.05 × 10^{-10} N.",
        "type": 1,
        "wrong_answers_1": "the force of attraction between the k^{+} and \\mathrm{o}^{2-} ions is 2.35 × 10^{-10} N.",
        "wrong_answers_2": "the force of attraction between the k^{+} and \\mathrm{o}^{2-} ions is 2.30 × 10^{-10} N.",
        "wrong_answers_3": "the force of attraction between the k^{+} and \\mathrm{o}^{2-} ions is 1.89 × 10^{-10} N."
    },
    {
        "idx": 628,
        "question": "Briefly cite the main differences between ionic, covalent, and metallic bonding.",
        "answer": "The main differences between the various forms of primary bonding are: Ionic--there is electrostatic attraction between oppositely charged ions. Covalent--there is electron sharing between two adjacent atoms such that each atom assumes a stable electron configuration. Metallic--the positively charged ion cores are shielded from one another, and also 'glued' together by the sea of valence electrons.",
        "type": 4,
        "wrong_answers_1": "With cements, the bonding process is a chemical, hydration reaction between the water that has been added and the various cement constituents. The cement particles are bonded together by reactions that occur at the particle surfaces.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) electron configuration is that of an alkaline earth metal because of two (s) electrons.",
        "wrong_answers_3": "When a current arises from a flow of electrons, the conduction is termed electronic; for \\\\ ionic conduction, the current results from the net motion of charged ions."
    },
    {
        "idx": 629,
        "question": "State the Pauli exclusion principle.",
        "answer": "The Pauli exclusion principle states that each electron state can hold no more than two electrons, which must have opposite spins.",
        "type": 4,
        "wrong_answers_1": "For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid.",
        "wrong_answers_2": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities.",
        "wrong_answers_3": "The mj quantum number designates the number of electron states in each electron subshell."
    },
    {
        "idx": 630,
        "question": "What type(s) of bonding would be expected for brass (a copper-zinc alloy)?",
        "answer": "For brass, the bonding is metallic since it is a metal alloy.",
        "type": 4,
        "wrong_answers_1": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin).",
        "wrong_answers_2": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin).",
        "wrong_answers_3": "For tungsten, the bonding is metallic since it is a metallic element from the periodic table."
    },
    {
        "idx": 631,
        "question": "What type(s) of bonding would be expected for rubber?",
        "answer": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)",
        "type": 4,
        "wrong_answers_1": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)",
        "wrong_answers_2": "For nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.)",
        "wrong_answers_3": "The intermolecular bonding for \\mathrm{HF} is hydrogen, whereas for \\mathrm{HCl}, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher melting temperature."
    },
    {
        "idx": 632,
        "question": "What type(s) of bonding would be expected for barium sulfide (BaS)?",
        "answer": "For BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of Ba and S in the periodic table.",
        "type": 4,
        "wrong_answers_1": "For AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of Al and P in the periodic table.",
        "wrong_answers_2": "For CaF2, the bonding is predominantly ionic (but with some slight covalent character) on the basis of the relative positions of Ca and F in the periodic table.",
        "wrong_answers_3": "For CdTe, the bonding is predominantly covalent (with some slight ionic character) on the basis of the relative positions of Cd and Te in the periodic table."
    },
    {
        "idx": 633,
        "question": "What type(s) of bonding would be expected for solid xenon?",
        "answer": "For solid xenon, the bonding is van der Waals since xenon is an inert gas.",
        "type": 4,
        "wrong_answers_1": "For solid xenon, the bonding is van der Waals since xenon is an inert gas.",
        "wrong_answers_2": "The intermolecular bonding for \\mathrm{HF} is hydrogen, whereas for \\mathrm{HCl}, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher melting temperature.",
        "wrong_answers_3": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)"
    },
    {
        "idx": 634,
        "question": "What type(s) of bonding would be expected for bronze?",
        "answer": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin).",
        "type": 4,
        "wrong_answers_1": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin).",
        "wrong_answers_2": "For brass, the bonding is metallic since it is a metal alloy.",
        "wrong_answers_3": "For tungsten, the bonding is metallic since it is a metallic element from the periodic table."
    },
    {
        "idx": 635,
        "question": "What type(s) of bonding would be expected for nylon?",
        "answer": "For nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.)",
        "type": 4,
        "wrong_answers_1": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)",
        "wrong_answers_2": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)",
        "wrong_answers_3": "The intermolecular bonding for \\mathrm{HF} is hydrogen, whereas for \\mathrm{HCl}, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher melting temperature."
    },
    {
        "idx": 636,
        "question": "What type(s) of bonding would be expected for aluminum phosphide (AlP)?",
        "answer": "For AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of Al and P in the periodic table.",
        "type": 4,
        "wrong_answers_1": "For BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of Ba and S in the periodic table.",
        "wrong_answers_2": "For CaF2, the bonding is predominantly ionic (but with some slight covalent character) on the basis of the relative positions of Ca and F in the periodic table.",
        "wrong_answers_3": "For CdTe, the bonding is predominantly covalent (with some slight ionic character) on the basis of the relative positions of Cd and Te in the periodic table."
    },
    {
        "idx": 637,
        "question": "What is the difference between atomic structure and crystal structure?",
        "answer": "Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material.",
        "type": 4,
        "wrong_answers_1": "The two characteristics of component ions that determine the crystal structure of a ceramic compound are: 1) the magnitude of the electrical charge on each ion, and 2) the relative sizes of the cations and anions.",
        "wrong_answers_2": "For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid.",
        "wrong_answers_3": "The mj quantum number designates the number of electron states in each electron subshell."
    },
    {
        "idx": 638,
        "question": "If the atomic radius of aluminum is 0.143nm, calculate the volume of its unit cell in cubic meters.",
        "answer": "the volume of the unit cell is 6.62 × 10^{-29} m^{3}.",
        "type": 1,
        "wrong_answers_1": "the volume of the unit cell is 6.07 × 10^{-29} m^{3}.",
        "wrong_answers_2": "the volume of the unit cell is 7.52 × 10^{-29} m^{3}.",
        "wrong_answers_3": "the volume of the unit cell is 6.04 × 10^{-29} m^{3}."
    },
    {
        "idx": 639,
        "question": "Show that the atomic packing factor for BCC is 0.68 .",
        "answer": "the atomic packing factor for bcc is 0.68.",
        "type": 1,
        "wrong_answers_1": "the atomic packing factor for bcc is 0.73.",
        "wrong_answers_2": "the atomic packing factor for bcc is 0.64.",
        "wrong_answers_3": "the atomic packing factor for bcc is 0.73."
    },
    {
        "idx": 640,
        "question": "Iron has a BCC crystal structure, an atomic radius of 0.124nm, and an atomic weight of 55.85g / mol. Compute and compare its theoretical density with the experimental value found inside the front cover.",
        "answer": "the theoretical density of iron is 7.90g / {cm}^{3}. the experimental value given inside the front cover is 7.87g / {cm}^{3}.",
        "type": 1,
        "wrong_answers_1": "the theoretical density of iron is 8.47g / {cm}^{3}. the experimental value given inside the front cover is 7.09g / {cm}^{3}.",
        "wrong_answers_2": "the theoretical density of iron is 9.03g / {cm}^{3}. the experimental value given inside the front cover is 7.24g / {cm}^{3}.",
        "wrong_answers_3": "the theoretical density of iron is 6.81g / {cm}^{3}. the experimental value given inside the front cover is 7.33g / {cm}^{3}."
    },
    {
        "idx": 641,
        "question": "Calculate the radius of an iridium atom, given that Ir has an FCC crystal structure, a density of 22.4g / {cm}^{3}, and an atomic weight of 192.2g / mol.",
        "answer": "the radius of an iridium atom is 0.136nm.",
        "type": 1,
        "wrong_answers_1": "the radius of an iridium atom is 0.148nm.",
        "wrong_answers_2": "the radius of an iridium atom is 0.120nm.",
        "wrong_answers_3": "the radius of an iridium atom is 0.141nm."
    },
    {
        "idx": 642,
        "question": "Calculate the radius of a vanadium atom, given thatV has a BCC crystal structure, a density of 5.96 g / {cm}^{3}, and an atomic weight of 50.9g / mol.",
        "answer": "the radius of a vanadium atom is 0.132 nm.",
        "type": 1,
        "wrong_answers_1": "the radius of a vanadium atom is 0.152 nm.",
        "wrong_answers_2": "the radius of a vanadium atom is 0.125 nm.",
        "wrong_answers_3": "the radius of a vanadium atom is 0.137 nm."
    },
    {
        "idx": 643,
        "question": "For alloy A with atomic weight 77.4 g/mol, density 8.22 g/cm³, and atomic radius 0.125 mm, determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination.",
        "answer": "for alloy a, its crystal structure is simple cubic.",
        "type": 1
    },
    {
        "idx": 644,
        "question": "For alloy B with atomic weight 107.6 g/mol, density 13.42 g/cm³, and atomic radius 0.133 mm, determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination.",
        "answer": "for alloy b, its crystal structure is fcc.",
        "type": 1
    },
    {
        "idx": 645,
        "question": "For alloy C with atomic weight 127.3 g/mol, density 9.23 g/cm³, and atomic radius 0.142 mm, determine whether its crystal structure is FCC, BCC, or simple cubic and then justify your determination.",
        "answer": "for alloy c, its crystal structure is simple cubic.",
        "type": 1
    },
    {
        "idx": 646,
        "question": "The unit cell for tin has tetragonal symmetry, with a and b lattice parameters of 0.583 and 0.318nm, respectively. If its density, atomic weight, and atomic radius are 7.30g / {cm}^{3}, 118.69g / mol, and 0.151nm, respectively, compute the atomic packing factor.",
        "answer": "the atomic packing factor (apf) for tin is 0.534.",
        "type": 1,
        "wrong_answers_1": "the atomic packing factor (apf) for tin is 0.569.",
        "wrong_answers_2": "the atomic packing factor (apf) for tin is 0.473.",
        "wrong_answers_3": "the atomic packing factor (apf) for tin is 0.596."
    },
    {
        "idx": 647,
        "question": "List the point coordinates of the titanium, barium, and oxygen ions for a unit cell of the perovskite crystal structure.",
        "answer": " the barium ions are situated at all corner positions. The point coordinates for these ions are as follows: 000,100,110,010,001,101,111, and 011 .\nThe oxygen ions are located at all face-centered positions; therefore, their coordinates are \\frac{1}{2} \\frac{1}{2} 0, \\frac{1}{2} \\frac{1}{2} 1, \\frac{1}{2} \\frac{1}{2}, \\frac{0}{2} \\frac{1}{2}, \\frac{1}{2} 0 \\frac{1}{2}, and \\frac{1}{2} 1 \\frac{1}{2}.\nAnd, finally, the titanium ion resides at the center of the cubic unit cell, with coordinates \\frac{1}{2} \\frac{1}{2} \\frac{1}{2}.",
        "type": 4,
        "wrong_answers_1": "The surface energy for a crystallographic plane will depend on its packing density-that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.55, the planar densities for BCC (100) and (110) are \\frac{3}{16 R^{2}} and \\frac{3}{8 R^{2} \\sqrt{2}}, respectively-that is \\frac{0.19}{R^{2}} and \\frac{0.27}{R^{2}}. Thus, since the planar density for (110) is greater, it will have the lower surface energy.",
        "wrong_answers_2": "The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar density (Section 3.11)] - that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.54, planar densities for FCC (100) and (111) planes are \\frac{1}{4 R^{2}} and \\frac{1}{2 R^{2} \\sqrt{3}}, respectively - that is \\frac{0.25}{R^{2}} and \\frac{0.29}{R^{2}} (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have the lower surface energy.",
        "wrong_answers_3": "The coordinates of these atoms are as follows: 000, 100, 110, 010, 001, 101, 111, and 011."
    },
    {
        "idx": 648,
        "question": "List the point coordinates of all carbon atoms that occupy the corner positions of the diamond cubic unit cell.",
        "answer": "The coordinates of these atoms are as follows: 000, 100, 110, 010, 001, 101, 111, and 011.",
        "type": 4,
        "wrong_answers_1": "The coordinates of these atoms are as follows: 3/4 1/4 1/4, 1/4 3/4 3/4, 1/4 1/4 3/4, and 3/4 3/4 1/4.",
        "wrong_answers_2": "The coordinates of these atoms are as follows: 1/2 1/2 0, 1/2 1/2 1, 1/2 0 1/2, 0 1/2 1/2, and 1/2 1/2 1/2.",
        "wrong_answers_3": " the barium ions are situated at all corner positions. The point coordinates for these ions are as follows: 000,100,110,010,001,101,111, and 011 .\nThe oxygen ions are located at all face-centered positions; therefore, their coordinates are \\frac{1}{2} \\frac{1}{2} 0, \\frac{1}{2} \\frac{1}{2} 1, \\frac{1}{2} \\frac{1}{2}, \\frac{0}{2} \\frac{1}{2}, \\frac{1}{2} 0 \\frac{1}{2}, and \\frac{1}{2} 1 \\frac{1}{2}.\nAnd, finally, the titanium ion resides at the center of the cubic unit cell, with coordinates \\frac{1}{2} \\frac{1}{2} \\frac{1}{2}."
    },
    {
        "idx": 649,
        "question": "List the point coordinates of all carbon atoms that reside on the face-centered positions of the diamond cubic unit cell.",
        "answer": "The coordinates of these atoms are as follows: 1/2 1/2 0, 1/2 1/2 1, 1/2 0 1/2, 0 1/2 1/2, and 1/2 1/2 1/2.",
        "type": 4,
        "wrong_answers_1": "The coordinates of these atoms are as follows: 3/4 1/4 1/4, 1/4 3/4 3/4, 1/4 1/4 3/4, and 3/4 3/4 1/4.",
        "wrong_answers_2": "The coordinates of these atoms are as follows: 000, 100, 110, 010, 001, 101, 111, and 011.",
        "wrong_answers_3": "First of all, open the \"Molecular Definition Utility\"; it may be found in either of \"Metallic Crystal Structures and Crystallography\" or \"Ceramic Crystal Structures\" modules.\nIn the \"Step 1\" window, it is necessary to define the atom types, colors for the spheres (atoms), and specify atom sizes. Let us enter \"Au\" as the name for the gold atoms (since \"Au\" the symbol for gold), and \"Cu\" as the name for the copper atoms. Next it is necessary to choose a color for each atom type from the selections that appear in the pull-down menu-for example, \"Yellow\" for Au and \"Red\" for Cu. In the \"Atom Size\" window, it is necessary to enter an atom/ion size. In the instructions for this step, it is suggested that the atom/ion diameter in nanometers be used. From the table found inside the front cover of the textbook, the atomic radii for gold and copper are 0.144nm and 0.128nm, respectively, and, therefore, their ionic diameters are twice these values (i.e., 0.288nm and 0.256nm ); therefore, we enter the values \" 0.288 \" and \" 0.256 \" for the two atom types. Now click on the \"Register\" button, followed by clicking on the \"Go to Step 2\" button.\nIn the \"Step 2\" window we specify positions for all of the atoms within the unit cell; their point coordinates are specified in the problem statement. Let's begin with gold. Click on the yellow sphere that is located to the right of the \"Molecule Definition Utility\" box. Again, Au atoms are situated at all eight corners of the cubic unit cell. One Au will be positioned at the origin of the coordinate system-i.e., its point coordinates are 000, and, therefore, we enter a \"0\" (zero) in each of the \" \\mathrm{x} \", \" \\mathrm{y} \", and \" \\mathrm{z} \" atom position boxes. Next we click on the \"Register Atom Position\" button. Now we enter the coordinates of another gold atom; let us arbitrarily select the one that resides at the corner of the unit cell that is one unit-cell length along the x-axis (i.e., at the 100 point coordinate). Inasmuch as it is located a distance of a units along the x-axis the value of \" 0.374 \" is entered in the \" \\mathrm{x} \" atom position box (since this is the value of a given in the problem statement); zeros are entered in each of the \" \\mathrm{y} \" and \" z \" position boxes. We repeat this procedure for the remaining six Au atoms.\nAfter this step has been completed, it is necessary to specify positions for the copper atoms, which are located at all six face-centered sites. To begin, we click on the red sphere that is located next to the \"Molecule Definition Utility\" box. The point coordinates for some of the Cu atoms are fractional ones; in these instances, the a unit cell length (i.e., 0.374) is multiplied by the fraction. For example, one Cu atom is located 1 \\frac{1}{2} \\frac{1}{2} coordinate. Therefore, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z} atoms positions are (1)(0.374) =0.374, \\frac{1}{2}(0.374)=0.187, and \\frac{1}{2}(0.374)=0.187, respectively.\nFor the gold atoms, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z}^{\\prime} atom position entries for all 8 sets of point coordinates are as follows: \\[\n\\begin{array}{l}\n0,0, \\text { and } 0 \\\\\n0.374,0, \\text { and } 0 \\\\\n0,0.374, \\text { and } 0 \\\\\n0,0, \\text { and } 0.374 \\\\\n0,0.374,0.374 \\\\\n0.374,0.374 \\\\\n0.374,0 .374,0 \\\\\n0.374,0.374,0.374\n\\end{array}\n\\]\nNow, for the copper atoms, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z} atom position entries for all 6 sets of point coordinates are as follows:\n\\[\n\\begin{array}{l}\n0.187,0.187,0 \\\\\n0.187,0,0.187 \\\\\n0,0.187,0.187 \\\\\n0.374,0.187,0.187 \\\\\n0.187,0.374,0.187 \\\\\n0.187,0.187,0.374\n\\end{array}\n\\]\nIn Step 3, we may specify which atoms are to be represented as being bonded to one another, and which type of bond(s) to use (single solid, single dashed, double, and triple are possibilities), or we may elect to not represent any bonds at all (in which case we are finished). If it is decided to show bonds, probably the best thing to do is to represent unit cell edges as bonds. This image may be rotated by using mouse click-and-drag\nYour image should appear as the following screen shot. Here the gold atoms appear lighter than the copper atoms.  [Note: Unfortunately, with this version of the Molecular Definition Utility, it is not possible to save either the data or the image that you have generated. You may use screen capture (or screen shot) software to record and store your image.]"
    },
    {
        "idx": 650,
        "question": "List the point coordinates of all carbon atoms that are positioned within the interior of the diamond cubic unit cell.",
        "answer": "The coordinates of these atoms are as follows: 3/4 1/4 1/4, 1/4 3/4 3/4, 1/4 1/4 3/4, and 3/4 3/4 1/4.",
        "type": 4,
        "wrong_answers_1": "The coordinates of these atoms are as follows: 1/2 1/2 0, 1/2 1/2 1, 1/2 0 1/2, 0 1/2 1/2, and 1/2 1/2 1/2.",
        "wrong_answers_2": "The coordinates of these atoms are as follows: 000, 100, 110, 010, 001, 101, 111, and 011.",
        "wrong_answers_3": "First of all, open the \"Molecular Definition Utility\"; it may be found in either of \"Metallic Crystal Structures and Crystallography\" or \"Ceramic Crystal Structures\" modules.\nIn the \"Step 1\" window, it is necessary to define the atom types, colors for the spheres (atoms), and specify atom sizes. Let us enter \"Au\" as the name for the gold atoms (since \"Au\" the symbol for gold), and \"Cu\" as the name for the copper atoms. Next it is necessary to choose a color for each atom type from the selections that appear in the pull-down menu-for example, \"Yellow\" for Au and \"Red\" for Cu. In the \"Atom Size\" window, it is necessary to enter an atom/ion size. In the instructions for this step, it is suggested that the atom/ion diameter in nanometers be used. From the table found inside the front cover of the textbook, the atomic radii for gold and copper are 0.144nm and 0.128nm, respectively, and, therefore, their ionic diameters are twice these values (i.e., 0.288nm and 0.256nm ); therefore, we enter the values \" 0.288 \" and \" 0.256 \" for the two atom types. Now click on the \"Register\" button, followed by clicking on the \"Go to Step 2\" button.\nIn the \"Step 2\" window we specify positions for all of the atoms within the unit cell; their point coordinates are specified in the problem statement. Let's begin with gold. Click on the yellow sphere that is located to the right of the \"Molecule Definition Utility\" box. Again, Au atoms are situated at all eight corners of the cubic unit cell. One Au will be positioned at the origin of the coordinate system-i.e., its point coordinates are 000, and, therefore, we enter a \"0\" (zero) in each of the \" \\mathrm{x} \", \" \\mathrm{y} \", and \" \\mathrm{z} \" atom position boxes. Next we click on the \"Register Atom Position\" button. Now we enter the coordinates of another gold atom; let us arbitrarily select the one that resides at the corner of the unit cell that is one unit-cell length along the x-axis (i.e., at the 100 point coordinate). Inasmuch as it is located a distance of a units along the x-axis the value of \" 0.374 \" is entered in the \" \\mathrm{x} \" atom position box (since this is the value of a given in the problem statement); zeros are entered in each of the \" \\mathrm{y} \" and \" z \" position boxes. We repeat this procedure for the remaining six Au atoms.\nAfter this step has been completed, it is necessary to specify positions for the copper atoms, which are located at all six face-centered sites. To begin, we click on the red sphere that is located next to the \"Molecule Definition Utility\" box. The point coordinates for some of the Cu atoms are fractional ones; in these instances, the a unit cell length (i.e., 0.374) is multiplied by the fraction. For example, one Cu atom is located 1 \\frac{1}{2} \\frac{1}{2} coordinate. Therefore, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z} atoms positions are (1)(0.374) =0.374, \\frac{1}{2}(0.374)=0.187, and \\frac{1}{2}(0.374)=0.187, respectively.\nFor the gold atoms, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z}^{\\prime} atom position entries for all 8 sets of point coordinates are as follows: \\[\n\\begin{array}{l}\n0,0, \\text { and } 0 \\\\\n0.374,0, \\text { and } 0 \\\\\n0,0.374, \\text { and } 0 \\\\\n0,0, \\text { and } 0.374 \\\\\n0,0.374,0.374 \\\\\n0.374,0.374 \\\\\n0.374,0 .374,0 \\\\\n0.374,0.374,0.374\n\\end{array}\n\\]\nNow, for the copper atoms, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z} atom position entries for all 6 sets of point coordinates are as follows:\n\\[\n\\begin{array}{l}\n0.187,0.187,0 \\\\\n0.187,0,0.187 \\\\\n0,0.187,0.187 \\\\\n0.374,0.187,0.187 \\\\\n0.187,0.374,0.187 \\\\\n0.187,0.187,0.374\n\\end{array}\n\\]\nIn Step 3, we may specify which atoms are to be represented as being bonded to one another, and which type of bond(s) to use (single solid, single dashed, double, and triple are possibilities), or we may elect to not represent any bonds at all (in which case we are finished). If it is decided to show bonds, probably the best thing to do is to represent unit cell edges as bonds. This image may be rotated by using mouse click-and-drag\nYour image should appear as the following screen shot. Here the gold atoms appear lighter than the copper atoms.  [Note: Unfortunately, with this version of the Molecular Definition Utility, it is not possible to save either the data or the image that you have generated. You may use screen capture (or screen shot) software to record and store your image.]"
    },
    {
        "idx": 651,
        "question": "Using the Molecule Definition Utility found in both \"Metallic Crystal Structures and Crystallography\" and \"Ceramic Crystal Structures\" modules of VMSE, located on the book's web site [www.wiley.com/college/Callister (Student Companion Site)], generate a three-dimensional unit cell for the intermetallic compound \\mathrm{AuCu}_{3} given the following: (1) the unit cell is cubic with an edge length of 0.374nm, (2) gold atoms are situated at all cube corners, and (3) copper atoms are positioned at the centers of all unit cell faces.",
        "answer": "First of all, open the \"Molecular Definition Utility\"; it may be found in either of \"Metallic Crystal Structures and Crystallography\" or \"Ceramic Crystal Structures\" modules.\nIn the \"Step 1\" window, it is necessary to define the atom types, colors for the spheres (atoms), and specify atom sizes. Let us enter \"Au\" as the name for the gold atoms (since \"Au\" the symbol for gold), and \"Cu\" as the name for the copper atoms. Next it is necessary to choose a color for each atom type from the selections that appear in the pull-down menu-for example, \"Yellow\" for Au and \"Red\" for Cu. In the \"Atom Size\" window, it is necessary to enter an atom/ion size. In the instructions for this step, it is suggested that the atom/ion diameter in nanometers be used. From the table found inside the front cover of the textbook, the atomic radii for gold and copper are 0.144nm and 0.128nm, respectively, and, therefore, their ionic diameters are twice these values (i.e., 0.288nm and 0.256nm ); therefore, we enter the values \" 0.288 \" and \" 0.256 \" for the two atom types. Now click on the \"Register\" button, followed by clicking on the \"Go to Step 2\" button.\nIn the \"Step 2\" window we specify positions for all of the atoms within the unit cell; their point coordinates are specified in the problem statement. Let's begin with gold. Click on the yellow sphere that is located to the right of the \"Molecule Definition Utility\" box. Again, Au atoms are situated at all eight corners of the cubic unit cell. One Au will be positioned at the origin of the coordinate system-i.e., its point coordinates are 000, and, therefore, we enter a \"0\" (zero) in each of the \" \\mathrm{x} \", \" \\mathrm{y} \", and \" \\mathrm{z} \" atom position boxes. Next we click on the \"Register Atom Position\" button. Now we enter the coordinates of another gold atom; let us arbitrarily select the one that resides at the corner of the unit cell that is one unit-cell length along the x-axis (i.e., at the 100 point coordinate). Inasmuch as it is located a distance of a units along the x-axis the value of \" 0.374 \" is entered in the \" \\mathrm{x} \" atom position box (since this is the value of a given in the problem statement); zeros are entered in each of the \" \\mathrm{y} \" and \" z \" position boxes. We repeat this procedure for the remaining six Au atoms.\nAfter this step has been completed, it is necessary to specify positions for the copper atoms, which are located at all six face-centered sites. To begin, we click on the red sphere that is located next to the \"Molecule Definition Utility\" box. The point coordinates for some of the Cu atoms are fractional ones; in these instances, the a unit cell length (i.e., 0.374) is multiplied by the fraction. For example, one Cu atom is located 1 \\frac{1}{2} \\frac{1}{2} coordinate. Therefore, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z} atoms positions are (1)(0.374) =0.374, \\frac{1}{2}(0.374)=0.187, and \\frac{1}{2}(0.374)=0.187, respectively.\nFor the gold atoms, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z}^{\\prime} atom position entries for all 8 sets of point coordinates are as follows: \\[\n\\begin{array}{l}\n0,0, \\text { and } 0 \\\\\n0.374,0, \\text { and } 0 \\\\\n0,0.374, \\text { and } 0 \\\\\n0,0, \\text { and } 0.374 \\\\\n0,0.374,0.374 \\\\\n0.374,0.374 \\\\\n0.374,0 .374,0 \\\\\n0.374,0.374,0.374\n\\end{array}\n\\]\nNow, for the copper atoms, the \\mathrm{x}, \\mathrm{y}, and \\mathrm{z} atom position entries for all 6 sets of point coordinates are as follows:\n\\[\n\\begin{array}{l}\n0.187,0.187,0 \\\\\n0.187,0,0.187 \\\\\n0,0.187,0.187 \\\\\n0.374,0.187,0.187 \\\\\n0.187,0.374,0.187 \\\\\n0.187,0.187,0.374\n\\end{array}\n\\]\nIn Step 3, we may specify which atoms are to be represented as being bonded to one another, and which type of bond(s) to use (single solid, single dashed, double, and triple are possibilities), or we may elect to not represent any bonds at all (in which case we are finished). If it is decided to show bonds, probably the best thing to do is to represent unit cell edges as bonds. This image may be rotated by using mouse click-and-drag\nYour image should appear as the following screen shot. Here the gold atoms appear lighter than the copper atoms.  [Note: Unfortunately, with this version of the Molecular Definition Utility, it is not possible to save either the data or the image that you have generated. You may use screen capture (or screen shot) software to record and store your image.]",
        "type": 4,
        "wrong_answers_1": "The coordinates of these atoms are as follows: 1/2 1/2 0, 1/2 1/2 1, 1/2 0 1/2, 0 1/2 1/2, and 1/2 1/2 1/2.",
        "wrong_answers_2": "The coordinates of these atoms are as follows: 3/4 1/4 1/4, 1/4 3/4 3/4, 1/4 1/4 3/4, and 3/4 3/4 1/4.",
        "wrong_answers_3": "of the possible copolymers, the only one for which there is a restriction on the ratio of repeat unit types is alternating; the ratio must be 1:1. therefore, on the basis of the result in part (a), the possibilities for this copolymer are random, graft, and block."
    },
    {
        "idx": 652,
        "question": "Convert the (010) plane into the four-index Miller-Bravais scheme for hexagonal unit cells.",
        "answer": "the (010) plane in the four-index miller-bravais scheme is (01 -1 0).",
        "type": 1,
        "wrong_answers_1": "the (11) plane in the four-index miller-bravais scheme is (1 -1 0).",
        "wrong_answers_2": "the (10) plane in the four-index miller-bravais scheme is (1 -1 0).",
        "wrong_answers_3": "the (9) plane in the four-index miller-bravais scheme is (1 -1 0)."
    },
    {
        "idx": 653,
        "question": "Convert the (101) plane into the four-index Miller-Bravais scheme for hexagonal unit cells.",
        "answer": "the (101) plane in the four-index miller-bravais scheme is (10 -1 1).",
        "type": 1,
        "wrong_answers_1": "the (90) plane in the four-index miller-bravais scheme is (11 -1 1).",
        "wrong_answers_2": "the (94) plane in the four-index miller-bravais scheme is (9 -1 1).",
        "wrong_answers_3": "the (96) plane in the four-index miller-bravais scheme is (10 -1 1)."
    },
    {
        "idx": 654,
        "question": "(a) Derive the planar density expression for the HCP (0001) plane in terms of the atomic radius R.",
        "answer": "the planar density expression for the hcp (0001) plane in terms of the atomic radius r is: pd_{0001} = 1 / (2 r^{2} sqrt(3))",
        "type": 1,
        "wrong_answers_1": "the planar density expression for the hcp (01) plane in terms of the atomic radius r is: pd_{01} = 1 / (2 r^{2} sqrt(3))",
        "wrong_answers_2": "the planar density expression for the hcp (01) plane in terms of the atomic radius r is: pd_{01} = 1 / (2 r^{2} sqrt(3))",
        "wrong_answers_3": "the planar density expression for the hcp (01) plane in terms of the atomic radius r is: pd_{01} = 1 / (2 r^{2} sqrt(3))"
    },
    {
        "idx": 655,
        "question": "(b) Compute the planar density value for this same plane for magnesium.",
        "answer": "the planar density value for the (0001) plane for magnesium is: pd_{0001}(mg) = 1.128 × 10^{19} m^{-2}",
        "type": 1,
        "wrong_answers_1": "the planar density value for the (01) plane for magnesium is: pd_{01}(mg) = 1.015 × 10^{19} m^{-2}",
        "wrong_answers_2": "the planar density value for the (01) plane for magnesium is: pd_{01}(mg) = 1.038 × 10^{19} m^{-2}",
        "wrong_answers_3": "the planar density value for the (01) plane for magnesium is: pd_{01}(mg) = 1.028 × 10^{19} m^{-2}"
    },
    {
        "idx": 656,
        "question": "Explain why the properties of polycrystalline materials are most often isotropic.",
        "answer": "Although each individual grain in a polycrystalline material may be anisotropic, if the grains have random orientations, then the solid aggregate of the many anisotropic grains will behave isotropically.",
        "type": 4,
        "wrong_answers_1": "During cold-working, the grain structure of the metal has been distorted to accommodate the deformation. Recrystallization produces grains that are equiaxed and smaller than the parent grains.",
        "wrong_answers_2": "Coring is the phenomenon whereby concentration gradients exist across grains in polycrystalline alloys, with higher concentrations of the component having the lower melting temperature at the grain boundaries. It occurs, during solidification, as a consequence of cooling rates that are too rapid to allow for the maintenance of the equilibrium composition of the solid phase.",
        "wrong_answers_3": "A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time."
    },
    {
        "idx": 657,
        "question": "The metal iridium has an FCC crystal structure. If the angle of diffraction for the (220) set of planes occurs at 69.22 degrees (first-order reflection) when monochromatic x-radiation having a wavelength of 0.1542 nm is used, compute the interplanar spacing for this set of planes.",
        "answer": "the interplanar spacing for the (220) set of planes for iridium is 0.1357 nm.",
        "type": 1,
        "wrong_answers_1": "the interplanar spacing for the (232) set of planes for iridium is 0.1525 nm.",
        "wrong_answers_2": "the interplanar spacing for the (187) set of planes for iridium is 0.1227 nm.",
        "wrong_answers_3": "the interplanar spacing for the (244) set of planes for iridium is 0.1297 nm."
    },
    {
        "idx": 658,
        "question": "The metal iridium has an FCC crystal structure. Given the interplanar spacing for the (220) set of planes is 0.1357 nm, compute the atomic radius for an iridium atom.",
        "answer": "the atomic radius for an iridium atom is 0.1357 nm.",
        "type": 1,
        "wrong_answers_1": "the atomic radius for an iridium atom is 0.1191 nm.",
        "wrong_answers_2": "the atomic radius for an iridium atom is 0.1166 nm.",
        "wrong_answers_3": "the atomic radius for an iridium atom is 0.1409 nm."
    },
    {
        "idx": 659,
        "question": "Which of these elements would you expect to form with copper a substitutional solid solution having complete solubility? Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.",
        "answer": "Ni, Pd, and Pt meet all of the criteria and thus form substitutional solid solutions having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures.",
        "type": 2
    },
    {
        "idx": 660,
        "question": "Which of these elements would you expect to form with copper a substitutional solid solution of incomplete solubility? Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.",
        "answer": "Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Cu are greater than ±15%, and/or have a valence different than 2+.",
        "type": 2
    },
    {
        "idx": 661,
        "question": "Which of these elements would you expect to form with copper an interstitial solid solution? Below, atomic radius, crystal structure, electronegativity, and the most common valence are tabulated, for several elements; for those that are nonmetals, only atomic radii are indicated.",
        "answer": "C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Cu.",
        "type": 2
    },
    {
        "idx": 662,
        "question": "What is the composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu ?",
        "answer": "the composition of the alloy is 29.4 \\, \\text{at}% \\, \\mathrm{zn} and 70.6 \\, \\text{at}% \\, \\mathrm{cu}.",
        "type": 1,
        "wrong_answers_1": "the composition of the alloy is 31.4 \\, \\text{at}% \\, \\mathrm{zn} and 78.7 \\, \\text{at}% \\, \\mathrm{cu}.",
        "wrong_answers_2": "the composition of the alloy is 30.8 \\, \\text{at}% \\, \\mathrm{zn} and 62.7 \\, \\text{at}% \\, \\mathrm{cu}.",
        "wrong_answers_3": "the composition of the alloy is 31.5 \\, \\text{at}% \\, \\mathrm{zn} and 68.4 \\, \\text{at}% \\, \\mathrm{cu}."
    },
    {
        "idx": 663,
        "question": "What is the composition, in weight percent, of an alloy that consists of 6 at% \\mathrm{Pb} and 94 \\mathrm{at}% \\mathrm{Sn} ?",
        "answer": "the composition of the alloy is: 1. 10.0 \\, \\text{wt% pb} 2. 90.0 \\, \\text{wt% sn}",
        "type": 1,
        "wrong_answers_1": "the composition of the alloy is: 1. 10.9 \\, \\text{wt% pb} 2. 80.3 \\, \\text{wt% sn}",
        "wrong_answers_2": "the composition of the alloy is: 1. 9.1 \\, \\text{wt% pb} 2. 85.7 \\, \\text{wt% sn}",
        "wrong_answers_3": "the composition of the alloy is: 1. 9.1 \\, \\text{wt% pb} 2. 81.0 \\, \\text{wt% sn}"
    },
    {
        "idx": 664,
        "question": "What is the composition, in atom percent, of an alloy that contains 99.7 lb_{m} copper, 102 lb_{m} zinc, and 2.1 lb_{m} lead?\n",
        "answer": "the composition of the alloy in atom percent is: \\[\nc_{\\mathrm{cu}} = 50.0 \\, \\text{at}%\n\\] \\[\nc_{\\mathrm{zn}} = 49.7 \\, \\text{at}%\n\\] \\[\nc_{\\mathrm{pb}} = 0.3 \\, \\text{at}%\n\\]",
        "type": 1,
        "wrong_answers_1": "the composition of the alloy in atom percent is: \\[\nc_{\\mathrm{cu}} = 53.3 \\, \\text{at}%\n\\] \\[\nc_{\\mathrm{zn}} = 52.4 \\, \\text{at}%\n\\] \\[\nc_{\\mathrm{pb}} = 0.3 \\, \\text{at}%\n\\]",
        "wrong_answers_2": "the composition of the alloy in atom percent is: \\[\nc_{\\mathrm{cu}} = 52.5 \\, \\text{at}%\n\\] \\[\nc_{\\mathrm{zn}} = 46.3 \\, \\text{at}%\n\\] \\[\nc_{\\mathrm{pb}} = 0.3 \\, \\text{at}%\n\\]",
        "wrong_answers_3": "the composition of the alloy in atom percent is: \\[\nc_{\\mathrm{cu}} = 47.4 \\, \\text{at}%\n\\] \\[\nc_{\\mathrm{zn}} = 52.9 \\, \\text{at}%\n\\] \\[\nc_{\\mathrm{pb}} = 0.3 \\, \\text{at}%\n\\]"
    },
    {
        "idx": 665,
        "question": "What is the composition, in atom percent, of an alloy that consists of 97 wt% Fe and 3 wt% Si?",
        "answer": "the composition of the alloy is 94.2 at% fe and 5.8 at% si.",
        "type": 1,
        "wrong_answers_1": "the composition of the alloy is 98.9 at% fe and 6.4 at% si.",
        "wrong_answers_2": "the composition of the alloy is 88.8 at% fe and 5.0 at% si.",
        "wrong_answers_3": "the composition of the alloy is 102.0 at% fe and 6.4 at% si."
    },
    {
        "idx": 666,
        "question": "Calculate the number of atoms per cubic meter in aluminum.",
        "answer": "6.05 × 10^{28} atoms/m^3",
        "type": 1,
        "wrong_answers_1": "5.52 × 10^{28} atoms/m^3",
        "wrong_answers_2": "5.64 × 10^{28} atoms/m^3",
        "wrong_answers_3": "5.81 × 10^{28} atoms/m^3"
    },
    {
        "idx": 667,
        "question": "The concentration of carbon in an iron-carbon alloy is 0.15 wt%. What is the concentration in kilograms of carbon per cubic meter of alloy?",
        "answer": "the concentration of carbon in the alloy is 11.8 \\mathrm{kg/m^{3}}.",
        "type": 1,
        "wrong_answers_1": "the concentration of carbon in the alloy is 12.2 \\mathrm{kg/m^{3}}.",
        "wrong_answers_2": "the concentration of carbon in the alloy is 13.4 \\mathrm{kg/m^{3}}.",
        "wrong_answers_3": "the concentration of carbon in the alloy is 12.3 \\mathrm{kg/m^{3}}."
    },
    {
        "idx": 668,
        "question": "Cite the relative Burgers vector-dislocation line orientations for edge, screw, and mixed dislocations.",
        "answer": "The Burgers vector and dislocation line are perpendicular for edge dislocations, parallel for screw dislocations, and neither perpendicular nor parallel for mixed dislocations.",
        "type": 4,
        "wrong_answers_1": "mixed dislocation--neither parallel nor perpendicular",
        "wrong_answers_2": "screw dislocation--perpendicular",
        "wrong_answers_3": "edge dislocation--parallel"
    },
    {
        "idx": 669,
        "question": "For an FCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (111) plane? Why? ",
        "answer": "The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar density (Section 3.11)] - that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.54, planar densities for FCC (100) and (111) planes are \\frac{1}{4 R^{2}} and \\frac{1}{2 R^{2} \\sqrt{3}}, respectively - that is \\frac{0.25}{R^{2}} and \\frac{0.29}{R^{2}} (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have the lower surface energy.",
        "type": 4,
        "wrong_answers_1": "The surface energy for a crystallographic plane will depend on its packing density-that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.55, the planar densities for BCC (100) and (110) are \\frac{3}{16 R^{2}} and \\frac{3}{8 R^{2} \\sqrt{2}}, respectively-that is \\frac{0.19}{R^{2}} and \\frac{0.27}{R^{2}}. Thus, since the planar density for (110) is greater, it will have the lower surface energy.",
        "wrong_answers_2": " the barium ions are situated at all corner positions. The point coordinates for these ions are as follows: 000,100,110,010,001,101,111, and 011 .\nThe oxygen ions are located at all face-centered positions; therefore, their coordinates are \\frac{1}{2} \\frac{1}{2} 0, \\frac{1}{2} \\frac{1}{2} 1, \\frac{1}{2} \\frac{1}{2}, \\frac{0}{2} \\frac{1}{2}, \\frac{1}{2} 0 \\frac{1}{2}, and \\frac{1}{2} 1 \\frac{1}{2}.\nAnd, finally, the titanium ion resides at the center of the cubic unit cell, with coordinates \\frac{1}{2} \\frac{1}{2} \\frac{1}{2}.",
        "wrong_answers_3": "The silicate materials have relatively low densities because the atomic bonds are primarily covalent in nature, and, therefore, directional. This limits the packing efficiency of the atoms, and therefore, the magnitude of the density."
    },
    {
        "idx": 670,
        "question": "For a BCC single crystal, would you expect the surface energy for a (100) plane to be greater or less than that for a (110) plane? Why? ",
        "answer": "The surface energy for a crystallographic plane will depend on its packing density-that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.55, the planar densities for BCC (100) and (110) are \\frac{3}{16 R^{2}} and \\frac{3}{8 R^{2} \\sqrt{2}}, respectively-that is \\frac{0.19}{R^{2}} and \\frac{0.27}{R^{2}}. Thus, since the planar density for (110) is greater, it will have the lower surface energy.",
        "type": 4,
        "wrong_answers_1": "The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar density (Section 3.11)] - that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.54, planar densities for FCC (100) and (111) planes are \\frac{1}{4 R^{2}} and \\frac{1}{2 R^{2} \\sqrt{3}}, respectively - that is \\frac{0.25}{R^{2}} and \\frac{0.29}{R^{2}} (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have the lower surface energy.",
        "wrong_answers_2": " the barium ions are situated at all corner positions. The point coordinates for these ions are as follows: 000,100,110,010,001,101,111, and 011 .\nThe oxygen ions are located at all face-centered positions; therefore, their coordinates are \\frac{1}{2} \\frac{1}{2} 0, \\frac{1}{2} \\frac{1}{2} 1, \\frac{1}{2} \\frac{1}{2}, \\frac{0}{2} \\frac{1}{2}, \\frac{1}{2} 0 \\frac{1}{2}, and \\frac{1}{2} 1 \\frac{1}{2}.\nAnd, finally, the titanium ion resides at the center of the cubic unit cell, with coordinates \\frac{1}{2} \\frac{1}{2} \\frac{1}{2}.",
        "wrong_answers_3": "The silicate materials have relatively low densities because the atomic bonds are primarily covalent in nature, and, therefore, directional. This limits the packing efficiency of the atoms, and therefore, the magnitude of the density."
    },
    {
        "idx": 671,
        "question": "For a given material, would you expect the surface energy to be greater than, the same as, or less than the grain boundary energy? Why?",
        "answer": "The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary.",
        "type": 4,
        "wrong_answers_1": "The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases.",
        "wrong_answers_2": "A twin boundary is an interface such that atoms on one side are located at mirror image positions of those atoms situated on the other boundary side. The region on one side of this boundary is called a twin.",
        "wrong_answers_3": "The small-angle grain boundary energy is lower than for a high-angle one because more atoms bond across the boundary for the small-angle, and, thus, there are fewer unsatisfied bonds."
    },
    {
        "idx": 672,
        "question": "The grain boundary energy of a small-angle grain boundary is less than for a high-angle one. Why is this so?",
        "answer": "The small-angle grain boundary energy is lower than for a high-angle one because more atoms bond across the boundary for the small-angle, and, thus, there are fewer unsatisfied bonds.",
        "type": 4,
        "wrong_answers_1": "Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.",
        "wrong_answers_2": "The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary.",
        "wrong_answers_3": "The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases."
    },
    {
        "idx": 673,
        "question": "(a) Briefly describe a twin and a twin boundary.",
        "answer": "A twin boundary is an interface such that atoms on one side are located at mirror image positions of those atoms situated on the other boundary side. The region on one side of this boundary is called a twin.",
        "type": 4,
        "wrong_answers_1": "The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary.",
        "wrong_answers_2": "One undesirable consequence of a cored structure is that, upon heating, the grain boundary regions will melt first and at a temperature below the equilibrium phase boundary from the phase diagram; this melting results in a loss in mechanical integrity of the alloy.",
        "wrong_answers_3": "The small-angle grain boundary energy is lower than for a high-angle one because more atoms bond across the boundary for the small-angle, and, thus, there are fewer unsatisfied bonds."
    },
    {
        "idx": 674,
        "question": "(b) Cite the difference between mechanical and annealing twins.",
        "answer": "Mechanical twins are produced as a result of mechanical deformation and generally occur in BCC and HCP metals. Annealing twins form during annealing heat treatments, most often in FCC metals.",
        "type": 4,
        "wrong_answers_1": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP.",
        "wrong_answers_2": "Metals are typically better thermal conductors than are ceramic materials because, for metals, most of the heat is transported by free electrons (of which there are relatively large numbers). In ceramic materials, the primary mode of thermal conduction is via phonons, and phonons are more easily scattered than are free electrons.",
        "wrong_answers_3": "the diffusion coefficient of carbon in bcc iron at 912^{\\circ} C is 1.51 × 10^{-6} cm^{2} / s, while in fcc iron it is (1.92 × 10^{-7} cm^{2} / s). the packing factor of the bcc lattice (0.68) is less than that of the fcc lattice; consequently, atoms are expected to be able to diffuse more rapidly in the bcc iron."
    },
    {
        "idx": 675,
        "question": "Briefly explain the concept of steady state as it applies to diffusion.",
        "answer": "Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species--i.e., the diffusion flux is independent of time.",
        "type": 4,
        "wrong_answers_1": "The driving force for steady-state diffusion is the concentration gradient.",
        "wrong_answers_2": "the diffusion rate of oxygen ions in \\mathrm{al}_{2}\\mathrm{o}_{3} at 1500^{\\circ}C is 3.47 × 10^{-16} {cm}^{2}/s, while the diffusion rate of aluminum ions is 2.48 × 10^{-13} {cm}^{2}/s. the difference in diffusion rates is due to the ionic radii: the oxygen ion has a larger radius (1.32 Å) compared to the aluminum ion (0.51 Å), making it easier for the smaller aluminum ion to diffuse in the ceramic.",
        "wrong_answers_3": "Self-diffusion is atomic migration in pure metals--i.e., when all atoms exchanging positions are of the same type. Interdiffusion is diffusion of atoms of one metal into another metal."
    },
    {
        "idx": 676,
        "question": "(a) Briefly explain the concept of a driving force.",
        "answer": "The driving force is that which compels a reaction to occur.",
        "type": 4,
        "wrong_answers_1": "These phenomena occur because right at the phase transition temperature, the driving force is not sufficient to cause the transformation to occur. The driving force is enhanced during superheating or supercooling.",
        "wrong_answers_2": "The driving force for steady-state diffusion is the concentration gradient.",
        "wrong_answers_3": "The driving force for the formation of spheroidite is the net reduction in ferrite-cementite phase boundary area."
    },
    {
        "idx": 677,
        "question": "(b) What is the driving force for steady-state diffusion?",
        "answer": "The driving force for steady-state diffusion is the concentration gradient.",
        "type": 4,
        "wrong_answers_1": "Increasing the concentration of oxygen in the ambient environment will increase the concentration gradient, thereby increasing the diffusion flux of oxygen.",
        "wrong_answers_2": "Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species--i.e., the diffusion flux is independent of time.",
        "wrong_answers_3": "The driving force is that which compels a reaction to occur."
    },
    {
        "idx": 678,
        "question": "For a steel alloy it has been determined that a carburizing heat treatment of 10-h duration will raise the carbon concentration to 0.45 wt% at a point 2.5mm from the surface. Estimate the time necessary to achieve the same concentration at a 5.0-mm position for an identical steel and at the same carburizing temperature.",
        "answer": "the time necessary to achieve a carbon concentration of 0.45 \\text{wt}% at a 5.0 \\text{mm} position is 40 \\text{h}.",
        "type": 1,
        "wrong_answers_1": "the time necessary to achieve a carbon concentration of 0.41 \\text{wt}% at a 5.2 \\text{mm} position is 41 \\text{h}.",
        "wrong_answers_2": "the time necessary to achieve a carbon concentration of 0.41 \\text{wt}% at a 5.5 \\text{mm} position is 44 \\text{h}.",
        "wrong_answers_3": "the time necessary to achieve a carbon concentration of 0.40 \\text{wt}% at a 5.4 \\text{mm} position is 35 \\text{h}."
    },
    {
        "idx": 679,
        "question": "The preexponential and activation energy for the diffusion of iron in cobalt are 1.1 × 10^{-5}{m}^{2} / s and 253,300 J/ mol, respectively. At what temperature will the diffusion coefficient have a value of 2.1 × 10^{-14}{m}^{2} / s ?",
        "answer": "the temperature at which the diffusion coefficient has a value of 2.1 × 10^{-14} m^{2}/s is 1518 k (1245^{\\circ} C).  alternatively, using the vmse software, the temperature is found to be 1519 k.",
        "type": 1,
        "wrong_answers_1": "the temperature at which the diffusion coefficient has a value of 2.3 × 10^{-14} m^{2}/s is 1469 k (1356^{\\circ} C).  alternatively, using the vmse software, the temperature is found to be 16510 k.",
        "wrong_answers_2": "the temperature at which the diffusion coefficient has a value of 2.4 × 10^{-14} m^{2}/s is 1439 k (1315^{\\circ} C).  alternatively, using the vmse software, the temperature is found to be 17210 k.",
        "wrong_answers_3": "the temperature at which the diffusion coefficient has a value of 2.3 × 10^{-14} m^{2}/s is 1567 k (1305^{\\circ} C).  alternatively, using the vmse software, the temperature is found to be 1318 k."
    },
    {
        "idx": 680,
        "question": "The activation energy for the diffusion of carbon in chromium is 111,000 J/ mol. Calculate the diffusion coefficient at 1100 K\\left(827^{\\circ} C\\right), given that D at 1400 K\\left(1127^{\\circ} C\\right) is 6.25 × 10^{-11}{m}^{2} / s.",
        "answer": "the diffusion coefficient at 1100k is 4.6 × 10^{-12}{m}^{2} / s.",
        "type": 1,
        "wrong_answers_1": "the diffusion coefficient at 1220k is 5.0 × 10^{-12}{m}^{2} / s.",
        "wrong_answers_2": "the diffusion coefficient at 1140k is 5.3 × 10^{-12}{m}^{2} / s.",
        "wrong_answers_3": "the diffusion coefficient at 1020k is 4.1 × 10^{-12}{m}^{2} / s."
    },
    {
        "idx": 681,
        "question": "The diffusion coefficients for iron in nickel are given at two temperatures: 1273 K with D = 9.4 x 10^-16 m^2/s and 1473 K with D = 2.4 x 10^-14 m^2/s. Determine the values of D0 and the activation energy Qd.",
        "answer": "the values are d0 = 2.2 x 10^-5 m^2/s and the activation energy qd = 252,400 j/mol.",
        "type": 1,
        "wrong_answers_1": "the values are d0 = 1.9 x 10^-5 m^2/s and the activation energy qd = 217240 j/mol.",
        "wrong_answers_2": "the values are d0 = 2.4 x 10^-5 m^2/s and the activation energy qd = 267115 j/mol.",
        "wrong_answers_3": "the values are d0 = 2.5 x 10^-5 m^2/s and the activation energy qd = 282562 j/mol."
    },
    {
        "idx": 682,
        "question": "The diffusion coefficients for iron in nickel are given at two temperatures: 1273 K with D = 9.4 x 10^-16 m^2/s and 1473 K with D = 2.4 x 10^-14 m^2/s. What is the magnitude of D at 1100°C (1373 K)?",
        "answer": "the magnitude of d at 1100°c (1373 k) is 5.4 x 10^-15 m^2/s.",
        "type": 1,
        "wrong_answers_1": "the magnitude of d at 1010°c (1313 k) is 5.8 x 10^-15 m^2/s.",
        "wrong_answers_2": "the magnitude of d at 1190°c (1473 k) is 4.6 x 10^-15 m^2/s.",
        "wrong_answers_3": "the magnitude of d at 1000°c (1173 k) is 5.8 x 10^-15 m^2/s."
    },
    {
        "idx": 683,
        "question": "The diffusion coefficients for silver in copper are given at two temperatures: T(°C) = 650, D(m2/s) = 5.5 × 10^-16 and T(°C) = 900, D(m2/s) = 1.3 × 10^-13. Determine the values of D0 and Qd.",
        "answer": "the values are d0 = 7.5 × 10^-5 m2/s and qd = 196,700 j/mol.",
        "type": 1,
        "wrong_answers_1": "the values are d0 = 6.6 × 10^-5 m2/s and qd = 173873 j/mol.",
        "wrong_answers_2": "the values are d0 = 7.2 × 10^-5 m2/s and qd = 174087 j/mol.",
        "wrong_answers_3": "the values are d0 = 6.6 × 10^-5 m2/s and qd = 203498 j/mol."
    },
    {
        "idx": 684,
        "question": "The diffusion coefficients for silver in copper are given at two temperatures: T(°C) = 650, D(m2/s) = 5.5 × 10^-16 and T(°C) = 900, D(m2/s) = 1.3 × 10^-13. What is the magnitude of D at 875°C?",
        "answer": "the magnitude of d at 875°c is 8.3 × 10^-14 m2/s.",
        "type": 1,
        "wrong_answers_1": "the magnitude of d at 835°c is 7.2 × 10^-14 m2/s.",
        "wrong_answers_2": "the magnitude of d at 762°c is 8.0 × 10^-14 m2/s.",
        "wrong_answers_3": "the magnitude of d at 945°c is 9.3 × 10^-14 m2/s."
    },
    {
        "idx": 685,
        "question": "Carbon is allowed to diffuse through a steel plate 15mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30kg C / m^{2} Fe, which are maintained constant. If the preexponential and activation energy are 6.2 × 10^{-7}{m}^{2} / s and 80,000 J/ mol, respectively, compute the temperature at which the diffusion flux is 1.43 × 10^{-9}kg / m^{2}-s.",
        "answer": "the temperature at which the diffusion flux is 1.43 × 10^{-9}kg / m^{2}-s is 1044k or 771^{\\circ} C.",
        "type": 1,
        "wrong_answers_1": "the temperature at which the diffusion flux is 1.64 × 10^{-9}kg / m^{2}-s is 1004k or 679^{\\circ} C.",
        "wrong_answers_2": "the temperature at which the diffusion flux is 1.47 × 10^{-9}kg / m^{2}-s is 1184k or 859^{\\circ} C.",
        "wrong_answers_3": "the temperature at which the diffusion flux is 1.56 × 10^{-9}kg / m^{2}-s is 1134k or 880^{\\circ} C."
    },
    {
        "idx": 686,
        "question": "The steady-state diffusion flux through a metal plate is 5.4 × 10^{-10}kg / m^{2}-s at a temperature of 727^{\\circ} C (1000 K) and when the concentration gradient is -350kg / m^{4}. Calculate the diffusion flux at 1027^{\\circ} C(1300 K) for the same concentration gradient and assuming an activation energy for diffusion of 125,000 J/ mol.",
        "answer": "the diffusion flux at 1027^{\\circ} C (1300 k) is 1.74 × 10^{-8} \\mathrm{kg/m^{2}-s}.",
        "type": 1,
        "wrong_answers_1": "the diffusion flux at 1088^{\\circ} C (1460 k) is 1.62 × 10^{-8} \\mathrm{kg/m^{2}-s}.",
        "wrong_answers_2": "the diffusion flux at 947^{\\circ} C (1410 k) is 1.99 × 10^{-8} \\mathrm{kg/m^{2}-s}.",
        "wrong_answers_3": "the diffusion flux at 1056^{\\circ} C (1480 k) is 1.49 × 10^{-8} \\mathrm{kg/m^{2}-s}."
    },
    {
        "idx": 687,
        "question": "For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 x 10^6 psi). What is the maximum load that may be applied to a specimen with a cross-sectional area of 325 mm^2 (0.5 in.^2) without plastic deformation?",
        "answer": "the maximum load that may be applied without plastic deformation is 89,375 n (20,000 lb_f).",
        "type": 1,
        "wrong_answers_1": "the maximum load that may be applied without plastic deformation is 97546 n (21758 lb_f).",
        "wrong_answers_2": "the maximum load that may be applied without plastic deformation is 77819 n (22782 lb_f).",
        "wrong_answers_3": "the maximum load that may be applied without plastic deformation is 98095 n (18788 lb_f)."
    },
    {
        "idx": 688,
        "question": "For a bronze alloy, the stress at which plastic deformation begins is 275 MPa (40,000 psi), and the modulus of elasticity is 115 GPa (16.7 x 10^6 psi). If the original specimen length is 115mm (4.5 in.), what is the maximum length to which it may be stretched without causing plastic deformation?",
        "answer": "the maximum length to which the specimen may be stretched without causing plastic deformation is 115.28mm (4.51 in).",
        "type": 1,
        "wrong_answers_1": "the maximum length to which the specimen may be stretched without causing plastic deformation is 122.22mm (3.84 in).",
        "wrong_answers_2": "the maximum length to which the specimen may be stretched without causing plastic deformation is 111.65mm (3.88 in).",
        "wrong_answers_3": "the maximum length to which the specimen may be stretched without causing plastic deformation is 122.94mm (4.75 in)."
    },
    {
        "idx": 689,
        "question": "A cylindrical rod of copper (E=110g P a, 16 × 10^{6} psi) having a yield strength of 240 MPa(35,000 psi) is to be subjected to a load of 6660N\\left(1500 lb p\\right). If the length of the rod is 380 mm(15.0 in.), what must be the diameter to allow an elongation of 0.50 mm(0.020 in.) ?",
        "answer": "the diameter must be 7.65mm (0.30 in).",
        "type": 1,
        "wrong_answers_1": "the diameter must be 6.64mm (0.34 in).",
        "wrong_answers_2": "the diameter must be 8.34mm (0.27 in).",
        "wrong_answers_3": "the diameter must be 8.28mm (0.26 in)."
    },
    {
        "idx": 690,
        "question": "Compute the elastic modulus for titanium, whose stress-strain behavior may be observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for the same metal?",
        "answer": "(a) the elastic modulus for titanium is 106.4 gpa, which is in very good agreement with the value of 107 gpa from table 6.1.",
        "type": 2
    },
    {
        "idx": 691,
        "question": "Compute the elastic modulus for tempered steel, whose stress-strain behavior may be observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for the same metal?",
        "answer": "(b) the elastic modulus for tempered steel is 204.7 gpa, which is in reasonably good agreement with the value of 207 gpa from table 6.1.",
        "type": 2
    },
    {
        "idx": 692,
        "question": "Compute the elastic modulus for aluminum, whose stress-strain behavior may be observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for the same metal?",
        "answer": "(c) the elastic modulus for aluminum is 69.1 gpa, which is in excellent agreement with the value of 69 gpa from table 6.1.",
        "type": 2
    },
    {
        "idx": 693,
        "question": "Compute the elastic modulus for carbon steel, whose stress-strain behavior may be observed in the 'Tensile Tests' module of Virtual Materials Science and Engineering (VMSE). How does this value compare with that presented in Table 6.1 for the same metal?",
        "answer": "(d) the elastic modulus for carbon steel is 215 gpa, which is in reasonable agreement with the value of 207 gpa from table 6.1.",
        "type": 2
    },
    {
        "idx": 694,
        "question": "Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 8.0 mm(0.31 in.). A tensile force of 1000N\\left(225 lb_{\\mathrm{f}}\\right) produces an elastic reduction in diameter of 2.8 × 10^{-4} mm\\left(1.10 × 10^{-5} in\\right..). Compute the modulus of elasticity for this alloy, given that Poisson's ratio is 0.30 .",
        "answer": "the modulus of elasticity for this alloy is 1.705 × 10^{11} pa or 170.5 gpa.",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity for this alloy is 1.593 × 10^{11} pa or 191.5 gpa.",
        "wrong_answers_2": "the modulus of elasticity for this alloy is 1.942 × 10^{11} pa or 176.6 gpa.",
        "wrong_answers_3": "the modulus of elasticity for this alloy is 1.633 × 10^{11} pa or 190.9 gpa."
    },
    {
        "idx": 695,
        "question": "A cylindrical rod 100mm long and having a diameter of 10.0mm is to be deformed using a tensile load of 27,500 N. It must not experience plastic deformation. Of the materials listed as follows, which are possible candidates? Justify your choice(s).",
        "answer": "Of the alloys listed, the steel alloy is a possible candidate because it meets the criterion of not experiencing plastic deformation.",
        "type": 4,
        "wrong_answers_1": "Of the alloys listed, the steel alloy is a possible candidate because it meets the criterion regarding diameter reduction. The titanium alloy is not a candidate as it fails this criterion.",
        "wrong_answers_2": "Aluminum alloy is not a candidate because its elongation of 1.64mm exceeds the limit.",
        "wrong_answers_3": "The first criterion is consequences of failure."
    },
    {
        "idx": 696,
        "question": "A cylindrical rod 100mm long and having a diameter of 10.0mm is to be deformed using a tensile load of 27,500 N. It must not experience a diameter reduction of more than 7.5 x 10^-3 mm. Of the materials listed as follows, which are possible candidates? Justify your choice(s).",
        "answer": "Of the alloys listed, the steel alloy is a possible candidate because it meets the criterion regarding diameter reduction. The titanium alloy is not a candidate as it fails this criterion.",
        "type": 4,
        "wrong_answers_1": "Of the alloys listed, the steel alloy is a possible candidate because it meets the criterion of not experiencing plastic deformation.",
        "wrong_answers_2": "Aluminum alloy is not a candidate because its elongation of 1.64mm exceeds the limit.",
        "wrong_answers_3": "The fourth criterion is economics."
    },
    {
        "idx": 697,
        "question": "A cylindrical metal specimen having an original diameter of 12.8mm (0.505 in.) and gauge length of 50.80mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.60mm (0.260 in.), and the fractured gauge length is 72.14mm (2.840 in.). Calculate the ductility in terms of percent reduction in area.",
        "answer": "the ductility in terms of percent reduction in area is 73.4%.",
        "type": 1,
        "wrong_answers_1": "the ductility in terms of percent reduction in area is 78.7%.",
        "wrong_answers_2": "the ductility in terms of percent reduction in area is 80.3%.",
        "wrong_answers_3": "the ductility in terms of percent reduction in area is 66.6%."
    },
    {
        "idx": 698,
        "question": "A cylindrical metal specimen having an original diameter of 12.8mm (0.505 in.) and gauge length of 50.80mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 6.60mm (0.260 in.), and the fractured gauge length is 72.14mm (2.840 in.). Calculate the ductility in terms of percent elongation.",
        "answer": "the ductility in terms of percent elongation is 42%.",
        "type": 1,
        "wrong_answers_1": "the ductility in terms of percent elongation is 38%.",
        "wrong_answers_2": "the ductility in terms of percent elongation is 47%.",
        "wrong_answers_3": "the ductility in terms of percent elongation is 44%."
    },
    {
        "idx": 699,
        "question": "For some metal alloy, a true stress of 415 MPa(60,175 psi) produces a plastic true strain of 0.475 . How much will a specimen of this material elongate when a true stress of 325 MPa(46,125 psi) is applied if the original length is 300mm (11.8 in.)? Assume a value of 0.25 for the strain-hardening exponent n.",
        "answer": "the specimen will elongate by 58.8 mm (2.31 in).",
        "type": 1,
        "wrong_answers_1": "the specimen will elongate by 67.3 mm (2.18 in).",
        "wrong_answers_2": "the specimen will elongate by 56.8 mm (2.06 in).",
        "wrong_answers_3": "the specimen will elongate by 67.2 mm (2.62 in)."
    },
    {
        "idx": 700,
        "question": "Cite five factors that lead to scatter in measured material properties.",
        "answer": "The five factors that lead to scatter in measured material properties are the following: (1) test method; (2) variation in specimen fabrication procedure; (3) operator bias; (4) apparatus calibration; and (5) material inhomogeneities and/or compositional differences.",
        "type": 4,
        "wrong_answers_1": "Five factors that lead to scatter in fatigue life data are (1) specimen fabrication and surface preparation, (2) metallurgical variables, (3) specimen alignment in the test apparatus, (4) variation in mean stress, and (5) variation in test cycle frequency.",
        "wrong_answers_2": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material.",
        "wrong_answers_3": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material."
    },
    {
        "idx": 701,
        "question": "To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 10^4 mm^-2. Suppose that all the dislocations in 1000 mm^3 (1 cm^3) were somehow removed and linked end to end. How far (in miles) would this chain extend?",
        "answer": "for a dislocation density of 10^4 mm^-2, the chain would extend 6.2 miles.",
        "type": 1,
        "wrong_answers_1": "for a dislocation density of 11^4 mm^-2, the chain would extend 5.4 miles.",
        "wrong_answers_2": "for a dislocation density of 11^4 mm^-2, the chain would extend 6.0 miles.",
        "wrong_answers_3": "for a dislocation density of 11^4 mm^-2, the chain would extend 5.6 miles."
    },
    {
        "idx": 702,
        "question": "Now suppose that the density is increased to 10^10 mm^-2 by cold working. What would be the chain length of dislocations in 1000 mm^3 of material?",
        "answer": "for a dislocation density of 10^10 mm^-2, the chain would extend 6.2 x 10^6 miles.",
        "type": 1,
        "wrong_answers_1": "for a dislocation density of 10^11 mm^-2, the chain would extend 6.9 x 10^6 miles.",
        "wrong_answers_2": "for a dislocation density of 9^9 mm^-2, the chain would extend 6.6 x 10^6 miles.",
        "wrong_answers_3": "for a dislocation density of 10^11 mm^-2, the chain would extend 7.0 x 10^6 miles."
    },
    {
        "idx": 703,
        "question": "For edge dislocation, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.",
        "answer": "edge dislocation--parallel",
        "type": 4,
        "wrong_answers_1": "The Burgers vector and dislocation line are perpendicular for edge dislocations, parallel for screw dislocations, and neither perpendicular nor parallel for mixed dislocations.",
        "wrong_answers_2": "mixed dislocation--neither parallel nor perpendicular",
        "wrong_answers_3": "screw dislocation--perpendicular"
    },
    {
        "idx": 704,
        "question": "For screw dislocation, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.",
        "answer": "screw dislocation--perpendicular",
        "type": 4,
        "wrong_answers_1": "The Burgers vector and dislocation line are perpendicular for edge dislocations, parallel for screw dislocations, and neither perpendicular nor parallel for mixed dislocations.",
        "wrong_answers_2": "mixed dislocation--neither parallel nor perpendicular",
        "wrong_answers_3": "Decrease in strength perpendicular to the direction of drawing."
    },
    {
        "idx": 705,
        "question": "For mixed dislocation, cite the relationship between the direction of the applied shear stress and the direction of dislocation line motion.",
        "answer": "mixed dislocation--neither parallel nor perpendicular",
        "type": 4,
        "wrong_answers_1": "The Burgers vector and dislocation line are perpendicular for edge dislocations, parallel for screw dislocations, and neither perpendicular nor parallel for mixed dislocations.",
        "wrong_answers_2": "screw dislocation--perpendicular",
        "wrong_answers_3": "edge dislocation--parallel"
    },
    {
        "idx": 706,
        "question": "Define a slip system.",
        "answer": "A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation motion (or slip) occurs.",
        "type": 4,
        "wrong_answers_1": "All metals do not have the same slip system. The reason for this is that for most metals, the slip system will consist of the most densely packed crystallographic plane, and within that plane the most closely packed direction. This plane and direction will vary from crystal structure to crystal structure.",
        "wrong_answers_2": "For slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples. Slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems.",
        "wrong_answers_3": "These composites are constructed in order to have a relatively high strength in virtually all directions within the plane of the laminate."
    },
    {
        "idx": 707,
        "question": "Do all metals have the same slip system? Why or why not?",
        "answer": "All metals do not have the same slip system. The reason for this is that for most metals, the slip system will consist of the most densely packed crystallographic plane, and within that plane the most closely packed direction. This plane and direction will vary from crystal structure to crystal structure.",
        "type": 4,
        "wrong_answers_1": "A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation motion (or slip) occurs.",
        "wrong_answers_2": "The stacking sequence of close-packed planes of anions for the zinc blende crystal structure will be the same as FCC (and not HCP) because the anion packing is FCC.",
        "wrong_answers_3": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP."
    },
    {
        "idx": 708,
        "question": "Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 43.1^{\\circ} and 47.9^{\\circ}, respectively, with the tensile axis. If the critical resolved shear stress is 20.7 MPa (3000 psi), will an applied stress of 45 MPa(6500 psi) cause the single crystal to yield? If not, what stress will be necessary?",
        "answer": "the resolved shear stress is 22.0 MPa (3181 psi). since this is greater than the critical resolved shear stress of 20.7 MPa (3000 psi), the single crystal will yield.",
        "type": 1,
        "wrong_answers_1": "the resolved shear stress is 24.3 MPa (2891 psi). since this is greater than the critical resolved shear stress of 18.1 MPa (3120 psi), the single crystal will yield.",
        "wrong_answers_2": "the resolved shear stress is 24.1 MPa (2741 psi). since this is greater than the critical resolved shear stress of 21.4 MPa (2840 psi), the single crystal will yield.",
        "wrong_answers_3": "the resolved shear stress is 20.8 MPa (3281 psi). since this is greater than the critical resolved shear stress of 17.8 MPa (2580 psi), the single crystal will yield."
    },
    {
        "idx": 709,
        "question": "A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes an angle of 28.1 degrees with the tensile axis. Three possible slip directions make angles of 62.4 degrees, 72.0 degrees, and 81.1 degrees with the same tensile axis. Which of these three slip directions is most favored?",
        "answer": "the most favored slip direction is at an angle of 62.4 degrees with the tensile axis.",
        "type": 2
    },
    {
        "idx": 710,
        "question": "A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes an angle of 28.1 degrees with the tensile axis. The most favored slip direction makes an angle of 62.4 degrees with the tensile axis. If plastic deformation begins at a tensile stress of 1.95 MPa (280 psi), determine the critical resolved shear stress for aluminum.",
        "answer": "the critical resolved shear stress for aluminum is 0.80 mpa (114 psi).",
        "type": 1,
        "wrong_answers_1": "the critical resolved shear stress for aluminum is 0.68 mpa (105 psi).",
        "wrong_answers_2": "the critical resolved shear stress for aluminum is 0.85 mpa (102 psi).",
        "wrong_answers_3": "the critical resolved shear stress for aluminum is 0.84 mpa (130 psi)."
    },
    {
        "idx": 711,
        "question": "The critical resolved shear stress for iron is 27 MPa(4000 psi). Determine the maximum possible yield strength for a single crystal of \\mathrm{Fe} pulled in tension.",
        "answer": "the maximum possible yield strength for a single crystal of \\mathrm{fe} pulled in tension is 54 MPa (8000 psi).",
        "type": 1,
        "wrong_answers_1": "the maximum possible yield strength for a single crystal of \\mathrm{fe} pulled in tension is 47 MPa (8720 psi).",
        "wrong_answers_2": "the maximum possible yield strength for a single crystal of \\mathrm{fe} pulled in tension is 51 MPa (7610 psi).",
        "wrong_answers_3": "the maximum possible yield strength for a single crystal of \\mathrm{fe} pulled in tension is 47 MPa (7350 psi)."
    },
    {
        "idx": 712,
        "question": "What is the difference between deformation by twinning and deformation by slip relative to mechanism?",
        "answer": "With slip deformation there is no crystallographic reorientation, whereas with twinning there is a reorientation.",
        "type": 4,
        "wrong_answers_1": "For slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples. Slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems.",
        "wrong_answers_2": "Normally slip results in relatively large deformations, whereas only small deformations result for twinning.",
        "wrong_answers_3": "A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation motion (or slip) occurs."
    },
    {
        "idx": 713,
        "question": "What is the difference between deformation by twinning and deformation by slip relative to conditions of occurrence?",
        "answer": "For slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples. Slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems.",
        "type": 4,
        "wrong_answers_1": "Normally slip results in relatively large deformations, whereas only small deformations result for twinning.",
        "wrong_answers_2": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP.",
        "wrong_answers_3": "A slip system is a crystallographic plane, and, within that plane, a direction along which dislocation motion (or slip) occurs."
    },
    {
        "idx": 714,
        "question": "What is the difference between deformation by twinning and deformation by slip relative to final result?",
        "answer": "Normally slip results in relatively large deformations, whereas only small deformations result for twinning.",
        "type": 4,
        "wrong_answers_1": "For slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples. Slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems.",
        "wrong_answers_2": "The advantages of hot working are: (1) Large deformations are possible, which may be repeated. (2) Deformation energy requirements are relatively low.",
        "wrong_answers_3": "The disadvantages of cold working are: (1) High deformation energy requirements. (2) Large deformations must be accomplished in steps, which may be expensive. (3) A loss of ductility."
    },
    {
        "idx": 715,
        "question": "Briefly explain why small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries.",
        "answer": "Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction.",
        "type": 4,
        "wrong_answers_1": "The small-angle grain boundary energy is lower than for a high-angle one because more atoms bond across the boundary for the small-angle, and, thus, there are fewer unsatisfied bonds.",
        "wrong_answers_2": "The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases.",
        "wrong_answers_3": "The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary."
    },
    {
        "idx": 716,
        "question": "Briefly explain why HCP metals are typically more brittle than FCC and BCC metals.",
        "answer": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP.",
        "type": 4,
        "wrong_answers_1": "Crystalline ceramics are harder yet more brittle than metals because they (ceramics) have fewer slip systems, and, therefore, dislocation motion is highly restricted.",
        "wrong_answers_2": "Mechanical twins are produced as a result of mechanical deformation and generally occur in BCC and HCP metals. Annealing twins form during annealing heat treatments, most often in FCC metals.",
        "wrong_answers_3": "The stacking sequence of close-packed planes of anions for the zinc blende crystal structure will be the same as FCC (and not HCP) because the anion packing is FCC."
    },
    {
        "idx": 717,
        "question": "Two previously undeformed cylindrical specimens of an alloy are to be strain hardened by reducing their cross-sectional areas (while maintaining their circular cross sections). For one specimen, the initial and deformed radii are 16mm and 11 mm, respectively. The second specimen, with an initial radius of 12 mm, must have the same deformed hardness as the first specimen; compute the second specimen's radius after deformation.",
        "answer": "the second specimen's radius after deformation is 8.25 mm.",
        "type": 1,
        "wrong_answers_1": "the second specimen's radius after deformation is 9.45 mm.",
        "wrong_answers_2": "the second specimen's radius after deformation is 9.25 mm.",
        "wrong_answers_3": "the second specimen's radius after deformation is 8.96 mm."
    },
    {
        "idx": 718,
        "question": "Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as such. Their original and deformed dimensions are as follows:\n\\begin{tabular}{ccc}\n\\hline & Circular (diameter, \\mathbf{m m} ) & Rectangular (\\mathbf{m m}) \\\\\n\\hline Original dimensions & 15.2 & 125 × 175 \\\\\nDeformed dimensions & 11.4 & 75 × 200 \\\\\n\\hline\n\\end{tabular}\nWhich of these specimens will be the hardest after plastic deformation, and why?",
        "answer": "the hardest specimen will be the deformed circular one with a cold work percentage of 43.8%.",
        "type": 2
    },
    {
        "idx": 719,
        "question": "A cylindrical specimen of cold-worked copper has a ductility (%EL) of 25%. If its cold-worked radius is 10mm (0.40 in.), what was its radius before deformation?",
        "answer": "the radius before deformation was 10.6mm (0.424 in.).",
        "type": 1,
        "wrong_answers_1": "the radius before deformation was 11.6mm (0.470 in.).",
        "wrong_answers_2": "the radius before deformation was 9.4mm (0.468 in.).",
        "wrong_answers_3": "the radius before deformation was 11.1mm (0.447 in.)."
    },
    {
        "idx": 720,
        "question": "Explain the differences in grain structure for a metal that has been cold worked and one that has been cold worked and then recrystallized.",
        "answer": "During cold-working, the grain structure of the metal has been distorted to accommodate the deformation. Recrystallization produces grains that are equiaxed and smaller than the parent grains.",
        "type": 4,
        "wrong_answers_1": "Although each individual grain in a polycrystalline material may be anisotropic, if the grains have random orientations, then the solid aggregate of the many anisotropic grains will behave isotropically.",
        "wrong_answers_2": "For recovery, there is some relief of internal strain energy by dislocation motion; however, there are virtually no changes in either the grain structure or mechanical characteristics. During recrystallization, on the other hand, a new set of strain-free grains forms, and the material becomes softer and more ductile.",
        "wrong_answers_3": "Coring is the phenomenon whereby concentration gradients exist across grains in polycrystalline alloys, with higher concentrations of the component having the lower melting temperature at the grain boundaries. It occurs, during solidification, as a consequence of cooling rates that are too rapid to allow for the maintenance of the equilibrium composition of the solid phase."
    },
    {
        "idx": 721,
        "question": "What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5 × 10^{-4} mm\\left(10^{-5} in.\\right) and a crack length of 2.5 × 10^{-2} mm\\left(10^{-3} in.\\right) when a tensile stress of 170 MPa(25,000 psi) is applied?",
        "answer": "the magnitude of the maximum stress at the tip of the internal crack is 2404 \\text{ mpa} (354,000 psi).",
        "type": 1,
        "wrong_answers_1": "the magnitude of the maximum stress at the tip of the internal crack is 2734 \\text{ mpa} (306321 psi).",
        "wrong_answers_2": "the magnitude of the maximum stress at the tip of the internal crack is 2074 \\text{ mpa} (384478 psi).",
        "wrong_answers_3": "the magnitude of the maximum stress at the tip of the internal crack is 2734 \\text{ mpa} (304383 psi)."
    },
    {
        "idx": 722,
        "question": "Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.25 mm(0.01 in.) and having a tip radius of curvature of 1.2 × 10^{-3} mm\\left(4.7 × 10^{-5} in\\right..) when a stress of 1200 MPa(174,000 psi) is applied.",
        "answer": "the theoretical fracture strength is 3.5 × 10^{4} MPa (5.1 × 10^{6} psi).",
        "type": 1,
        "wrong_answers_1": "the theoretical fracture strength is 4.0 × 10^{4} MPa (5.6 × 10^{6} psi).",
        "wrong_answers_2": "the theoretical fracture strength is 3.1 × 10^{4} MPa (5.5 × 10^{6} psi).",
        "wrong_answers_3": "the theoretical fracture strength is 3.8 × 10^{4} MPa (4.5 × 10^{6} psi)."
    },
    {
        "idx": 723,
        "question": "If the specific surface energy for soda-lime glass is 0.30 J/ m^{2}, using data contained in Table 12.5, compute the critical stress required for the propagation of a surface crack of length 0.05 mm.",
        "answer": "the critical stress required for the propagation of a surface crack of length 0.05 mm is 16.2 MPa.",
        "type": 1,
        "wrong_answers_1": "the critical stress required for the propagation of a surface crack of length 0.04 mm is 18.6 MPa.",
        "wrong_answers_2": "the critical stress required for the propagation of a surface crack of length 0.05 mm is 17.8 MPa.",
        "wrong_answers_3": "the critical stress required for the propagation of a surface crack of length 0.05 mm is 17.6 MPa."
    },
    {
        "idx": 724,
        "question": "A polystyrene component must not fail when a tensile stress of 1.25 MPa(180 psi) is applied. Determine the maximum allowable surface crack length if the surface energy of polystyrene is 0.50 J/ m^{2}\\left(2.86 × 10^{-3}\\right. in.-\\left.lb_{\\mathrm{f}} / in^{2}\\right). Assume a modulus of elasticity of 3.0 \\mathrm{GPa}\\left(0.435 × 10^{6} psi\\right).",
        "answer": "the maximum allowable surface crack length is 6.1 × 10^{-4} \\, m (0.61mm or 0.024 in.).",
        "type": 1,
        "wrong_answers_1": "the maximum allowable surface crack length is 6.8 × 10^{-4} \\, m (0.69mm or 0.026 in.).",
        "wrong_answers_2": "the maximum allowable surface crack length is 6.6 × 10^{-4} \\, m (0.52mm or 0.027 in.).",
        "wrong_answers_3": "the maximum allowable surface crack length is 5.8 × 10^{-4} \\, m (0.59mm or 0.022 in.)."
    },
    {
        "idx": 725,
        "question": "A specimen of a 4340 steel alloy having a plane strain fracture toughness of 45 MPa \\sqrt{m}(41 ksi \\sqrt{in}.) is exposed to a stress of 1000 MPa(145,000 psi). Will this specimen experience fracture if it is known that the largest surface crack is 0.75 mm(0.03 in.) long? Why or why not? Assume that the parameter Y has a value of 1.0 .",
        "answer": "the specimen will experience fracture because it can only tolerate a stress of 927 mpa (133,500 \\text{psi}) before fracture, which is less than the applied stress of 1000 mpa (145,000 \\text{psi}).",
        "type": 1,
        "wrong_answers_1": "the specimen will experience fracture because it can only tolerate a stress of 972 mpa (139976 \\text{psi}) before fracture, which is less than the applied stress of 910 mpa (149819 \\text{psi}).",
        "wrong_answers_2": "the specimen will experience fracture because it can only tolerate a stress of 859 mpa (113522 \\text{psi}) before fracture, which is less than the applied stress of 900 mpa (151614 \\text{psi}).",
        "wrong_answers_3": "the specimen will experience fracture because it can only tolerate a stress of 855 mpa (142022 \\text{psi}) before fracture, which is less than the applied stress of 920 mpa (152506 \\text{psi})."
    },
    {
        "idx": 726,
        "question": "Some aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 35 MPa \\sqrt{m} (31.9 ksi \\sqrt{m}. It has been determined that fracture results at a stress of 250 MPa (36,250 psi) when the maximum (or critical) internal crack length is 2.0 mm(0.08 in.). For this same component and alloy, will fracture occur at a stress level of 325 MPa(47,125 psi) when the maximum internal crack length is 1.0 mm(0.04 in).? Why or why not?",
        "answer": "fracture will not occur since the value 32.2 MPa \\sqrt{m} is less than the fracture toughness k_{ic} of the material, 35 MPa \\sqrt{m}.",
        "type": 1,
        "wrong_answers_1": "fracture will not occur since the value 31.2 MPa \\sqrt{m} is less than the fracture toughness k_{ic} of the material, 39 MPa \\sqrt{m}.",
        "wrong_answers_2": "fracture will not occur since the value 35.7 MPa \\sqrt{m} is less than the fracture toughness k_{ic} of the material, 37 MPa \\sqrt{m}.",
        "wrong_answers_3": "fracture will not occur since the value 36.7 MPa \\sqrt{m} is less than the fracture toughness k_{ic} of the material, 38 MPa \\sqrt{m}."
    },
    {
        "idx": 727,
        "question": "Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPa \\sqrt{m} (36.4 k \\left.\\mathrm{si} \\sqrt{m}\\right). It has been determined that fracture results at a stress of 365 MPa(53,000 psi) when the maximum internal crack length is 2.5 mm(0.10 in.). For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 4.0 mm(0.16 in.).",
        "answer": "the stress level at which fracture will occur for a critical internal crack length of 4.0mm is 288 MPa (41,500 psi).",
        "type": 1,
        "wrong_answers_1": "the stress level at which fracture will occur for a critical internal crack length of 3.4mm is 315 MPa (46057 psi).",
        "wrong_answers_2": "the stress level at which fracture will occur for a critical internal crack length of 3.4mm is 317 MPa (45792 psi).",
        "wrong_answers_3": "the stress level at which fracture will occur for a critical internal crack length of 3.7mm is 276 MPa (45330 psi)."
    },
    {
        "idx": 728,
        "question": "A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 55 MPa \\sqrt{m}(50 ksi \\sqrt{in}.). If, during service use, the plate is exposed to a tensile stress of 200 MPa(29,000 psi), determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.",
        "answer": "the minimum length of a surface crack that will lead to fracture is 0.024 \\, m (24 mm).",
        "type": 1,
        "wrong_answers_1": "the minimum length of a surface crack that will lead to fracture is 0.027 \\, m (26 mm).",
        "wrong_answers_2": "the minimum length of a surface crack that will lead to fracture is 0.026 \\, m (25 mm).",
        "wrong_answers_3": "the minimum length of a surface crack that will lead to fracture is 0.021 \\, m (27 mm)."
    },
    {
        "idx": 729,
        "question": "Calculate the maximum internal crack length allowable for a 7075-T651 aluminum alloy (Table 8.1) component that is loaded to a stress one half of its yield strength. Assume that the value of Y is 1.35 .",
        "answer": "the maximum internal crack length allowable is 0.0033 \\, m (3.3 mm).",
        "type": 1,
        "wrong_answers_1": "the maximum internal crack length allowable is 0.0034 \\, m (3.7 mm).",
        "wrong_answers_2": "the maximum internal crack length allowable is 0.0037 \\, m (3.0 mm).",
        "wrong_answers_3": "the maximum internal crack length allowable is 0.0038 \\, m (3.0 mm)."
    },
    {
        "idx": 730,
        "question": "A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 77.0 MPa \\sqrt{m}(70.1 ksi \\sqrt{m}.) and a yield strength of 1400 MPa(205,000 psi). The flaw size resolution limit of the flaw detection apparatus is 4.0 mm(0.16 in.) If the design stress is one half of the yield strength and the value of Y is 1.0 , determine whether or not a critical flaw for this plate is subject to detection.",
        "answer": "the critical flaw size a_c is 3.9 mm. therefore, the critical flaw is not subject to detection since it is less than the 4.0 mm resolution limit.",
        "type": 1,
        "wrong_answers_1": "the critical flaw size a_c is 4.3 mm. therefore, the critical flaw is not subject to detection since it is less than the 4.5 mm resolution limit.",
        "wrong_answers_2": "the critical flaw size a_c is 3.6 mm. therefore, the critical flaw is not subject to detection since it is less than the 4.1 mm resolution limit.",
        "wrong_answers_3": "the critical flaw size a_c is 3.4 mm. therefore, the critical flaw is not subject to detection since it is less than the 4.2 mm resolution limit."
    },
    {
        "idx": 731,
        "question": "Cite five factors that may lead to scatter in fatigue life data.",
        "answer": "Five factors that lead to scatter in fatigue life data are (1) specimen fabrication and surface preparation, (2) metallurgical variables, (3) specimen alignment in the test apparatus, (4) variation in mean stress, and (5) variation in test cycle frequency.",
        "type": 4,
        "wrong_answers_1": "The five factors that lead to scatter in measured material properties are the following: (1) test method; (2) variation in specimen fabrication procedure; (3) operator bias; (4) apparatus calibration; and (5) material inhomogeneities and/or compositional differences.",
        "wrong_answers_2": "For thermoplastic polymers, five factors that favor brittle fracture are as follows: (1) a reduction in temperature, (2) an increase in strain rate, (3) the presence of a sharp notch, (4) increased specimen thickness, and (5) modifications of the polymer structure.",
        "wrong_answers_3": "With regard to origin, beachmarks result from interruptions in the stress cycles; each fatigue striation corresponds to the advance of a fatigue crack during a single load cycle."
    },
    {
        "idx": 732,
        "question": "Briefly explain the difference between fatigue striations and beachmarks in terms of size.",
        "answer": "With regard to size, beachmarks are normally of macroscopic dimensions and may be observed with the naked eye; fatigue striations are of microscopic size and it is necessary to observe them using electron microscopy.",
        "type": 4,
        "wrong_answers_1": "With regard to origin, beachmarks result from interruptions in the stress cycles; each fatigue striation corresponds to the advance of a fatigue crack during a single load cycle.",
        "wrong_answers_2": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probably of the existence of a flaw of that is capable of initiating a crack diminishes.",
        "wrong_answers_3": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probability of the existence of a flaw that is capable of initiating a crack diminishes."
    },
    {
        "idx": 733,
        "question": "Briefly explain the difference between fatigue striations and beachmarks in terms of origin.",
        "answer": "With regard to origin, beachmarks result from interruptions in the stress cycles; each fatigue striation corresponds to the advance of a fatigue crack during a single load cycle.",
        "type": 4,
        "wrong_answers_1": "With regard to size, beachmarks are normally of macroscopic dimensions and may be observed with the naked eye; fatigue striations are of microscopic size and it is necessary to observe them using electron microscopy.",
        "wrong_answers_2": "Five factors that lead to scatter in fatigue life data are (1) specimen fabrication and surface preparation, (2) metallurgical variables, (3) specimen alignment in the test apparatus, (4) variation in mean stress, and (5) variation in test cycle frequency.",
        "wrong_answers_3": "A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time."
    },
    {
        "idx": 734,
        "question": "List four measures that may be taken to increase the resistance to fatigue of a metal alloy.",
        "answer": "Four measures that may be taken to increase the fatigue resistance of a metal alloy are:\n(1) Polish the surface to remove stress amplification sites.\n(2) Reduce the number of internal defects (pores, etc.) by means of altering processing and fabrication techniques.\n(3) Modify the design to eliminate notches and sudden contour changes.\n(4) Harden the outer surface of the structure by case hardening (carburizing, nitriding) or shot peening.",
        "type": 4,
        "wrong_answers_1": "Density will increase with degree of vitrification since the total remaining pore volume decreases. Strength will also increase with degree of vitrification inasmuch as more of the liquid phase forms, which fills in a greater fraction of pore volume. Upon cooling, the liquid forms a glass matrix of relatively high strength. Corrosion resistance normally increases also, especially at service temperatures below that at which the glass phase begins to soften. The rate of corrosion is dependent on the amount of surface area exposed to the corrosive medium; hence, decreasing the total surface area by filling in some of the surface pores, diminishes the corrosion rate. Thermal conductivity will increase with degree of vitrification. The glass phase has a higher conductivity than the pores that it has filled.",
        "wrong_answers_2": "Thermal stresses may be introduced into a structure by rapid heating or cooling because temperature gradients are established across the cross section due to more rapid temperature changes at the surface than within the interior; thus, the surface will expand or contract at a different rate than the interior and since this surface expansion or contraction will be restrained by the interior, stresses will be introduced.",
        "wrong_answers_3": "Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material."
    },
    {
        "idx": 735,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for nickel.",
        "answer": "for ni: 418 degrees c (785 degrees f)",
        "type": 1,
        "wrong_answers_1": "for ni: 390 degrees c (695 degrees f)",
        "wrong_answers_2": "for ni: 468 degrees c (819 degrees f)",
        "wrong_answers_3": "for ni: 387 degrees c (828 degrees f)"
    },
    {
        "idx": 736,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for copper.",
        "answer": "for cu: 270 degrees c (518 degrees f)",
        "type": 4,
        "wrong_answers_1": "for w: 1200 degrees c (2190 degrees f)",
        "wrong_answers_2": "for al: 100 degrees c (212 degrees f)",
        "wrong_answers_3": "for fe: 450 degrees c (845 degrees f)"
    },
    {
        "idx": 737,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for iron.",
        "answer": "for fe: 450 degrees c (845 degrees f)",
        "type": 4,
        "wrong_answers_1": "for fe, 451°c or 845°f.",
        "wrong_answers_2": "for w: 1200 degrees c (2190 degrees f)",
        "wrong_answers_3": "for al: 100 degrees c (212 degrees f)"
    },
    {
        "idx": 738,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for tungsten.",
        "answer": "for w: 1200 degrees c (2190 degrees f)",
        "type": 4,
        "wrong_answers_1": "for al: 100 degrees c (212 degrees f)",
        "wrong_answers_2": "for cu: 270 degrees c (518 degrees f)",
        "wrong_answers_3": "for fe: 450 degrees c (845 degrees f)"
    },
    {
        "idx": 739,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for lead.",
        "answer": "for pb: -33 degrees c (-27 degrees f)",
        "type": 1,
        "wrong_answers_1": "for pb: -35 degrees c (-30 degrees f)",
        "wrong_answers_2": "for pb: -34 degrees c (-30 degrees f)",
        "wrong_answers_3": "for pb: -37 degrees c (-26 degrees f)"
    },
    {
        "idx": 740,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for aluminum.",
        "answer": "for al: 100 degrees c (212 degrees f)",
        "type": 4,
        "wrong_answers_1": "for w: 1200 degrees c (2190 degrees f)",
        "wrong_answers_2": "for cu: 270 degrees c (518 degrees f)",
        "wrong_answers_3": "for fe: 450 degrees c (845 degrees f)"
    },
    {
        "idx": 741,
        "question": "Cite three variables that determine the microstructure of an alloy.",
        "answer": "Three variables that determine the microstructure of an alloy are (1) the alloying elements present, (2) the concentrations of these alloying elements, and (3) the heat treatment of the alloy.",
        "type": 4,
        "wrong_answers_1": "The presence of alloying elements (other than carbon) causes a much more gradual decrease in hardness with position from the quenched end for a hardenability curve. The reason for this effect is that alloying elements retard the formation of pearlitic and bainitic structures which are not as hard as martensite.",
        "wrong_answers_2": "The purpose of a spheroidizing heat treatment is to produce a very soft and ductile steel alloy having a spheroiditic microstructure.",
        "wrong_answers_3": "The alloying elements in tool steels (e.g., \\mathrm{Cr}, V, \\mathrm{W}, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds."
    },
    {
        "idx": 742,
        "question": "What thermodynamic condition must be met for a state of equilibrium to exist?",
        "answer": "In order for a system to exist in a state of equilibrium the free energy must be a minimum for some specified combination of temperature, pressure, and composition.",
        "type": 4,
        "wrong_answers_1": "Coring is the phenomenon whereby concentration gradients exist across grains in polycrystalline alloys, with higher concentrations of the component having the lower melting temperature at the grain boundaries. It occurs, during solidification, as a consequence of cooling rates that are too rapid to allow for the maintenance of the equilibrium composition of the solid phase.",
        "wrong_answers_2": "Two molecular characteristics essential for elastomers are: (1) they must be amorphous, having chains that are extensively coiled and kinked in the unstressed state; and (2) there must be some crosslinking.",
        "wrong_answers_3": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities."
    },
    {
        "idx": 743,
        "question": "For alloys of two hypothetical metals A and B, there exist an α, A-rich phase and a β, B-rich phase. From the mass fractions of both phases for two different alloys provided in the table below, (which are at the same temperature), determine the composition of the phase boundary (or solubility limit) for the α phase at this temperature.",
        "answer": "the composition of the phase boundary for the α phase is 90 wt% A-10 wt% B.",
        "type": 1,
        "wrong_answers_1": "the composition of the phase boundary for the α phase is 96 wt% A-11 wt% B.",
        "wrong_answers_2": "the composition of the phase boundary for the α phase is 77 wt% A-11 wt% B.",
        "wrong_answers_3": "the composition of the phase boundary for the α phase is 82 wt% A-9 wt% B."
    },
    {
        "idx": 744,
        "question": "For alloys of two hypothetical metals A and B, there exist an α, A-rich phase and a β, B-rich phase. From the mass fractions of both phases for two different alloys provided in the table below, (which are at the same temperature), determine the composition of the phase boundary (or solubility limit) for the β phase at this temperature.",
        "answer": "the composition of the phase boundary for the β phase is 20.2 wt% A-79.8 wt% B.",
        "type": 1,
        "wrong_answers_1": "the composition of the phase boundary for the β phase is 18.7 wt% A-83.2 wt% B.",
        "wrong_answers_2": "the composition of the phase boundary for the β phase is 18.4 wt% A-86.0 wt% B.",
        "wrong_answers_3": "the composition of the phase boundary for the β phase is 21.4 wt% A-88.7 wt% B."
    },
    {
        "idx": 745,
        "question": "A hypothetical A-B alloy of composition 55 wt% B-45 wt% A at some temperature is found to consist of mass fractions of 0.5 for both \\alpha and \\beta phases. If the composition of the \\beta phase is 90 wt% \\mathrm{~B}-10 wt% \\mathrm{~A}, what is the composition of the \\alpha phase?",
        "answer": "the composition of the \\alpha phase is 20 \\text{ wt}% \\text{ b- } 80 \\text{ wt}% \\text{ a}.",
        "type": 1,
        "wrong_answers_1": "the composition of the \\alpha phase is 19 \\text{ wt}% \\text{ b- } 90 \\text{ wt}% \\text{ a}.",
        "wrong_answers_2": "the composition of the \\alpha phase is 22 \\text{ wt}% \\text{ b- } 84 \\text{ wt}% \\text{ a}.",
        "wrong_answers_3": "the composition of the \\alpha phase is 18 \\text{ wt}% \\text{ b- } 76 \\text{ wt}% \\text{ a}."
    },
    {
        "idx": 746,
        "question": "For 11.20 kg of a magnesium-lead alloy of composition 30 wt% Pb-70 wt% Mg, is it possible, at equilibrium, to have α and Mg2Pb phases having respective masses of 7.39 kg and 3.81 kg?",
        "answer": "yes, it is possible to have a 30 wt% pb-70 wt% mg alloy with masses of 7.39 kg and 3.81 kg for the α and mg2pb phases, respectively.",
        "type": 3
    },
    {
        "idx": 747,
        "question": "What will be the approximate temperature of the alloy?",
        "answer": "the approximate temperature of the alloy is 190 degrees c.",
        "type": 4,
        "wrong_answers_1": "for w: 1200 degrees c (2190 degrees f)",
        "wrong_answers_2": "for al: 100 degrees c (212 degrees f)",
        "wrong_answers_3": "for cu: 270 degrees c (518 degrees f)"
    },
    {
        "idx": 748,
        "question": "Briefly describe the phenomenon of coring and why it occurs.",
        "answer": "Coring is the phenomenon whereby concentration gradients exist across grains in polycrystalline alloys, with higher concentrations of the component having the lower melting temperature at the grain boundaries. It occurs, during solidification, as a consequence of cooling rates that are too rapid to allow for the maintenance of the equilibrium composition of the solid phase.",
        "type": 4,
        "wrong_answers_1": "One undesirable consequence of a cored structure is that, upon heating, the grain boundary regions will melt first and at a temperature below the equilibrium phase boundary from the phase diagram; this melting results in a loss in mechanical integrity of the alloy.",
        "wrong_answers_2": "Upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers\nof the two solid phases because during the solidification atomic diffusion must occur, and with this layered configuration the diffusion path length for the atoms is a minimum.",
        "wrong_answers_3": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The polystyrene has a bulkier side group than the polypropylene; on the basis of this effect alone, the polystyrene should have the greater Tm. However, the polystyrene has more branching and a lower degree of polymerization; both of these factors lead to a lowering of the melting temperature."
    },
    {
        "idx": 749,
        "question": "Cite one undesirable consequence of coring.",
        "answer": "One undesirable consequence of a cored structure is that, upon heating, the grain boundary regions will melt first and at a temperature below the equilibrium phase boundary from the phase diagram; this melting results in a loss in mechanical integrity of the alloy.",
        "type": 4,
        "wrong_answers_1": "The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases.",
        "wrong_answers_2": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_3": "The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary."
    },
    {
        "idx": 750,
        "question": "Briefly explain why, upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers of the two solid phases.",
        "answer": "Upon solidification, an alloy of eutectic composition forms a microstructure consisting of alternating layers\nof the two solid phases because during the solidification atomic diffusion must occur, and with this layered configuration the diffusion path length for the atoms is a minimum.",
        "type": 4,
        "wrong_answers_1": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_2": "Coring is the phenomenon whereby concentration gradients exist across grains in polycrystalline alloys, with higher concentrations of the component having the lower melting temperature at the grain boundaries. It occurs, during solidification, as a consequence of cooling rates that are too rapid to allow for the maintenance of the equilibrium composition of the solid phase.",
        "wrong_answers_3": "Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species--i.e., the diffusion flux is independent of time."
    },
    {
        "idx": 751,
        "question": "What is the difference between a phase and a microconstituent?",
        "answer": "A \"phase\" is a homogeneous portion of the system having uniform physical and chemical characteristics, whereas a \"microconstituent\" is an identifiable element of the microstructure (that may consist of more than one phase).",
        "type": 4,
        "wrong_answers_1": "The matrix phase is a continuous phase that surrounds the noncontinuous dispersed phase.",
        "wrong_answers_2": "Tempered martensite is harder and stronger inasmuch as there is much more ferrite-cementite phase boundary area for the smaller particles; thus, there is greater reinforcement of the ferrite phase, and more phase boundary barriers to dislocation motion.",
        "wrong_answers_3": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite."
    },
    {
        "idx": 752,
        "question": "Two intermetallic compounds, A B and A B_{2}, exist for elements A and B. If the compositions for A B and A B_{2} are 34.3 wt% \\mathrm{~A}-65.7 wt% \\mathrm{~B} and 20.7 wt% \\mathrm{~A}-79.3 wt% \\mathrm{~B}, respectively, and element A is potassium, identify element B.",
        "answer": "element  b  is arsenic (as), with an atomic weight of  74.92 \\, \\text{g/mol} . the two intermetallic compounds are  \\text{kas}  and  \\text{kas}_2 .",
        "type": 4,
        "wrong_answers_1": "the alloy is hypereutectoid since c_0 is greater than 0.76 \\, \\text{wt% c}.",
        "wrong_answers_2": "cu dissolves ni until the cu contains enough ni that it solidifies completely. when 10% ni is dissolved, freezing begins with x = 55.5 \\text{g ni}. when 18% ni is dissolved, the bath is completely solid with x = 109.8 \\text{g ni}.",
        "wrong_answers_3": "Atomic mass is the mass of an individual atom, whereas atomic weight is the average (weighted) of the atomic masses of an atom's naturally occurring isotopes."
    },
    {
        "idx": 753,
        "question": "What is the principal difference between congruent and incongruent phase transformations?",
        "answer": "The principal difference between congruent and incongruent phase transformations is that for congruent no compositional changes occur with any of the phases that are involved in the transformation. For incongruent there will be compositional alterations of the phases.",
        "type": 4,
        "wrong_answers_1": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_2": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.",
        "wrong_answers_3": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers, which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded by an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles."
    },
    {
        "idx": 754,
        "question": "Compute the mass fraction of α ferrite in pearlite.",
        "answer": "the mass fraction of α ferrite in pearlite is 0.89.",
        "type": 1,
        "wrong_answers_1": "the mass fraction of α ferrite in pearlite is 1.01.",
        "wrong_answers_2": "the mass fraction of α ferrite in pearlite is 0.95.",
        "wrong_answers_3": "the mass fraction of α ferrite in pearlite is 0.95."
    },
    {
        "idx": 755,
        "question": "Compute the mass fraction of cementite in pearlite.",
        "answer": "the mass fraction of cementite in pearlite is 0.11.",
        "type": 1,
        "wrong_answers_1": "the mass fraction of cementite in pearlite is 0.12.",
        "wrong_answers_2": "the mass fraction of cementite in pearlite is 0.09.",
        "wrong_answers_3": "the mass fraction of cementite in pearlite is 0.10."
    },
    {
        "idx": 756,
        "question": "What is the distinction between hypoeutectoid and hypereutectoid steels?",
        "answer": "A \"hypoeutectoid\" steel has a carbon concentration less than the eutectoid; on the other hand, a \"hypereutectoid\" steel has a carbon content greater than the eutectoid.",
        "type": 4,
        "wrong_answers_1": "For a hypoeutectoid steel, the proeutectoid ferrite is a microconstituent that formed above the eutectoid temperature. The eutectoid ferrite is one of the constituents of pearlite that formed at a temperature below the eutectoid. The carbon concentration for both ferrites is 0.022 wt% C.",
        "wrong_answers_2": "For the low-carbon steel-copper couple, corrosion is possible, and the low-carbon steel will corrode.",
        "wrong_answers_3": "It is normally used on medium- and high-carbon steels, which, by virtue of carbon content, are relatively hard and strong."
    },
    {
        "idx": 757,
        "question": "In a hypoeutectoid steel, both eutectoid and proeutectoid ferrite exist. Explain the difference between them. What will be the carbon concentration in each?",
        "answer": "For a hypoeutectoid steel, the proeutectoid ferrite is a microconstituent that formed above the eutectoid temperature. The eutectoid ferrite is one of the constituents of pearlite that formed at a temperature below the eutectoid. The carbon concentration for both ferrites is 0.022 wt% C.",
        "type": 4,
        "wrong_answers_1": "A \"hypoeutectoid\" steel has a carbon concentration less than the eutectoid; on the other hand, a \"hypereutectoid\" steel has a carbon content greater than the eutectoid.",
        "wrong_answers_2": "ferrite is the proeutectoid phase.",
        "wrong_answers_3": "the proeutectoid phase is \\alpha ferrite because the alloy composition c_{0} is less than 0.76 wt% C."
    },
    {
        "idx": 758,
        "question": "What is the carbon concentration of an iron-carbon alloy for which the fraction of total ferrite is 0.94 ?",
        "answer": "the carbon concentration of the iron-carbon alloy is 0.42 \\text{ wt% c}.",
        "type": 1,
        "wrong_answers_1": "the carbon concentration of the iron-carbon alloy is 0.44 \\text{ wt% c}.",
        "wrong_answers_2": "the carbon concentration of the iron-carbon alloy is 0.38 \\text{ wt% c}.",
        "wrong_answers_3": "the carbon concentration of the iron-carbon alloy is 0.44 \\text{ wt% c}."
    },
    {
        "idx": 759,
        "question": "What is the proeutectoid phase for an iron-carbon alloy in which the mass fractions of total ferrite and total cementite are 0.92 and 0.08 , respectively? Why?",
        "answer": "the proeutectoid phase is \\alpha ferrite because the alloy composition c_{0} is less than 0.76 wt% C.",
        "type": 4,
        "wrong_answers_1": "ferrite is the proeutectoid phase.",
        "wrong_answers_2": "the alloy is hypereutectoid since c_0 is greater than 0.76 \\, \\text{wt% c}.",
        "wrong_answers_3": "the proeutectoid phase is fe3c."
    },
    {
        "idx": 760,
        "question": "Consider 1.0 kg of austenite containing 1.15 wt % C, cooled to below 727 C (1341 F). What is the proeutectoid phase?",
        "answer": "the proeutectoid phase is fe3c.",
        "type": 4,
        "wrong_answers_1": "ferrite is the proeutectoid phase.",
        "wrong_answers_2": "primary Fe3C",
        "wrong_answers_3": "the proeutectoid phase is \\alpha ferrite because the alloy composition c_{0} is less than 0.76 wt% C."
    },
    {
        "idx": 761,
        "question": "Consider 1.0 kg of austenite containing 1.15 wt % C, cooled to below 727 C (1341 F). How many kilograms each of total ferrite and cementite form?",
        "answer": "the total ferrite formed is 0.83 kg, and the total cementite formed is 0.17 kg.",
        "type": 1,
        "wrong_answers_1": "the total ferrite formed is 0.88 kg, and the total cementite formed is 0.14 kg.",
        "wrong_answers_2": "the total ferrite formed is 0.72 kg, and the total cementite formed is 0.15 kg.",
        "wrong_answers_3": "the total ferrite formed is 0.89 kg, and the total cementite formed is 0.18 kg."
    },
    {
        "idx": 762,
        "question": "Consider 1.0 kg of austenite containing 1.15 wt % C, cooled to below 727 C (1341 F). How many kilograms each of pearlite and the proeutectoid phase form?",
        "answer": "the pearlite formed is 0.93 kg, and the proeutectoid phase (cementite) formed is 0.07 kg.",
        "type": 1,
        "wrong_answers_1": "the pearlite formed is 0.98 kg, and the proeutectoid phase (cementite) formed is 0.08 kg.",
        "wrong_answers_2": "the pearlite formed is 1.03 kg, and the proeutectoid phase (cementite) formed is 0.08 kg.",
        "wrong_answers_3": "the pearlite formed is 0.82 kg, and the proeutectoid phase (cementite) formed is 0.07 kg."
    },
    {
        "idx": 763,
        "question": "Consider 2.5 kg of austenite containing 0.65 wt % C, cooled to below 727 C (1341 F). What is the proeutectoid phase?",
        "answer": "ferrite is the proeutectoid phase.",
        "type": 4,
        "wrong_answers_1": "the proeutectoid phase is fe3c.",
        "wrong_answers_2": "martensite, proeutectoid ferrite, and bainite form.",
        "wrong_answers_3": "martensite, proeutectoid ferrite, pearlite, and bainite form."
    },
    {
        "idx": 764,
        "question": "Consider 2.5 kg of austenite containing 0.65 wt % C, cooled to below 727 C (1341 F). How many kilograms each of total ferrite and cementite form?",
        "answer": "the total ferrite formed is 2.27 kg, and the total cementite formed is 0.23 kg.",
        "type": 1,
        "wrong_answers_1": "the total ferrite formed is 2.58 kg, and the total cementite formed is 0.20 kg.",
        "wrong_answers_2": "the total ferrite formed is 2.44 kg, and the total cementite formed is 0.22 kg.",
        "wrong_answers_3": "the total ferrite formed is 2.46 kg, and the total cementite formed is 0.26 kg."
    },
    {
        "idx": 765,
        "question": "Consider 2.5 kg of austenite containing 0.65 wt % C, cooled to below 727 C (1341 F). How many kilograms each of pearlite and the proeutectoid phase form?",
        "answer": "the pearlite formed is 2.12 kg, and the proeutectoid ferrite formed is 0.38 kg.",
        "type": 1,
        "wrong_answers_1": "the pearlite formed is 2.35 kg, and the proeutectoid ferrite formed is 0.40 kg.",
        "wrong_answers_2": "the pearlite formed is 2.31 kg, and the proeutectoid ferrite formed is 0.34 kg.",
        "wrong_answers_3": "the pearlite formed is 2.36 kg, and the proeutectoid ferrite formed is 0.33 kg."
    },
    {
        "idx": 766,
        "question": "Compute the mass fraction of proeutectoid ferrite that forms in an iron-carbon alloy containing 0.25 wt% C.",
        "answer": "the mass fraction of proeutectoid ferrite is 0.69.",
        "type": 1,
        "wrong_answers_1": "the mass fraction of proeutectoid ferrite is 0.79.",
        "wrong_answers_2": "the mass fraction of proeutectoid ferrite is 0.61.",
        "wrong_answers_3": "the mass fraction of proeutectoid ferrite is 0.77."
    },
    {
        "idx": 767,
        "question": "Compute the mass fraction of pearlite that forms in an iron-carbon alloy containing 0.25 wt% C.",
        "answer": "the mass fraction of pearlite is 0.31.",
        "type": 1,
        "wrong_answers_1": "the mass fraction of pearlite is 0.36.",
        "wrong_answers_2": "the mass fraction of pearlite is 0.29.",
        "wrong_answers_3": "the mass fraction of pearlite is 0.28."
    },
    {
        "idx": 768,
        "question": "The microstructure of an iron-carbon alloy consists of proeutectoid ferrite and pearlite; the mass fractions of these two microconstituents are 0.286 and 0.714 , respectively. Determine the concentration of carbon in this alloy.\n\\[\n\\text {",
        "answer": "0.55% \\text{c}",
        "type": 1,
        "wrong_answers_1": "0.62% \\text{c}",
        "wrong_answers_2": "0.49% \\text{c}",
        "wrong_answers_3": "0.49% \\text{c}"
    },
    {
        "idx": 769,
        "question": "The mass fractions of total ferrite and total cementite in an iron-carbon alloy are 0.88 and 0.12 , respectively. Is this a hypoeutectoid or hypereutectoid alloy? Why?",
        "answer": "the alloy is hypereutectoid since c_0 is greater than 0.76 \\, \\text{wt% c}.",
        "type": 4,
        "wrong_answers_1": "the proeutectoid phase is \\alpha ferrite because the alloy composition c_{0} is less than 0.76 wt% C.",
        "wrong_answers_2": "element  b  is arsenic (as), with an atomic weight of  74.92 \\, \\text{g/mol} . the two intermetallic compounds are  \\text{kas}  and  \\text{kas}_2 .",
        "wrong_answers_3": "White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification. Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700°C may be necessary to produce a ferritic matrix."
    },
    {
        "idx": 770,
        "question": "The microstructure of an iron-carbon alloy consists of proeutectoid ferrite and pearlite; the mass fractions of these microconstituents are 0.20 and 0.80 , respectively. Determine the concentration of carbon in this alloy.",
        "answer": "the concentration of carbon in this alloy is 0.61% \\text{c}.",
        "type": 1,
        "wrong_answers_1": "the concentration of carbon in this alloy is 0.69% \\text{c}.",
        "wrong_answers_2": "the concentration of carbon in this alloy is 0.57% \\text{c}.",
        "wrong_answers_3": "the concentration of carbon in this alloy is 0.70% \\text{c}."
    },
    {
        "idx": 771,
        "question": "Consider 2.0 kg of a 99.6 wt % Fe-0.4 wt % C alloy that is cooled to a temperature just below the eutectoid. How many kilograms of proeutectoid ferrite form?",
        "answer": "0.98 kg of proeutectoid ferrite forms.",
        "type": 1,
        "wrong_answers_1": "0.91 kg of proeutectoid ferrite forms.",
        "wrong_answers_2": "0.92 kg of proeutectoid ferrite forms.",
        "wrong_answers_3": "1.03 kg of proeutectoid ferrite forms."
    },
    {
        "idx": 772,
        "question": "Consider 2.0 kg of a 99.6 wt % Fe-0.4 wt % C alloy that is cooled to a temperature just below the eutectoid. How many kilograms of eutectoid ferrite form?",
        "answer": "0.90 kg of eutectoid ferrite forms.",
        "type": 1,
        "wrong_answers_1": "1.02 kg of eutectoid ferrite forms.",
        "wrong_answers_2": "0.79 kg of eutectoid ferrite forms.",
        "wrong_answers_3": "1.03 kg of eutectoid ferrite forms."
    },
    {
        "idx": 773,
        "question": "Consider 2.0 kg of a 99.6 wt % Fe-0.4 wt % C alloy that is cooled to a temperature just below the eutectoid. How many kilograms of cementite form?",
        "answer": "0.114 kg of cementite forms.",
        "type": 1,
        "wrong_answers_1": "0.128 kg of cementite forms.",
        "wrong_answers_2": "0.122 kg of cementite forms.",
        "wrong_answers_3": "0.124 kg of cementite forms."
    },
    {
        "idx": 774,
        "question": "Compute the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid ironcarbon alloy.",
        "answer": "the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron-carbon alloy is 0.232.",
        "type": 1,
        "wrong_answers_1": "the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron-carbon alloy is 0.209.",
        "wrong_answers_2": "the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron-carbon alloy is 0.247.",
        "wrong_answers_3": "the maximum mass fraction of proeutectoid cementite possible for a hypereutectoid iron-carbon alloy is 0.216."
    },
    {
        "idx": 775,
        "question": "Is it possible to have an iron-carbon alloy for which the mass fractions of total ferrite and proeutectoid cementite are 0.846 and 0.049 , respectively? Why or why not?",
        "answer": "yes, it is possible to have an iron-carbon alloy with mass fractions of total ferrite w_{\\alpha}=0.846 and proeutectoid cementite w_{\\mathrm{fe}_{3} C^{\\prime}}=0.049.  the alloy composition for both conditions is 1.05 wt% C. since the composition values are equal, such an alloy exists.",
        "type": 3
    },
    {
        "idx": 776,
        "question": "Is it possible to have an iron-carbon alloy for which the mass fractions of total cementite and pearlite are 0.039 and 0.417 , respectively? Why or why not?",
        "answer": "no, such an iron-carbon alloy is not possible because the required carbon compositions for the given mass fractions of total cementite (0.28 wt% C) and pearlite (0.33 wt% C) are different.",
        "type": 3
    },
    {
        "idx": 777,
        "question": "Compute the mass fraction of eutectoid ferrite in an iron-carbon alloy that contains 0.43 wt% C.",
        "answer": "the mass fraction of eutectoid ferrite in the iron-carbon alloy is 0.493.",
        "type": 1,
        "wrong_answers_1": "the mass fraction of eutectoid ferrite in the iron-carbon alloy is 0.442.",
        "wrong_answers_2": "the mass fraction of eutectoid ferrite in the iron-carbon alloy is 0.422.",
        "wrong_answers_3": "the mass fraction of eutectoid ferrite in the iron-carbon alloy is 0.508."
    },
    {
        "idx": 778,
        "question": "Is it possible to determine the composition of an iron-carbon alloy if the mass fraction of eutectoid cementite is 0.104?",
        "answer": "yes, it is possible to determine the alloy composition, and there are two possible answers.",
        "type": 3
    },
    {
        "idx": 779,
        "question": "What is the composition of the alloy if eutectoid cementite exists in addition to proeutectoid cementite?",
        "answer": "the alloy composition is 1.11 wt% c.",
        "type": 1,
        "wrong_answers_1": "the alloy composition is 1.15 wt% c.",
        "wrong_answers_2": "the alloy composition is 1.21 wt% c.",
        "wrong_answers_3": "the alloy composition is 0.96 wt% c."
    },
    {
        "idx": 780,
        "question": "What is the composition of a hypoeutectoid alloy where all cementite is eutectoid cementite and the mass fraction of total cementite is 0.104?",
        "answer": "the composition can be determined using the lever rule.",
        "type": 1
    },
    {
        "idx": 781,
        "question": "Is it possible to determine the composition of an iron-carbon alloy if the mass fraction of eutectoid ferrite is 0.82?",
        "answer": "yes, it is possible to determine the alloy composition; and, in fact, there are two possible answers.",
        "type": 3
    },
    {
        "idx": 782,
        "question": "What is the first possible composition of the alloy if the mass fraction of eutectoid ferrite is 0.82?",
        "answer": "for the first case, the composition is c0 = 0.70 wt % c.",
        "type": 1,
        "wrong_answers_1": "for the first case, the composition is c0 = 0.73 wt % c.",
        "wrong_answers_2": "for the first case, the composition is c0 = 0.76 wt % c.",
        "wrong_answers_3": "for the first case, the composition is c0 = 0.64 wt % c."
    },
    {
        "idx": 783,
        "question": "What is the second possible composition of the alloy if the mass fraction of eutectoid ferrite is 0.82?",
        "answer": "for the second case, the composition is c0 = 1.22 wt % c.",
        "type": 1,
        "wrong_answers_1": "for the second case, the composition is c0 = 1.39 wt % c.",
        "wrong_answers_2": "for the second case, the composition is c0 = 1.34 wt % c.",
        "wrong_answers_3": "for the second case, the composition is c0 = 1.12 wt % c."
    },
    {
        "idx": 784,
        "question": "What is the first stage involved in the formation of particles of a new phase? Briefly describe it.",
        "answer": "The first stage is nucleation. The nucleation process involves the formation of normally very small particles of the new phase(s) which are stable and capable of continued growth.",
        "type": 4,
        "wrong_answers_1": "The second stage is growth. The growth stage is simply the increase in size of the new phase particles.",
        "wrong_answers_2": "The driving force for the formation of spheroidite is the net reduction in ferrite-cementite phase boundary area.",
        "wrong_answers_3": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles."
    },
    {
        "idx": 785,
        "question": "What is the second stage involved in the formation of particles of a new phase? Briefly describe it.",
        "answer": "The second stage is growth. The growth stage is simply the increase in size of the new phase particles.",
        "type": 4,
        "wrong_answers_1": "The first stage is nucleation. The nucleation process involves the formation of normally very small particles of the new phase(s) which are stable and capable of continued growth.",
        "wrong_answers_2": "The hardness of tempered martensite diminishes with increasing temperature at constant tempering time because higher temperatures accelerate the rate of cementite particle growth. This growth reduces the ferrite-cementite phase boundary area, which acts as barriers to dislocation motion. Consequently, the reduction in phase boundary area results in a decrease in hardness.",
        "wrong_answers_3": "In ionic materials, the larger the size of the component ions the greater the degree of electronic polarization."
    },
    {
        "idx": 786,
        "question": "If copper (which has a melting point of 1085^{\\circ} C ) homogeneously nucleates at 849^{\\circ} C, calculate the critical radius given values of -1.77 × 10^{9} J/ m^{3} and 0.200 J/ m^{2}, respectively, for the latent heat of fusion and the surface free energy.",
        "answer": "1.30 nm",
        "type": 1,
        "wrong_answers_1": "1.26 nm",
        "wrong_answers_2": "1.26 nm",
        "wrong_answers_3": "1.37 nm"
    },
    {
        "idx": 787,
        "question": "The kinetics of the austenite-to-pearlite transformation obey the Avrami relationship. Using the fraction transformed-time data given here, determine the total time required for 95% of the austenite to transform to pearlite:\n\\begin{tabular}{cc}\n\\hline Fraction Transformed & Time (s) \\\\\n\\hline 0.2 & 12.6 \\\\\n0.8 & 28.2 \\\\\n\\hline\n\\end{tabular}",
        "answer": "35.7(s)",
        "type": 1,
        "wrong_answers_1": "37.5(s)",
        "wrong_answers_2": "39.2(s)",
        "wrong_answers_3": "38.0(s)"
    },
    {
        "idx": 788,
        "question": "The fraction recrystallized-time data for the recrystallization at 600^{\\circ} C of a previously deformed steel are tabulated here. Assuming that the kinetics of this process obey the Avrami relationship, determine the fraction recrystallized after a total time of 22.8min.\n\\begin{tabular}{cc}\n\\hline \\begin{tabular}{c} \nFraction \\\\\nRecrystallized\n\\end{tabular} & Time (min) \\\\\n\\hline 0.20 & 13.1 \\\\\n0.70 & 29.1 \\\\\n\\hline\n\\end{tabular}",
        "answer": "0.51",
        "type": 1,
        "wrong_answers_1": "0.46",
        "wrong_answers_2": "0.44",
        "wrong_answers_3": "0.46"
    },
    {
        "idx": 789,
        "question": "In terms of heat treatment and the development of microstructure, what is one major limitation of the iron-iron carbide phase diagram related to nonequilibrium phases?",
        "answer": "The nonequilibrium martensite does not appear on the diagram.",
        "type": 4,
        "wrong_answers_1": "only martensite forms.",
        "wrong_answers_2": "No, it is not possible. Linear polymers have higher degrees of crystallization (and higher tensile moduli) than branched polymers - on this basis, the PVC material should have the higher value of E. On the other hand, PVC has a more complex repeat unit structure than does polyethylene, which means that, on this basis, the PE would have a higher degree of crystallinity and also a greater tensile modulus. Also, tensile modulus is relatively independent of number-average molecular weight. Therefore, this determination is not possible since it is not possible to determine which of the two materials has the greater degree of crystallinity.",
        "wrong_answers_3": "both martensite and bainite form."
    },
    {
        "idx": 790,
        "question": "In terms of heat treatment and the development of microstructure, what is one major limitation of the iron-iron carbide phase diagram related to time-temperature relationships?",
        "answer": "The diagram provides no indication as to the time-temperature relationships for the formation of pearlite, bainite, and spheroidite, all of which are composed of the equilibrium ferrite and cementite phases.",
        "type": 4,
        "wrong_answers_1": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.",
        "wrong_answers_2": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers, which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded by an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.",
        "wrong_answers_3": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite."
    },
    {
        "idx": 791,
        "question": "Briefly describe the phenomena of superheating and supercooling.",
        "answer": "Superheating and supercooling correspond, respectively, to heating or cooling above or below a phase transition temperature without the occurrence of the transformation.",
        "type": 4,
        "wrong_answers_1": "These phenomena occur because right at the phase transition temperature, the driving force is not sufficient to cause the transformation to occur. The driving force is enhanced during superheating or supercooling.",
        "wrong_answers_2": "There is no bainite transformation region on the continuous cooling transformation diagram for an ironcarbon alloy of eutectoid composition because by the time a cooling curve has passed into the bainite region, the entirety of the alloy specimen will have transformed to pearlite.",
        "wrong_answers_3": "This question asks us to name which, of several polymers, would be suitable for the fabrication of cups to contain hot coffee. At its glass transition temperature, an amorphous polymer begins to soften. The maximum temperature of hot coffee is probably slightly below 100^{\\circ} C\\left(212^{\\circ} F\\right). Of the polymers listed, only polystyrene and polycarbonate have glass transition temperatures of 100^{\\circ} C or above, and would be suitable for this application."
    },
    {
        "idx": 792,
        "question": "Why do these phenomena occur?",
        "answer": "These phenomena occur because right at the phase transition temperature, the driving force is not sufficient to cause the transformation to occur. The driving force is enhanced during superheating or supercooling.",
        "type": 4,
        "wrong_answers_1": "The driving force is that which compels a reaction to occur.",
        "wrong_answers_2": "Superheating and supercooling correspond, respectively, to heating or cooling above or below a phase transition temperature without the occurrence of the transformation.",
        "wrong_answers_3": "The driving force for the formation of spheroidite is the net reduction in ferrite-cementite phase boundary area."
    },
    {
        "idx": 793,
        "question": "Briefly cite the differences between pearlite, bainite, and spheroidite relative to microstructure.",
        "answer": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.",
        "type": 4,
        "wrong_answers_1": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers, which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded by an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.",
        "wrong_answers_2": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_3": "The diagram provides no indication as to the time-temperature relationships for the formation of pearlite, bainite, and spheroidite, all of which are composed of the equilibrium ferrite and cementite phases."
    },
    {
        "idx": 794,
        "question": "Briefly cite the differences between pearlite, bainite, and spheroidite relative to mechanical properties.",
        "answer": "Bainite is harder and stronger than pearlite, which, in turn, is harder and stronger than spheroidite.",
        "type": 4,
        "wrong_answers_1": "Bainite is harder and stronger than pearlite, which, in turn, is harder and stronger than spheroidite.",
        "wrong_answers_2": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_3": "Tempered martensite is harder and stronger inasmuch as there is much more ferrite-cementite phase boundary area for the smaller particles; thus, there is greater reinforcement of the ferrite phase, and more phase boundary barriers to dislocation motion."
    },
    {
        "idx": 795,
        "question": "What is the driving force for the formation of spheroidite?",
        "answer": "The driving force for the formation of spheroidite is the net reduction in ferrite-cementite phase boundary area.",
        "type": 4,
        "wrong_answers_1": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_2": "The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases.",
        "wrong_answers_3": "The hardness of tempered martensite diminishes with tempering time at constant temperature because the microstructure consists of small sphere-like particles of cementite embedded within a ferrite matrix. As tempering time increases, the cementite particles grow, which reduces the ferrite-cementite phase boundary area. Since these phase boundaries act as barriers to dislocation motion, the reduction in phase boundary area leads to a decrease in hardness."
    },
    {
        "idx": 796,
        "question": "Name the microstructural products of eutectoid iron-carbon alloy (0.76 wt% C) specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 200°C/s.",
        "answer": "At a rate of 200°C/s, only martensite forms.",
        "type": 4,
        "wrong_answers_1": "only martensite forms.",
        "wrong_answers_2": "At a rate of 20°C/s, only fine pearlite forms.",
        "wrong_answers_3": "At a rate of 100°C/s, both martensite and pearlite form."
    },
    {
        "idx": 797,
        "question": "Name the microstructural products of eutectoid iron-carbon alloy (0.76 wt% C) specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 100°C/s.",
        "answer": "At a rate of 100°C/s, both martensite and pearlite form.",
        "type": 4,
        "wrong_answers_1": "both martensite and bainite form.",
        "wrong_answers_2": "martensite, proeutectoid ferrite, pearlite, and bainite form.",
        "wrong_answers_3": "At a rate of 200°C/s, only martensite forms."
    },
    {
        "idx": 798,
        "question": "Name the microstructural products of eutectoid iron-carbon alloy (0.76 wt% C) specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 20°C/s.",
        "answer": "At a rate of 20°C/s, only fine pearlite forms.",
        "type": 4,
        "wrong_answers_1": "At a rate of 200°C/s, only martensite forms.",
        "wrong_answers_2": "only martensite forms.",
        "wrong_answers_3": "At a rate of 100°C/s, both martensite and pearlite form."
    },
    {
        "idx": 799,
        "question": "What is one important difference between continuous cooling transformation diagrams for plain carbon and alloy steels regarding the presence of a bainite nose?",
        "answer": "For an alloy steel, a bainite nose will be present, which nose will be absent for plain carbon alloys.",
        "type": 4,
        "wrong_answers_1": "The pearlite-proeutectoid noses for plain carbon steel alloys are positioned at shorter times than for the alloy steels.",
        "wrong_answers_2": "For the low-carbon steel-copper couple, corrosion is possible, and the low-carbon steel will corrode.",
        "wrong_answers_3": "There is no bainite transformation region on the continuous cooling transformation diagram for an ironcarbon alloy of eutectoid composition because by the time a cooling curve has passed into the bainite region, the entirety of the alloy specimen will have transformed to pearlite."
    },
    {
        "idx": 800,
        "question": "What is one important difference between continuous cooling transformation diagrams for plain carbon and alloy steels regarding the position of the pearlite-proeutectoid noses?",
        "answer": "The pearlite-proeutectoid noses for plain carbon steel alloys are positioned at shorter times than for the alloy steels.",
        "type": 4,
        "wrong_answers_1": "The four classifications of steels are: Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, and High Alloy Steels (Stainless and Tool).",
        "wrong_answers_2": "Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, High Alloy Steels (Stainless and Tool)",
        "wrong_answers_3": "For an alloy steel, a bainite nose will be present, which nose will be absent for plain carbon alloys."
    },
    {
        "idx": 801,
        "question": "Briefly explain why there is no bainite transformation region on the continuous cooling transformation diagram for an iron-carbon alloy of eutectoid composition.",
        "answer": "There is no bainite transformation region on the continuous cooling transformation diagram for an ironcarbon alloy of eutectoid composition because by the time a cooling curve has passed into the bainite region, the entirety of the alloy specimen will have transformed to pearlite.",
        "type": 4,
        "wrong_answers_1": "The diagram provides no indication as to the time-temperature relationships for the formation of pearlite, bainite, and spheroidite, all of which are composed of the equilibrium ferrite and cementite phases.",
        "wrong_answers_2": "For a hypoeutectoid steel, the proeutectoid ferrite is a microconstituent that formed above the eutectoid temperature. The eutectoid ferrite is one of the constituents of pearlite that formed at a temperature below the eutectoid. The carbon concentration for both ferrites is 0.022 wt% C.",
        "wrong_answers_3": "Superheating and supercooling correspond, respectively, to heating or cooling above or below a phase transition temperature without the occurrence of the transformation."
    },
    {
        "idx": 802,
        "question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 10°C/s.",
        "answer": "only martensite forms.",
        "type": 4,
        "wrong_answers_1": "At a rate of 200°C/s, only martensite forms.",
        "wrong_answers_2": "At a rate of 20°C/s, only fine pearlite forms.",
        "wrong_answers_3": "both martensite and bainite form."
    },
    {
        "idx": 803,
        "question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 1°C/s.",
        "answer": "both martensite and bainite form.",
        "type": 4,
        "wrong_answers_1": "martensite, proeutectoid ferrite, and bainite form.",
        "wrong_answers_2": "martensite, proeutectoid ferrite, pearlite, and bainite form.",
        "wrong_answers_3": "At a rate of 100°C/s, both martensite and pearlite form."
    },
    {
        "idx": 804,
        "question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 0.1°C/s.",
        "answer": "martensite, proeutectoid ferrite, and bainite form.",
        "type": 4,
        "wrong_answers_1": "martensite, proeutectoid ferrite, pearlite, and bainite form.",
        "wrong_answers_2": "both martensite and bainite form.",
        "wrong_answers_3": "ferrite is the proeutectoid phase."
    },
    {
        "idx": 805,
        "question": "Name the microstructural products of 4340 alloy steel specimens that are first completely transformed to austenite, then cooled to room temperature at a rate of 0.01°C/s.",
        "answer": "martensite, proeutectoid ferrite, pearlite, and bainite form.",
        "type": 4,
        "wrong_answers_1": "martensite, proeutectoid ferrite, and bainite form.",
        "wrong_answers_2": "both martensite and bainite form.",
        "wrong_answers_3": "ferrite is the proeutectoid phase."
    },
    {
        "idx": 806,
        "question": "On the basis of diffusion considerations, explain why fine pearlite forms for the moderate cooling of austenite through the eutectoid temperature, whereas coarse pearlite is the product for relatively slow cooling rates.",
        "answer": "For moderately rapid cooling, the time allowed for carbon diffusion is not as great as for slower cooling rates. Therefore, the diffusion distance is shorter, and thinner layers of ferrite and cementite form (i.e., fine pearlite forms).",
        "type": 4,
        "wrong_answers_1": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_2": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.",
        "wrong_answers_3": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers, which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded by an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles."
    },
    {
        "idx": 807,
        "question": "Briefly explain why fine pearlite is harder and stronger than coarse pearlite, which in turn is harder and stronger than spheroidite.",
        "answer": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "type": 4,
        "wrong_answers_1": "Tempered martensite is harder and stronger inasmuch as there is much more ferrite-cementite phase boundary area for the smaller particles; thus, there is greater reinforcement of the ferrite phase, and more phase boundary barriers to dislocation motion.",
        "wrong_answers_2": "The hardness of tempered martensite diminishes with tempering time at constant temperature because the microstructure consists of small sphere-like particles of cementite embedded within a ferrite matrix. As tempering time increases, the cementite particles grow, which reduces the ferrite-cementite phase boundary area. Since these phase boundaries act as barriers to dislocation motion, the reduction in phase boundary area leads to a decrease in hardness.",
        "wrong_answers_3": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles."
    },
    {
        "idx": 808,
        "question": "Cite two reasons why martensite is so hard and brittle.",
        "answer": "Two reasons why martensite is so hard and brittle are: (1) there are relatively few operable slip systems for the body-centered tetragonal crystal structure, and (2) virtually all of the carbon is in solid solution, which produces a solid-solution hardening effect.",
        "type": 4,
        "wrong_answers_1": "It is normally used on medium- and high-carbon steels, which, by virtue of carbon content, are relatively hard and strong.",
        "wrong_answers_2": "For slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples. Slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems.",
        "wrong_answers_3": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP."
    },
    {
        "idx": 809,
        "question": "Rank the following iron-carbon alloys and associated microstructures from the highest to the lowest tensile strength:\n(a) 0.25 wt% C with spheroidite,\n(b) 0.25 wt% C with coarse pearlite,\n(c) 0.60 wt% C with fine pearlite, and\n(d) 0.60 wt% C with coarse pearlit.",
        "answer": "(c)(d)(b)(a)",
        "type": 2
    },
    {
        "idx": 810,
        "question": "Briefly explain why the hardness of tempered martensite diminishes with tempering time at constant temperature.",
        "answer": "The hardness of tempered martensite diminishes with tempering time at constant temperature because the microstructure consists of small sphere-like particles of cementite embedded within a ferrite matrix. As tempering time increases, the cementite particles grow, which reduces the ferrite-cementite phase boundary area. Since these phase boundaries act as barriers to dislocation motion, the reduction in phase boundary area leads to a decrease in hardness.",
        "type": 4,
        "wrong_answers_1": "The hardness of tempered martensite diminishes with increasing temperature at constant tempering time because higher temperatures accelerate the rate of cementite particle growth. This growth reduces the ferrite-cementite phase boundary area, which acts as barriers to dislocation motion. Consequently, the reduction in phase boundary area results in a decrease in hardness.",
        "wrong_answers_2": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_3": "Tempered martensite is harder and stronger inasmuch as there is much more ferrite-cementite phase boundary area for the smaller particles; thus, there is greater reinforcement of the ferrite phase, and more phase boundary barriers to dislocation motion."
    },
    {
        "idx": 811,
        "question": "Briefly explain why the hardness of tempered martensite diminishes with increasing temperature at constant tempering time.",
        "answer": "The hardness of tempered martensite diminishes with increasing temperature at constant tempering time because higher temperatures accelerate the rate of cementite particle growth. This growth reduces the ferrite-cementite phase boundary area, which acts as barriers to dislocation motion. Consequently, the reduction in phase boundary area results in a decrease in hardness.",
        "type": 4,
        "wrong_answers_1": "The hardness of tempered martensite diminishes with tempering time at constant temperature because the microstructure consists of small sphere-like particles of cementite embedded within a ferrite matrix. As tempering time increases, the cementite particles grow, which reduces the ferrite-cementite phase boundary area. Since these phase boundaries act as barriers to dislocation motion, the reduction in phase boundary area leads to a decrease in hardness.",
        "wrong_answers_2": "Tempered martensite is harder and stronger inasmuch as there is much more ferrite-cementite phase boundary area for the smaller particles; thus, there is greater reinforcement of the ferrite phase, and more phase boundary barriers to dislocation motion.",
        "wrong_answers_3": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite."
    },
    {
        "idx": 812,
        "question": "Briefly describe the microstructural difference between spheroidite and tempered martensite.",
        "answer": "Both tempered martensite and spheroidite have sphere-like cementite particles within a ferrite matrix; however, these particles are much larger for spheroidite.",
        "type": 4,
        "wrong_answers_1": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.",
        "wrong_answers_2": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers, which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded by an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.",
        "wrong_answers_3": "The hardness of tempered martensite diminishes with tempering time at constant temperature because the microstructure consists of small sphere-like particles of cementite embedded within a ferrite matrix. As tempering time increases, the cementite particles grow, which reduces the ferrite-cementite phase boundary area. Since these phase boundaries act as barriers to dislocation motion, the reduction in phase boundary area leads to a decrease in hardness."
    },
    {
        "idx": 813,
        "question": "Explain why tempered martensite is much harder and stronger.",
        "answer": "Tempered martensite is harder and stronger inasmuch as there is much more ferrite-cementite phase boundary area for the smaller particles; thus, there is greater reinforcement of the ferrite phase, and more phase boundary barriers to dislocation motion.",
        "type": 4,
        "wrong_answers_1": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_2": "The hardness of tempered martensite diminishes with tempering time at constant temperature because the microstructure consists of small sphere-like particles of cementite embedded within a ferrite matrix. As tempering time increases, the cementite particles grow, which reduces the ferrite-cementite phase boundary area. Since these phase boundaries act as barriers to dislocation motion, the reduction in phase boundary area leads to a decrease in hardness.",
        "wrong_answers_3": "The hardness of tempered martensite diminishes with increasing temperature at constant tempering time because higher temperatures accelerate the rate of cementite particle growth. This growth reduces the ferrite-cementite phase boundary area, which acts as barriers to dislocation motion. Consequently, the reduction in phase boundary area results in a decrease in hardness."
    },
    {
        "idx": 814,
        "question": "List the four classifications of steels.",
        "answer": "Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, High Alloy Steels (Stainless and Tool)",
        "type": 4,
        "wrong_answers_1": "The four classifications of steels are: Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, and High Alloy Steels (Stainless and Tool).",
        "wrong_answers_2": "It is normally used on medium- and high-carbon steels, which, by virtue of carbon content, are relatively hard and strong.",
        "wrong_answers_3": "Stainless steels and aluminum alloys often passivate."
    },
    {
        "idx": 815,
        "question": "For Low Carbon Steels, briefly describe the properties and typical applications.",
        "answer": "Properties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable. Typical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans.",
        "type": 4,
        "wrong_answers_1": "Properties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable. Typical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans.",
        "wrong_answers_2": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades.",
        "wrong_answers_3": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades."
    },
    {
        "idx": 816,
        "question": "For Medium Carbon Steels, briefly describe the properties and typical applications.",
        "answer": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.",
        "type": 4,
        "wrong_answers_1": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.",
        "wrong_answers_2": "The chief difference between heat-treatable and nonheat-treatable alloys is that heat-treatable alloys may be strengthened by a heat treatment wherein a precipitate phase is formed (precipitation hardening) or a martensitic transformation occurs. Nonheat-treatable alloys are not amenable to strengthening by such treatments.",
        "wrong_answers_3": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades."
    },
    {
        "idx": 817,
        "question": "For High Carbon Steels, briefly describe the properties and typical applications.",
        "answer": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades.",
        "type": 4,
        "wrong_answers_1": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades.",
        "wrong_answers_2": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.",
        "wrong_answers_3": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts."
    },
    {
        "idx": 818,
        "question": "For High Alloy Steels (Stainless and Tool), briefly describe the properties and typical applications.",
        "answer": "Properties: hard and wear resistant; resistant to corrosion in a large variety of environments. Typical applications: cutting tools, drills, cutlery, food processing, and surgical tools.",
        "type": 4,
        "wrong_answers_1": "Properties: hard and wear resistant; resistant to corrosion in a large variety of environments. Typical applications: cutting tools, drills, cutlery, food processing, and surgical tools.",
        "wrong_answers_2": "The alloying elements in tool steels (e.g., \\mathrm{Cr}, V, \\mathrm{W}, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds.",
        "wrong_answers_3": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades."
    },
    {
        "idx": 819,
        "question": "Cite three reasons why ferrous alloys are used so extensively.",
        "answer": "Ferrous alloys are used extensively because: (1) Iron ores exist in abundant quantities. (2) Economical extraction, refining, and fabrication techniques are available. (3) The alloys may be tailored to have a wide range of properties.",
        "type": 4,
        "wrong_answers_1": "The principal difference between wrought and cast alloys is as follows: wrought alloys are ductile enough so as to be hot or cold worked during fabrication, whereas cast alloys are brittle to the degree that shaping by deformation is not possible and they must be fabricated by casting.",
        "wrong_answers_2": "Disadvantages of ferrous alloys are: (1) They are susceptible to corrosion. (2) They have a relatively high density. (3) They have relatively low electrical conductivities.",
        "wrong_answers_3": "Four situations in which casting is the preferred fabrication technique are:\n(1) For large pieces and/or complicated shapes.\n(2) When mechanical strength is not an important consideration.\n(3) For alloys having low ductilities.\n(4) When it is the most economical fabrication technique."
    },
    {
        "idx": 820,
        "question": "Cite three characteristics of ferrous alloys that limit their utilization.",
        "answer": "Disadvantages of ferrous alloys are: (1) They are susceptible to corrosion. (2) They have a relatively high density. (3) They have relatively low electrical conductivities.",
        "type": 4,
        "wrong_answers_1": "Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of crystallinity.",
        "wrong_answers_2": "Several limitations of these composites are: (1) care must be exercised in handling the fibers inasmuch as they are susceptible to surface damage; (2) they are lacking in stiffness in comparison to other fibrous composites; and (3) they are limited as to maximum temperature use.",
        "wrong_answers_3": "Rivets of a 2017 aluminum alloy must be refrigerated before they are used because, after being solution heat treated, they precipitation harden at room temperature. Once precipitation hardened, they are too strong and brittle to be driven."
    },
    {
        "idx": 821,
        "question": "What is the function of alloying elements in tool steels?",
        "answer": "The alloying elements in tool steels (e.g., \\mathrm{Cr}, V, \\mathrm{W}, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds.",
        "type": 4,
        "wrong_answers_1": "Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, High Alloy Steels (Stainless and Tool)",
        "wrong_answers_2": "The four classifications of steels are: Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, and High Alloy Steels (Stainless and Tool).",
        "wrong_answers_3": "The presence of alloying elements (other than carbon) causes a much more gradual decrease in hardness with position from the quenched end for a hardenability curve. The reason for this effect is that alloying elements retard the formation of pearlitic and bainitic structures which are not as hard as martensite."
    },
    {
        "idx": 822,
        "question": "On the basis of microstructure, briefly explain why gray iron is brittle and weak in tension.",
        "answer": "Gray iron is weak and brittle in tension because the tips of the graphite flakes act as points of stress concentration.",
        "type": 4,
        "wrong_answers_1": "Gray iron--Relatively weak and brittle in tension; good capacity for damping vibrations. Malleable iron--Moderate strength and ductility.",
        "wrong_answers_2": "Gray iron--Graphite flakes are embedded in a ferrite or pearlite matrix. Malleable iron--Graphite clusters are embedded in a ferrite or pearlite matrix.",
        "wrong_answers_3": "White iron--There are regions of cementite interspersed within pearlite. Nodular cast iron--Nodules of graphite are embedded in a ferrite or pearlite matrix."
    },
    {
        "idx": 823,
        "question": "Compare white and nodular cast irons with respect to composition and heat treatment.",
        "answer": "White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification. Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700°C may be necessary to produce a ferritic matrix.",
        "type": 4,
        "wrong_answers_1": "Gray iron--2.5 to 4.0 wt% C and 1.0 to 3.0 wt% Si. For most gray irons there is no heat treatment after solidification. Malleable iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. White iron is heated in a nonoxidizing atmosphere and at a temperature between 800 and 900°C for an extended time period.",
        "wrong_answers_2": "White iron--Extremely hard and brittle. Nodular cast iron--Moderate strength and ductility.",
        "wrong_answers_3": "White iron--There are regions of cementite interspersed within pearlite. Nodular cast iron--Nodules of graphite are embedded in a ferrite or pearlite matrix."
    },
    {
        "idx": 824,
        "question": "Compare white and nodular cast irons with respect to microstructure.",
        "answer": "White iron--There are regions of cementite interspersed within pearlite. Nodular cast iron--Nodules of graphite are embedded in a ferrite or pearlite matrix.",
        "type": 4,
        "wrong_answers_1": "Gray iron--Graphite flakes are embedded in a ferrite or pearlite matrix. Malleable iron--Graphite clusters are embedded in a ferrite or pearlite matrix.",
        "wrong_answers_2": "White iron--Extremely hard and brittle. Nodular cast iron--Moderate strength and ductility.",
        "wrong_answers_3": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles."
    },
    {
        "idx": 825,
        "question": "Compare white and nodular cast irons with respect to mechanical characteristics.",
        "answer": "White iron--Extremely hard and brittle. Nodular cast iron--Moderate strength and ductility.",
        "type": 4,
        "wrong_answers_1": "Gray iron--Relatively weak and brittle in tension; good capacity for damping vibrations. Malleable iron--Moderate strength and ductility.",
        "wrong_answers_2": "White iron--There are regions of cementite interspersed within pearlite. Nodular cast iron--Nodules of graphite are embedded in a ferrite or pearlite matrix.",
        "wrong_answers_3": "White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification. Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700°C may be necessary to produce a ferritic matrix."
    },
    {
        "idx": 826,
        "question": "Is it possible to produce malleable cast iron in pieces having large cross-sectional dimensions? Why or why not?",
        "answer": "It is not possible to produce malleable iron in pieces having large cross-sectional dimensions. White cast iron is the precursor of malleable iron, and a rapid cooling rate is necessary for the formation of white iron, which may not be accomplished at interior regions of thick cross-sections.",
        "type": 3
    },
    {
        "idx": 827,
        "question": "What is the principal difference between wrought and cast alloys?",
        "answer": "The principal difference between wrought and cast alloys is as follows: wrought alloys are ductile enough so as to be hot or cold worked during fabrication, whereas cast alloys are brittle to the degree that shaping by deformation is not possible and they must be fabricated by casting.",
        "type": 4,
        "wrong_answers_1": "The chief difference between heat-treatable and nonheat-treatable alloys is that heat-treatable alloys may be strengthened by a heat treatment wherein a precipitate phase is formed (precipitation hardening) or a martensitic transformation occurs. Nonheat-treatable alloys are not amenable to strengthening by such treatments.",
        "wrong_answers_2": "Fiber materials that are drawn must be thermoplastic because during drawing, mechanical elongation must be possible; inasmuch as thermosetting materials are, in general, hard and relatively brittle, they are not easily elongated.",
        "wrong_answers_3": "Ferrous alloys are used extensively because: (1) Iron ores exist in abundant quantities. (2) Economical extraction, refining, and fabrication techniques are available. (3) The alloys may be tailored to have a wide range of properties."
    },
    {
        "idx": 828,
        "question": "Why must rivets of a 2017 aluminum alloy be refrigerated before they are used?",
        "answer": "Rivets of a 2017 aluminum alloy must be refrigerated before they are used because, after being solution heat treated, they precipitation harden at room temperature. Once precipitation hardened, they are too strong and brittle to be driven.",
        "type": 4,
        "wrong_answers_1": "Disadvantages of ferrous alloys are: (1) They are susceptible to corrosion. (2) They have a relatively high density. (3) They have relatively low electrical conductivities.",
        "wrong_answers_2": "Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of crystallinity.",
        "wrong_answers_3": "Several limitations of these composites are: (1) care must be exercised in handling the fibers inasmuch as they are susceptible to surface damage; (2) they are lacking in stiffness in comparison to other fibrous composites; and (3) they are limited as to maximum temperature use."
    },
    {
        "idx": 829,
        "question": "What is the chief difference between heat-treatable and non-heat-treatable alloys?",
        "answer": "The chief difference between heat-treatable and nonheat-treatable alloys is that heat-treatable alloys may be strengthened by a heat treatment wherein a precipitate phase is formed (precipitation hardening) or a martensitic transformation occurs. Nonheat-treatable alloys are not amenable to strengthening by such treatments.",
        "type": 4,
        "wrong_answers_1": "The two differences are: (1) the hardening/strengthening effect is not retained at elevated temperatures for precipitation hardening--however, it is retained for dispersion strengthening; and (2) the strength is developed by a heat treatment for precipitation hardening--such is not the case for dispersion strengthening.",
        "wrong_answers_2": "The principal difference between wrought and cast alloys is as follows: wrought alloys are ductile enough so as to be hot or cold worked during fabrication, whereas cast alloys are brittle to the degree that shaping by deformation is not possible and they must be fabricated by casting.",
        "wrong_answers_3": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts."
    },
    {
        "idx": 830,
        "question": "Cite advantages of cold working.",
        "answer": "The advantages of cold working are: (1) A high quality surface finish. (2) The mechanical properties may be varied. (3) Close dimensional tolerances.",
        "type": 4,
        "wrong_answers_1": "The disadvantages of hot working are: (1) A poor surface finish. (2) A variety of mechanical properties is not possible.",
        "wrong_answers_2": "The advantages of hot working are: (1) Large deformations are possible, which may be repeated. (2) Deformation energy requirements are relatively low.",
        "wrong_answers_3": "The disadvantages of cold working are: (1) High deformation energy requirements. (2) Large deformations must be accomplished in steps, which may be expensive. (3) A loss of ductility."
    },
    {
        "idx": 831,
        "question": "Cite disadvantages of cold working.",
        "answer": "The disadvantages of cold working are: (1) High deformation energy requirements. (2) Large deformations must be accomplished in steps, which may be expensive. (3) A loss of ductility.",
        "type": 4,
        "wrong_answers_1": "The advantages of hot working are: (1) Large deformations are possible, which may be repeated. (2) Deformation energy requirements are relatively low.",
        "wrong_answers_2": "The advantages of cold working are: (1) A high quality surface finish. (2) The mechanical properties may be varied. (3) Close dimensional tolerances.",
        "wrong_answers_3": "Normally slip results in relatively large deformations, whereas only small deformations result for twinning."
    },
    {
        "idx": 832,
        "question": "Cite advantages of hot working.",
        "answer": "The advantages of hot working are: (1) Large deformations are possible, which may be repeated. (2) Deformation energy requirements are relatively low.",
        "type": 4,
        "wrong_answers_1": "The disadvantages of cold working are: (1) High deformation energy requirements. (2) Large deformations must be accomplished in steps, which may be expensive. (3) A loss of ductility.",
        "wrong_answers_2": "Normally slip results in relatively large deformations, whereas only small deformations result for twinning.",
        "wrong_answers_3": "The disadvantages of hot working are: (1) A poor surface finish. (2) A variety of mechanical properties is not possible."
    },
    {
        "idx": 833,
        "question": "Cite disadvantages of hot working.",
        "answer": "The disadvantages of hot working are: (1) A poor surface finish. (2) A variety of mechanical properties is not possible.",
        "type": 4,
        "wrong_answers_1": "The advantages of cold working are: (1) A high quality surface finish. (2) The mechanical properties may be varied. (3) Close dimensional tolerances.",
        "wrong_answers_2": "The advantages of hot working are: (1) Large deformations are possible, which may be repeated. (2) Deformation energy requirements are relatively low.",
        "wrong_answers_3": "The disadvantages of cold working are: (1) High deformation energy requirements. (2) Large deformations must be accomplished in steps, which may be expensive. (3) A loss of ductility."
    },
    {
        "idx": 834,
        "question": "Cite advantages of forming metals by extrusion as opposed to rolling.",
        "answer": "The advantages of extrusion as opposed to rolling are as follows: (1) Pieces having more complicated cross-sectional geometries may be formed. (2) Seamless tubing may be produced.",
        "type": 4,
        "wrong_answers_1": "The disadvantages of extrusion over rolling are as follows: (1) Nonuniform deformation over the cross-section. (2) A variation in properties may result over a cross-section of an extruded piece.",
        "wrong_answers_2": "The coordinates of these atoms are as follows: 1/2 1/2 0, 1/2 1/2 1, 1/2 0 1/2, 0 1/2 1/2, and 1/2 1/2 1/2.",
        "wrong_answers_3": "The coordinates of these atoms are as follows: 3/4 1/4 1/4, 1/4 3/4 3/4, 1/4 1/4 3/4, and 3/4 3/4 1/4."
    },
    {
        "idx": 835,
        "question": "Cite some disadvantages of forming metals by extrusion as opposed to rolling.",
        "answer": "The disadvantages of extrusion over rolling are as follows: (1) Nonuniform deformation over the cross-section. (2) A variation in properties may result over a cross-section of an extruded piece.",
        "type": 4,
        "wrong_answers_1": "The advantages of extrusion as opposed to rolling are as follows: (1) Pieces having more complicated cross-sectional geometries may be formed. (2) Seamless tubing may be produced.",
        "wrong_answers_2": "For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and mechanical properties are relatively uniform throughout the cross-section.",
        "wrong_answers_3": "For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and mechanical properties are relatively uniform throughout the cross-section."
    },
    {
        "idx": 836,
        "question": "List four situations in which casting is the preferred fabrication technique.",
        "answer": "Four situations in which casting is the preferred fabrication technique are:\n(1) For large pieces and/or complicated shapes.\n(2) When mechanical strength is not an important consideration.\n(3) For alloys having low ductilities.\n(4) When it is the most economical fabrication technique.",
        "type": 4,
        "wrong_answers_1": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast.",
        "wrong_answers_2": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast.",
        "wrong_answers_3": "Four factors that determine what fabrication technique is used to form polymeric materials are: (1) whether the polymer is thermoplastic or thermosetting; (2) if thermoplastic, the softening temperature; (3) atmospheric stability; and (4) the geometry and size of the finished product."
    },
    {
        "idx": 837,
        "question": "Compare sand casting technique",
        "answer": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast.",
        "type": 4,
        "wrong_answers_1": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast.",
        "wrong_answers_2": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "wrong_answers_3": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast."
    },
    {
        "idx": 838,
        "question": "Compare die casting technique",
        "answer": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "type": 4,
        "wrong_answers_1": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "wrong_answers_2": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast.",
        "wrong_answers_3": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast."
    },
    {
        "idx": 839,
        "question": "Compare investment casting technique",
        "answer": "For investment casting, a single-piece mold is used, which is not reusable; it results in high dimensional accuracy, good reproduction of detail, and a fine surface finish; and casting rates are low.",
        "type": 4,
        "wrong_answers_1": "For investment casting, a single-piece mold is used, which is not reusable; it results in high dimensional accuracy, good reproduction of detail, and a fine surface finish; and casting rates are low.",
        "wrong_answers_2": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "wrong_answers_3": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast."
    },
    {
        "idx": 840,
        "question": "Compare lost foam casting technique",
        "answer": "For lost foam casting, the pattern is polystyrene foam, whereas the mold material is sand. Complex geometries and tight tolerances are possible. Casting rates are higher than for investment, and there are few environmental wastes.",
        "type": 4,
        "wrong_answers_1": "For lost foam casting, the pattern is polystyrene foam, whereas the mold material is sand. Complex geometries and tight tolerances are possible. Casting rates are higher than for investment, and there are few environmental wastes.",
        "wrong_answers_2": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast.",
        "wrong_answers_3": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast."
    },
    {
        "idx": 841,
        "question": "Compare continuous casting technique",
        "answer": "For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and mechanical properties are relatively uniform throughout the cross-section.",
        "type": 4,
        "wrong_answers_1": "For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and mechanical properties are relatively uniform throughout the cross-section.",
        "wrong_answers_2": "The disadvantages of extrusion over rolling are as follows: (1) Nonuniform deformation over the cross-section. (2) A variation in properties may result over a cross-section of an extruded piece.",
        "wrong_answers_3": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast."
    },
    {
        "idx": 842,
        "question": "Describe one problem that might exist with a steel weld that was cooled very rapidly.",
        "answer": "If a steel weld is cooled very rapidly, martensite may form, which is very brittle. In some situations, cracks may form in the weld region as it cools.",
        "type": 4,
        "wrong_answers_1": "both martensite and bainite form.",
        "wrong_answers_2": "martensite, proeutectoid ferrite, and bainite form.",
        "wrong_answers_3": "At a rate of 100°C/s, both martensite and pearlite form."
    },
    {
        "idx": 843,
        "question": "Cite three sources of internal residual stresses in metal components.",
        "answer": "Three sources of residual stresses in metal components are plastic deformation processes, nonuniform cooling of a piece that was cooled from an elevated temperature, and a phase transformation in which parent and product phases have different densities.",
        "type": 4,
        "wrong_answers_1": "Residual thermal stresses are introduced into a glass piece when it is cooled because surface and interior regions cool at different rates, and, therefore, contract different amounts; since the material will experience very little, if any deformation, stresses are established.",
        "wrong_answers_2": "The principal difference between congruent and incongruent phase transformations is that for congruent no compositional changes occur with any of the phases that are involved in the transformation. For incongruent there will be compositional alterations of the phases.",
        "wrong_answers_3": "Superheating and supercooling correspond, respectively, to heating or cooling above or below a phase transition temperature without the occurrence of the transformation."
    },
    {
        "idx": 844,
        "question": "What are two possible adverse consequences of internal residual stresses in metal components?",
        "answer": "Two adverse consequences of these stresses are distortion (or warpage) and fracture.",
        "type": 4,
        "wrong_answers_1": "The first criterion is consequences of failure.",
        "wrong_answers_2": "Residual thermal stresses are introduced into a glass piece when it is cooled because surface and interior regions cool at different rates, and, therefore, contract different amounts; since the material will experience very little, if any deformation, stresses are established.",
        "wrong_answers_3": "The coordinates of these atoms are as follows: 1/2 1/2 0, 1/2 1/2 1, 1/2 0 1/2, 0 1/2 1/2, and 1/2 1/2 1/2."
    },
    {
        "idx": 845,
        "question": "Give the approximate minimum temperature at which it is possible to austenitize a 0.20 wt% C iron-carbon alloy during a normalizing heat treatment.",
        "answer": "At least 905°C (1660°F)",
        "type": 1,
        "wrong_answers_1": "At least 1032°C (1510°F)",
        "wrong_answers_2": "At least 847°C (1740°F)",
        "wrong_answers_3": "At least 967°C (1440°F)"
    },
    {
        "idx": 846,
        "question": "Give the approximate minimum temperature at which it is possible to austenitize a 0.76 wt% C iron-carbon alloy during a normalizing heat treatment.",
        "answer": "At least 782°C (1440°F)",
        "type": 1,
        "wrong_answers_1": "At least 884°C (1310°F)",
        "wrong_answers_2": "At least 891°C (1580°F)",
        "wrong_answers_3": "At least 680°C (1500°F)"
    },
    {
        "idx": 847,
        "question": "Give the approximate minimum temperature at which it is possible to austenitize a 0.95 wt% C iron-carbon alloy during a normalizing heat treatment.",
        "answer": "At least 840°C (1545°F)",
        "type": 1,
        "wrong_answers_1": "At least 716°C (1384°F)",
        "wrong_answers_2": "At least 779°C (1465°F)",
        "wrong_answers_3": "At least 880°C (1465°F)"
    },
    {
        "idx": 848,
        "question": "Give the approximate temperature at which it is desirable to heat a 0.25 wt% C iron-carbon alloy during a full anneal heat treatment.",
        "answer": "About 880°C (1510°F)",
        "type": 4,
        "wrong_answers_1": "About 777°C (1430°F)",
        "wrong_answers_2": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration.",
        "wrong_answers_3": "White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification. Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700°C may be necessary to produce a ferritic matrix."
    },
    {
        "idx": 849,
        "question": "Give the approximate temperature at which it is desirable to heat a 0.45 wt% C iron-carbon alloy during a full anneal heat treatment.",
        "answer": "About 830°C (1525°F)",
        "type": 1,
        "wrong_answers_1": "About 908°C (1425°F)",
        "wrong_answers_2": "About 881°C (1345°F)",
        "wrong_answers_3": "About 786°C (1696°F)"
    },
    {
        "idx": 850,
        "question": "Give the approximate temperature at which it is desirable to heat a 0.85 wt% C iron-carbon alloy during a full anneal heat treatment.",
        "answer": "About 777°C (1430°F)",
        "type": 1,
        "wrong_answers_1": "About 680°C (1490°F)",
        "wrong_answers_2": "About 727°C (1500°F)",
        "wrong_answers_3": "About 855°C (1630°F)"
    },
    {
        "idx": 851,
        "question": "Give the approximate temperature at which it is desirable to heat a 1.10 wt% C iron-carbon alloy during a full anneal heat treatment.",
        "answer": "About 777°C (1430°F)",
        "type": 4,
        "wrong_answers_1": "About 880°C (1510°F)",
        "wrong_answers_2": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration.",
        "wrong_answers_3": "White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification. Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700°C may be necessary to produce a ferritic matrix."
    },
    {
        "idx": 852,
        "question": "What is the purpose of a spheroidizing heat treatment?",
        "answer": "The purpose of a spheroidizing heat treatment is to produce a very soft and ductile steel alloy having a spheroiditic microstructure.",
        "type": 4,
        "wrong_answers_1": "Three variables that determine the microstructure of an alloy are (1) the alloying elements present, (2) the concentrations of these alloying elements, and (3) the heat treatment of the alloy.",
        "wrong_answers_2": "White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification. Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700°C may be necessary to produce a ferritic matrix.",
        "wrong_answers_3": "The matrix phase must be ductile and is usually relatively soft, whereas the fiber phase must be stiff and strong."
    },
    {
        "idx": 853,
        "question": "On what classes of alloys is a spheroidizing heat treatment normally used?",
        "answer": "It is normally used on medium- and high-carbon steels, which, by virtue of carbon content, are relatively hard and strong.",
        "type": 4,
        "wrong_answers_1": "Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, High Alloy Steels (Stainless and Tool)",
        "wrong_answers_2": "The four classifications of steels are: Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, and High Alloy Steels (Stainless and Tool).",
        "wrong_answers_3": "A \"hypoeutectoid\" steel has a carbon concentration less than the eutectoid; on the other hand, a \"hypereutectoid\" steel has a carbon content greater than the eutectoid."
    },
    {
        "idx": 854,
        "question": "Briefly explain the difference between hardness and hardenability.",
        "answer": "Hardness is a measure of a material's resistance to localized surface deformation, whereas hardenability is a measure of the depth to which a ferrous alloy may be hardened by the formation of martensite. Hardenability is determined from hardness tests.",
        "type": 4,
        "wrong_answers_1": "The presence of alloying elements (other than carbon) causes a much more gradual decrease in hardness with position from the quenched end for a hardenability curve. The reason for this effect is that alloying elements retard the formation of pearlitic and bainitic structures which are not as hard as martensite.",
        "wrong_answers_2": "The mobility is the proportionality constant between the drift velocity and the electric field. It is also a measure of the frequency of scattering events (and is inversely proportional to the frequency of scattering).",
        "wrong_answers_3": "The hardness of tempered martensite diminishes with increasing temperature at constant tempering time because higher temperatures accelerate the rate of cementite particle growth. This growth reduces the ferrite-cementite phase boundary area, which acts as barriers to dislocation motion. Consequently, the reduction in phase boundary area results in a decrease in hardness."
    },
    {
        "idx": 855,
        "question": "What influence does the presence of alloying elements (other than carbon) have on the shape of a herdenability curve? Briefly explain this effect.",
        "answer": "The presence of alloying elements (other than carbon) causes a much more gradual decrease in hardness with position from the quenched end for a hardenability curve. The reason for this effect is that alloying elements retard the formation of pearlitic and bainitic structures which are not as hard as martensite.",
        "type": 4,
        "wrong_answers_1": "Three variables that determine the microstructure of an alloy are (1) the alloying elements present, (2) the concentrations of these alloying elements, and (3) the heat treatment of the alloy.",
        "wrong_answers_2": "The alloying elements in tool steels (e.g., \\mathrm{Cr}, V, \\mathrm{W}, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds.",
        "wrong_answers_3": "Hardness is a measure of a material's resistance to localized surface deformation, whereas hardenability is a measure of the depth to which a ferrous alloy may be hardened by the formation of martensite. Hardenability is determined from hardness tests."
    },
    {
        "idx": 856,
        "question": "Name two thermal properties of a liquid medium that will influence its quenching effectiveness.",
        "answer": "The two thermal properties of a liquid medium that influence its quenching effectiveness are thermal conductivity and heat capacity.",
        "type": 4,
        "wrong_answers_1": "Increasing the degree of crystallinity of a semicrystalline polymer enhances its thermal conductivity; the vibrations, rotations, etc. of the molecular chains are more effective modes of thermal transport when a crystalline structure prevails.",
        "wrong_answers_2": "Porosity decreases the thermal conductivity of ceramic and polymeric materials because the thermal conductivity of a gas phase that occupies pore space is extremely small relative to that of the solid material. Furthermore, contributions from gaseous convection are generally insignificant.",
        "wrong_answers_3": "For refractory ceramic materials, three characteristics that improve with increasing porosity are (1) decreased thermal expansion and contraction upon thermal cycling, (2) improved thermal insulation, and (3) improved resistance to thermal shock."
    },
    {
        "idx": 857,
        "question": "For a ceramic compound, what are the two characteristics of the component ions that determine the crystal structure?",
        "answer": "The two characteristics of component ions that determine the crystal structure of a ceramic compound are: 1) the magnitude of the electrical charge on each ion, and 2) the relative sizes of the cations and anions.",
        "type": 4,
        "wrong_answers_1": "In ionic materials, the larger the size of the component ions the greater the degree of electronic polarization.",
        "wrong_answers_2": "Stoichiometric means having exactly the ratio of anions to cations as specified by the chemical formula for the compound.",
        "wrong_answers_3": "Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material."
    },
    {
        "idx": 858,
        "question": "The zinc blende crystal structure is one that may be generated from close-packed planes of anions. Will the stacking sequence for this structure be FCC or HCP? Why?",
        "answer": "The stacking sequence of close-packed planes of anions for the zinc blende crystal structure will be the same as FCC (and not HCP) because the anion packing is FCC.",
        "type": 4,
        "wrong_answers_1": "BeO has the zinc blende structure.",
        "wrong_answers_2": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP.",
        "wrong_answers_3": "All metals do not have the same slip system. The reason for this is that for most metals, the slip system will consist of the most densely packed crystallographic plane, and within that plane the most closely packed direction. This plane and direction will vary from crystal structure to crystal structure."
    },
    {
        "idx": 859,
        "question": "The zinc blende crystal structure is one that may be generated from close-packed planes of anions. Will cations fill tetrahedral or octahedral positions? Why?",
        "answer": "The cations will fill tetrahedral positions since the coordination number for cations is four.",
        "type": 4,
        "wrong_answers_1": "Stoichiometric means having exactly the ratio of anions to cations as specified by the chemical formula for the compound.",
        "wrong_answers_2": "Only one-half of the tetrahedral positions will be occupied because there are two tetrahedral sites per anion, and yet only one cation per anion.",
        "wrong_answers_3": "The two characteristics of component ions that determine the crystal structure of a ceramic compound are: 1) the magnitude of the electrical charge on each ion, and 2) the relative sizes of the cations and anions."
    },
    {
        "idx": 860,
        "question": "The zinc blende crystal structure is one that may be generated from close-packed planes of anions. What fraction of the positions will be occupied?",
        "answer": "Only one-half of the tetrahedral positions will be occupied because there are two tetrahedral sites per anion, and yet only one cation per anion.",
        "type": 4,
        "wrong_answers_1": "Since both Fe2+ and Ti4+ ions occupy octahedral sites, no tetrahedral sites will be occupied.",
        "wrong_answers_2": "The cations will fill tetrahedral positions since the coordination number for cations is four.",
        "wrong_answers_3": "only martensite forms."
    },
    {
        "idx": 861,
        "question": "Cadmium sulfide (CdS) has a cubic unit cell, and from x-ray diffraction data it is known that the cell edge length is 0.582nm. If the measured density is 4.82g / {cm}^{3}, how many \\mathrm{Cd}^{2+} and \\mathrm{S}^{2-} ions are there per unit cell?",
        "answer": "there are four \\mathrm{cd}^{2+} and four s^{2-} ions per unit cell.",
        "type": 1,
        "wrong_answers_1": "there are four \\mathrm{cd}^{2+} and four s^{2-} ions per unit cell.",
        "wrong_answers_2": "there are four \\mathrm{cd}^{2+} and four s^{2-} ions per unit cell.",
        "wrong_answers_3": "there are four \\mathrm{cd}^{2+} and four s^{2-} ions per unit cell."
    },
    {
        "idx": 862,
        "question": "A hypothetical A X type of ceramic material is known to have a density of 2.65g / {cm}^{3} and a unit cell of cubic symmetry with a cell edge length of 0.43nm. The atomic weights of the A and X elements are 86.6 and 40.3g / mol, respectively. On the basis of this information, which of the following crystal structures is (are) possible for this material: rock salt, cesium chloride, or zinc blende? Justify your choice(s).",
        "answer": "the only possible crystal structure for this material is cesium chloride, which has one formula unit per unit cell.",
        "type": 2
    },
    {
        "idx": 863,
        "question": "In terms of bonding, explain why silicate materials have relatively low densities.",
        "answer": "The silicate materials have relatively low densities because the atomic bonds are primarily covalent in nature, and, therefore, directional. This limits the packing efficiency of the atoms, and therefore, the magnitude of the density.",
        "type": 4,
        "wrong_answers_1": "The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar density (Section 3.11)] - that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.54, planar densities for FCC (100) and (111) planes are \\frac{1}{4 R^{2}} and \\frac{1}{2 R^{2} \\sqrt{3}}, respectively - that is \\frac{0.25}{R^{2}} and \\frac{0.29}{R^{2}} (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have the lower surface energy.",
        "wrong_answers_2": "The surface energy for a crystallographic plane will depend on its packing density-that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.55, the planar densities for BCC (100) and (110) are \\frac{3}{16 R^{2}} and \\frac{3}{8 R^{2} \\sqrt{2}}, respectively-that is \\frac{0.19}{R^{2}} and \\frac{0.27}{R^{2}}. Thus, since the planar density for (110) is greater, it will have the lower surface energy.",
        "wrong_answers_3": "The two characteristics of component ions that determine the crystal structure of a ceramic compound are: 1) the magnitude of the electrical charge on each ion, and 2) the relative sizes of the cations and anions."
    },
    {
        "idx": 864,
        "question": "Would you expect Frenkel defects for anions to exist in ionic ceramics in relatively large concentrations? Why or why not?",
        "answer": "Frenkel defects for anions would not exist in appreciable concentrations because the anion is quite large and is highly unlikely to exist as an interstitial.",
        "type": 4,
        "wrong_answers_1": "Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller, are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent to a host (or substitutional impurity) atom.",
        "wrong_answers_2": "The stacking sequence of close-packed planes of anions for the zinc blende crystal structure will be the same as FCC (and not HCP) because the anion packing is FCC.",
        "wrong_answers_3": "In general, the matrix phase is relatively weak, has a low elastic modulus, but is quite ductile. On the other hand, the fiber phase is normally quite strong, stiff, and brittle."
    },
    {
        "idx": 865,
        "question": "Calculate the fraction of lattice sites that are Schottky defects for sodium chloride at its melting temperature \\left(801^{\\circ} C\\right). Assume an energy for defect formation of 2.3 \\mathrm{eV}.",
        "answer": "the fraction of lattice sites that are schottky defects for sodium chloride at its melting temperature is 4.03 × 10^{-6}.",
        "type": 1,
        "wrong_answers_1": "the fraction of lattice sites that are schottky defects for sodium chloride at its melting temperature is 4.44 × 10^{-6}.",
        "wrong_answers_2": "the fraction of lattice sites that are schottky defects for sodium chloride at its melting temperature is 4.47 × 10^{-6}.",
        "wrong_answers_3": "the fraction of lattice sites that are schottky defects for sodium chloride at its melting temperature is 4.33 × 10^{-6}."
    },
    {
        "idx": 866,
        "question": "(a) Suppose that Li2O is added as an impurity to CaO. If the Li+ substitutes for Ca2+, what kind of vacancies would you expect to form? How many of these vacancies are created for every Li+ added?",
        "answer": "oxygen vacancies. Two Li+ ions added, a single oxygen vacancy is formed.",
        "type": 4,
        "wrong_answers_1": "One magnesium vacancy would be formed for every two Al3+ ions added.",
        "wrong_answers_2": "diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen.",
        "wrong_answers_3": "For Ca2+ substituting for Li+ in Li2O, lithium vacancies would be created. For each Ca2+ substituting for Li+, one positive charge is added; in order to maintain charge neutrality, a single positive charge may be removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca2+ ion added, a single lithium vacancy is formed."
    },
    {
        "idx": 867,
        "question": "(b) Suppose that CaCl2 is added as an impurity to CaO. If the Cl substitutes for O2-, what kind of vacancies would you expect to form? How many of the vacancies are created for every Cl- added?",
        "answer": "calcium vacancies. Two Cl- ions will lead to the formation of one calcium vacancy.",
        "type": 4,
        "wrong_answers_1": "For O2- substituting for Cl- in CaCl2, chlorine vacancies would be created. For each O2- substituting for a Cl-, one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to maintain charge neutrality, one O2- ion will lead to the formation of one chlorine vacancy.",
        "wrong_answers_2": "oxygen vacancies. Two Li+ ions added, a single oxygen vacancy is formed.",
        "wrong_answers_3": "One magnesium vacancy would be formed for every two Al3+ ions added."
    },
    {
        "idx": 868,
        "question": "What point defects are possible for Al2O3 as an impurity in MgO?",
        "answer": "For every Al3+ ion that substitutes for Mg2+ in MgO, a single positive charge is added. Thus, in order to maintain charge neutrality, either a positive charge must be removed or a negative charge must be added. Negative charges are added by forming O2- interstitials, which are not likely to form. Positive charges may be removed by forming Mg2+ vacancies.",
        "type": 4,
        "wrong_answers_1": "For a Cu2+ O2- compound in which a small fraction of the copper ions exist as Cu+, for each Cu+ formed there is one less positive charge introduced (or one more negative charge). In order to maintain charge neutrality, we must either add an additional positive charge or subtract a negative charge. This may be accomplished be either creating Cu2+ interstitials or O2- vacancies.",
        "wrong_answers_2": "For Ca2+ substituting for Li+ in Li2O, lithium vacancies would be created. For each Ca2+ substituting for Li+, one positive charge is added; in order to maintain charge neutrality, a single positive charge may be removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca2+ ion added, a single lithium vacancy is formed.",
        "wrong_answers_3": "For O2- substituting for Cl- in CaCl2, chlorine vacancies would be created. For each O2- substituting for a Cl-, one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to maintain charge neutrality, one O2- ion will lead to the formation of one chlorine vacancy."
    },
    {
        "idx": 869,
        "question": "How many Al2+ ions must be added to form each of these defects?",
        "answer": "One magnesium vacancy would be formed for every two Al3+ ions added.",
        "type": 4,
        "wrong_answers_1": "oxygen vacancies. Two Li+ ions added, a single oxygen vacancy is formed.",
        "wrong_answers_2": "For Ca2+ substituting for Li+ in Li2O, lithium vacancies would be created. For each Ca2+ substituting for Li+, one positive charge is added; in order to maintain charge neutrality, a single positive charge may be removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca2+ ion added, a single lithium vacancy is formed.",
        "wrong_answers_3": "calcium vacancies. Two Cl- ions will lead to the formation of one calcium vacancy."
    },
    {
        "idx": 870,
        "question": "When kaolinite clay [Al2(Si2O5)(OH)4] is heated to a sufficiently high temperature, chemical water is driven off. Under these circumstances, what is the composition of the remaining product (in weight percent Al2O3)?",
        "answer": "the composition of the remaining product is 45.9% Al2O3 by weight.",
        "type": 1,
        "wrong_answers_1": "the composition of the remaining product is 52.7% Al2O3 by weight.",
        "wrong_answers_2": "the composition of the remaining product is 47.7% Al2O3 by weight.",
        "wrong_answers_3": "the composition of the remaining product is 51.5% Al2O3 by weight."
    },
    {
        "idx": 871,
        "question": "When kaolinite clay [Al2(Si2O5)(OH)4] is heated to a sufficiently high temperature, chemical water is driven off. What are the liquidus and solidus temperatures of this material?",
        "answer": "the liquidus and solidus temperatures of this material are 1825 degrees C and 1587 degrees C, respectively.",
        "type": 4,
        "wrong_answers_1": "for w: 1200 degrees c (2190 degrees f)",
        "wrong_answers_2": "for al: 100 degrees c (212 degrees f)",
        "wrong_answers_3": "for cu: 270 degrees c (518 degrees f)"
    },
    {
        "idx": 872,
        "question": "Why may there be significant scatter in the fracture strength for some given ceramic material?",
        "answer": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material.",
        "type": 4,
        "wrong_answers_1": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material.",
        "wrong_answers_2": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probability of the existence of a flaw that is capable of initiating a crack diminishes.",
        "wrong_answers_3": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probably of the existence of a flaw of that is capable of initiating a crack diminishes."
    },
    {
        "idx": 873,
        "question": "Why does fracture strength increase with decreasing specimen size?",
        "answer": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probability of the existence of a flaw that is capable of initiating a crack diminishes.",
        "type": 4,
        "wrong_answers_1": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probably of the existence of a flaw of that is capable of initiating a crack diminishes.",
        "wrong_answers_2": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material.",
        "wrong_answers_3": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material."
    },
    {
        "idx": 874,
        "question": "The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter surface crack geometry (i.e., reduce crack length and increase the tip radius). Compute the ratio of the original and etched crack tip radii for an eightfold increase in fracture strength if two-thirds of the crack length is removed.\n\\[\n\\text {",
        "answer": "the ratio of the original and etched crack tip radii is 21.3.",
        "type": 1,
        "wrong_answers_1": "the ratio of the original and etched crack tip radii is 22.3.",
        "wrong_answers_2": "the ratio of the original and etched crack tip radii is 19.5.",
        "wrong_answers_3": "the ratio of the original and etched crack tip radii is 22.3."
    },
    {
        "idx": 875,
        "question": "A circular specimen of MgO is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 425N\\left(95.5 lb \\mathrm{b}_{1}\\right), the flexural strength is 105 MPa(15,000 psi), and the separation between load points is 50 mm(2.0 in).",
        "answer": "the minimum possible radius of the specimen without fracture is 4.0mm (0.16 in).",
        "type": 1,
        "wrong_answers_1": "the minimum possible radius of the specimen without fracture is 4.6mm (0.17 in).",
        "wrong_answers_2": "the minimum possible radius of the specimen without fracture is 4.5mm (0.15 in).",
        "wrong_answers_3": "the minimum possible radius of the specimen without fracture is 3.6mm (0.18 in)."
    },
    {
        "idx": 876,
        "question": "A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 390 MPa (56,600 psi). If the specimen radius is 2.5mm (0.10 in.) and the support point separation distance is 30mm (1.2 in.), predict whether or not you would expect the specimen to fracture when a load of 620 N (140 lbf) is applied. Justify your prediction.",
        "answer": "the calculated stress is 379 mpa (53,500 psi), which is less than the flexural strength of 390 mpa. therefore, fracture is not predicted.",
        "type": 1,
        "wrong_answers_1": "the calculated stress is 327 mpa (48749 psi), which is less than the flexural strength of 434 mpa. therefore, fracture is not predicted.",
        "wrong_answers_2": "the calculated stress is 346 mpa (49407 psi), which is less than the flexural strength of 425 mpa. therefore, fracture is not predicted.",
        "wrong_answers_3": "the calculated stress is 411 mpa (57189 psi), which is less than the flexural strength of 370 mpa. therefore, fracture is not predicted."
    },
    {
        "idx": 877,
        "question": "Would you be 100% certain of the prediction in part (a)? Why or why not?",
        "answer": "the certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials. since the calculated stress 379 mpa is relatively close to the flexural strength 390 mpa, there is some chance that fracture will occur.",
        "type": 4,
        "wrong_answers_1": "the certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials, and the calculated stress is relatively close to the flexural strength, so there is some chance that fracture will occur.",
        "wrong_answers_2": "the applied stress is 61.5 MPa, which is less than the flexural strength of 85 MPa; the polymer is not expected to fracture.",
        "wrong_answers_3": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material."
    },
    {
        "idx": 878,
        "question": "Cite one reason why ceramic materials are, in general, harder yet more brittle than metals.",
        "answer": "Crystalline ceramics are harder yet more brittle than metals because they (ceramics) have fewer slip systems, and, therefore, dislocation motion is highly restricted.",
        "type": 4,
        "wrong_answers_1": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP.",
        "wrong_answers_2": "Thermal conductivities are higher for crystalline than for noncrystalline ceramics because, for noncrystalline, phonon scattering, and thus the resistance to heat transport, is much more effective due to the highly disordered and irregular atomic structure.",
        "wrong_answers_3": "For slip, the atomic displacements occur in atomic spacing multiples, whereas for twinning, these displacements may be other than by atomic spacing multiples. Slip occurs in metals having many slip systems, whereas twinning occurs in metals having relatively few slip systems."
    },
    {
        "idx": 879,
        "question": "The modulus of elasticity for beryllium oxide (BeO) having 5 vol% porosity is 310 GPa (45 x 10^6 psi). Compute the modulus of elasticity for the nonporous material.",
        "answer": "the modulus of elasticity for the nonporous material is 342 gpa (49.6 x 10^6 psi).",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity for the nonporous material is 360 gpa (44.4 x 10^6 psi).",
        "wrong_answers_2": "the modulus of elasticity for the nonporous material is 389 gpa (52.0 x 10^6 psi).",
        "wrong_answers_3": "the modulus of elasticity for the nonporous material is 362 gpa (44.3 x 10^6 psi)."
    },
    {
        "idx": 880,
        "question": "The modulus of elasticity for beryllium oxide (BeO) having 5 vol% porosity is 310 GPa (45 x 10^6 psi). Compute the modulus of elasticity for 10 vol% porosity.",
        "answer": "the modulus of elasticity for 10 vol% porosity is 280 gpa (40.6 x 10^6 psi).",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity for 9 vol% porosity is 270 gpa (35.2 x 10^6 psi).",
        "wrong_answers_2": "the modulus of elasticity for 9 vol% porosity is 307 gpa (35.3 x 10^6 psi).",
        "wrong_answers_3": "the modulus of elasticity for 11 vol% porosity is 311 gpa (34.8 x 10^6 psi)."
    },
    {
        "idx": 881,
        "question": "The modulus of elasticity for boron carbide (B4C) having 5 vol% porosity is 290 GPa (42 x 10^6 psi). Compute the modulus of elasticity for the nonporous material.",
        "answer": "320 GPa (46.3 x 10^6 psi)",
        "type": 1,
        "wrong_answers_1": "307 GPa (41.9 x 10^6 psi)",
        "wrong_answers_2": "279 GPa (51.0 x 10^6 psi)",
        "wrong_answers_3": "337 GPa (48.0 x 10^6 psi)"
    },
    {
        "idx": 882,
        "question": "The modulus of elasticity for boron carbide (B4C) having 5 vol% porosity is 290 GPa (42 x 10^6 psi). At what volume percent porosity will the modulus of elasticity be 235 GPa (34 x 10^6 psi)?",
        "answer": "the modulus of elasticity will be 235 GPa (34 x 10^6 psi) at a porosity of 15.1 vol%",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity will be 215 GPa (29 x 10^6 psi) at a porosity of 17.2 vol%",
        "wrong_answers_2": "the modulus of elasticity will be 259 GPa (37 x 10^6 psi) at a porosity of 17.2 vol%",
        "wrong_answers_3": "the modulus of elasticity will be 214 GPa (32 x 10^6 psi) at a porosity of 16.9 vol%"
    },
    {
        "idx": 883,
        "question": "Cite the two desirable characteristics of glasses.",
        "answer": "Two desirable characteristics of glasses are optical transparency and ease of fabrication.",
        "type": 4,
        "wrong_answers_1": "Two desirable characteristics of clay minerals relative to fabrication processes are (1) they become hydroplastic (and therefore formable) when mixed with water; and (2) during firing, clays melt over a range of temperatures, which allows some fusion and bonding of the ware without complete melting and a loss of mechanical integrity and shape.",
        "wrong_answers_2": "The two characteristics of component ions that determine the crystal structure of a ceramic compound are: 1) the magnitude of the electrical charge on each ion, and 2) the relative sizes of the cations and anions.",
        "wrong_answers_3": "Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of crystallinity."
    },
    {
        "idx": 884,
        "question": "What is crystallization?",
        "answer": "Crystallization is the process whereby a glass material is caused to transform to a crystalline solid, usually as a result of a heat treatment.",
        "type": 4,
        "wrong_answers_1": "Atomic structure relates to the number of protons and neutrons in the nucleus of an atom, as well as the number and probability distributions of the constituent electrons. On the other hand, crystal structure pertains to the arrangement of atoms in the crystalline solid material.",
        "wrong_answers_2": "White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification. Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700°C may be necessary to produce a ferritic matrix.",
        "wrong_answers_3": "The purpose of a spheroidizing heat treatment is to produce a very soft and ductile steel alloy having a spheroiditic microstructure."
    },
    {
        "idx": 885,
        "question": "Cite two properties that may be improved by crystallization.",
        "answer": "Two properties that may be improved by crystallization are (1) a lower coefficient of thermal expansion, and (2) higher strength.",
        "type": 4,
        "wrong_answers_1": "For refractory ceramic materials, three characteristics that improve with increasing porosity are (1) decreased thermal expansion and contraction upon thermal cycling, (2) improved thermal insulation, and (3) improved resistance to thermal shock.",
        "wrong_answers_2": "The two thermal properties of a liquid medium that influence its quenching effectiveness are thermal conductivity and heat capacity.",
        "wrong_answers_3": "Borosilicate glasses and fused silica are resistant to thermal shock because they have relatively low coefficients of thermal expansion; therefore, upon heating or cooling, the difference in the degree of expansion or contraction across a cross-section of a ware that is constructed from these materials will be relatively low."
    },
    {
        "idx": 886,
        "question": "For refractory ceramic materials, cite three characteristics that improve with increasing porosity.",
        "answer": "For refractory ceramic materials, three characteristics that improve with increasing porosity are (1) decreased thermal expansion and contraction upon thermal cycling, (2) improved thermal insulation, and (3) improved resistance to thermal shock.",
        "type": 4,
        "wrong_answers_1": "Thermal insulation increases.",
        "wrong_answers_2": "Two properties that may be improved by crystallization are (1) a lower coefficient of thermal expansion, and (2) higher strength.",
        "wrong_answers_3": "Borosilicate glasses and fused silica are resistant to thermal shock because they have relatively low coefficients of thermal expansion; therefore, upon heating or cooling, the difference in the degree of expansion or contraction across a cross-section of a ware that is constructed from these materials will be relatively low."
    },
    {
        "idx": 887,
        "question": "For refractory ceramic materials, cite two characteristics that are adversely affected by increasing porosity.",
        "answer": "For refractory ceramic materials, two characteristics that are adversely affected by increasing porosity are (1) load-bearing capacity and (2) resistance to attack by corrosive materials.",
        "type": 4,
        "wrong_answers_1": "For refractory ceramic materials, three characteristics that improve with increasing porosity are (1) decreased thermal expansion and contraction upon thermal cycling, (2) improved thermal insulation, and (3) improved resistance to thermal shock.",
        "wrong_answers_2": "Five important characteristics for polymers that are to be used in thin-film applications are: (1) low density; (2) high flexibility; (3) high tensile and tear strengths; (4) resistance to moisture/chemical attack; and (5) low gas permeability.\n}",
        "wrong_answers_3": "Metals are typically better thermal conductors than are ceramic materials because, for metals, most of the heat is transported by free electrons (of which there are relatively large numbers). In ceramic materials, the primary mode of thermal conduction is via phonons, and phonons are more easily scattered than are free electrons."
    },
    {
        "idx": 888,
        "question": "How do the aggregate particles become bonded together in clay-based mixtures during firing?",
        "answer": "For clay-based aggregates, a liquid phase forms during firing, which infiltrates the pores between the unmelted particles; upon cooling, this liquid becomes a glass, that serves as the bonding phase.",
        "type": 4,
        "wrong_answers_1": "Density will increase with degree of vitrification since the total remaining pore volume decreases. Strength will also increase with degree of vitrification inasmuch as more of the liquid phase forms, which fills in a greater fraction of pore volume. Upon cooling, the liquid forms a glass matrix of relatively high strength. Corrosion resistance normally increases also, especially at service temperatures below that at which the glass phase begins to soften. The rate of corrosion is dependent on the amount of surface area exposed to the corrosive medium; hence, decreasing the total surface area by filling in some of the surface pores, diminishes the corrosion rate. Thermal conductivity will increase with degree of vitrification. The glass phase has a higher conductivity than the pores that it has filled.",
        "wrong_answers_2": "The matrix phase is a continuous phase that surrounds the noncontinuous dispersed phase.",
        "wrong_answers_3": "(1) composition (2) the temperature of firing(3) the time at the firing temperature."
    },
    {
        "idx": 889,
        "question": "How do the aggregate particles become bonded together in cements during setting?",
        "answer": "With cements, the bonding process is a chemical, hydration reaction between the water that has been added and the various cement constituents. The cement particles are bonded together by reactions that occur at the particle surfaces.",
        "type": 4,
        "wrong_answers_1": "Concrete consists of an aggregate of particles that are bonded together by a cement.",
        "wrong_answers_2": "Clays become hydroplastic when water is added because the water molecules occupy regions between the layered molecular sheets; these water molecules essentially eliminate the secondary molecular bonds between adjacent sheets, and also form a thin film around the clay particles. The net result is that the clay particles are relatively free to move past one another, which is manifested as the hydroplasticity phenomenon.",
        "wrong_answers_3": "The driving force is that which compels a reaction to occur."
    },
    {
        "idx": 890,
        "question": "What is the glass transition temperature for a noncrystalline ceramic?",
        "answer": "The glass transition temperature is, for a noncrystalline ceramic, that temperature at which there is a change of slope for the specific volume versus temperature curve.",
        "type": 4,
        "wrong_answers_1": "The melting temperature is, for a crystalline material and upon cooling, that temperature at which there is a sudden and discontinuous decrease in the specific-volume-versus-temperature curve.",
        "wrong_answers_2": "(1) composition (2) the temperature of firing(3) the time at the firing temperature.",
        "wrong_answers_3": "For some ceramic materials, the thermal conductivity first decreases with rising temperature because the scattering of lattice vibrations increases with temperature. At higher temperatures, the thermal conductivity will increase for some ceramics that are porous because radiant heat transfer across pores may become important, which process increases with rising temperature."
    },
    {
        "idx": 891,
        "question": "What is the melting temperature for a crystalline material?",
        "answer": "The melting temperature is, for a crystalline material and upon cooling, that temperature at which there is a sudden and discontinuous decrease in the specific-volume-versus-temperature curve.",
        "type": 4,
        "wrong_answers_1": "The glass transition temperature is, for a noncrystalline ceramic, that temperature at which there is a change of slope for the specific volume versus temperature curve.",
        "wrong_answers_2": "(1) composition (2) the temperature of firing(3) the time at the firing temperature.",
        "wrong_answers_3": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The syndiotactic polypropylene will have a higher degree of crystallinity than the atactic material. On the basis of this effect alone, the syndiotactic PP should have the greater Tm, since melting temperature increases with degree of crystallinity. However, the molecular weight for the syndiotactic polypropylene (500,000 g/mol) is less than for the atactic material (750,000 g/mol); a lowering of molecular weight generally results in a reduction of melting temperature."
    },
    {
        "idx": 892,
        "question": "Compare the softening points for 96% silica, borosilicate, and soda-lime glasses.",
        "answer": "The softening point of a glass is that temperature at which the viscosity is 4 × 10^{6} \\mathrm{~Pa}·s; from Figure 13.7, these temperatures for the 96% silica, borosilicate, and soda-lime glasses are 1540^{\\circ} C\\left(2800^{\\circ} F\\right), 830^{\\circ} C\\left(1525^{\\circ} F\\right), and 700^{\\circ} C\\left(1290^{\\circ} F\\right), respectively.",
        "type": 4,
        "wrong_answers_1": "Austenitize at approximately 750^{\\circ} C,\nQuench to below 130^{\\circ} C (the M_{\\gamma} temperature)\nTemper at 620^{\\circ} C or less.",
        "wrong_answers_2": "This question asks us to name which, of several polymers, would be suitable for the fabrication of cups to contain hot coffee. At its glass transition temperature, an amorphous polymer begins to soften. The maximum temperature of hot coffee is probably slightly below 100^{\\circ} C\\left(212^{\\circ} F\\right). Of the polymers listed, only polystyrene and polycarbonate have glass transition temperatures of 100^{\\circ} C or above, and would be suitable for this application.",
        "wrong_answers_3": "Inasmuch as \\left[\\mathrm{M}_{\\mathrm{L}}^{2+}\\right] \\prec\\left[\\mathrm{M}_{H}^{2+}\\right] then the natural logarithm of the \\left[\\mathrm{M}^{2+}\\right] ratio is negative, which yields a positive value for \\Delta V. This means that the electrochemical reaction is spontaneous as written, or that oxidation occurs at the electrode having the lower \\mathrm{M}^{2+} concentration."
    },
    {
        "idx": 893,
        "question": "Explain why residual thermal stresses are introduced into a glass piece when it is cooled.",
        "answer": "Residual thermal stresses are introduced into a glass piece when it is cooled because surface and interior regions cool at different rates, and, therefore, contract different amounts; since the material will experience very little, if any deformation, stresses are established.",
        "type": 4,
        "wrong_answers_1": "Thermal stresses may be introduced into a structure by rapid heating or cooling because temperature gradients are established across the cross section due to more rapid temperature changes at the surface than within the interior; thus, the surface will expand or contract at a different rate than the interior and since this surface expansion or contraction will be restrained by the interior, stresses will be introduced.",
        "wrong_answers_2": "Three sources of residual stresses in metal components are plastic deformation processes, nonuniform cooling of a piece that was cooled from an elevated temperature, and a phase transformation in which parent and product phases have different densities.",
        "wrong_answers_3": "For cooling, the surface stresses will be tensile in nature since the interior contracts to a lesser degree than the cooler surface."
    },
    {
        "idx": 894,
        "question": "Are thermal stresses introduced upon heating? Why or why not?",
        "answer": "Yes, thermal stresses will be introduced because of thermal expansion upon heating for the same reason as for thermal contraction upon cooling.",
        "type": 3
    },
    {
        "idx": 895,
        "question": "Borosilicate glasses and fused silica are resistant to thermal shock. Why is this so?",
        "answer": "Borosilicate glasses and fused silica are resistant to thermal shock because they have relatively low coefficients of thermal expansion; therefore, upon heating or cooling, the difference in the degree of expansion or contraction across a cross-section of a ware that is constructed from these materials will be relatively low.",
        "type": 4,
        "wrong_answers_1": "For refractory ceramic materials, three characteristics that improve with increasing porosity are (1) decreased thermal expansion and contraction upon thermal cycling, (2) improved thermal insulation, and (3) improved resistance to thermal shock.",
        "wrong_answers_2": "Bimetallic materials are produced by bonding two materials having different coefficients of thermal expansion to one another, forming a laminar composite. When the temperature changes, one of the materials will expand or contract more than the other material. This difference in expansion or contraction causes the bimetallic material to change shape; if the original shape is that of a coil, then the device will coil or uncoil, depending on the direction of the temperature change.",
        "wrong_answers_3": "Thermal stresses may be introduced into a structure by rapid heating or cooling because temperature gradients are established across the cross section due to more rapid temperature changes at the surface than within the interior; thus, the surface will expand or contract at a different rate than the interior and since this surface expansion or contraction will be restrained by the interior, stresses will be introduced."
    },
    {
        "idx": 896,
        "question": "Glass pieces may also be strengthened by chemical tempering. With this procedure, the glass surface is put in a state of compression by exchanging some of the cations near the surface with other cations having a larger diameter. Suggest one type of cation that, by replacing \\mathrm{Na}^{+}, will induce chemical tempering in a soda-lime glass.",
        "answer": "Chemical tempering will be accomplished by substitution, for \\mathrm{Na}^{+}, another monovalent cation with a slightly larger diameter. Both K^{+}and \\mathrm{Cs}^{+}fill these criteria, having ionic radii of 0.138 and 0.170 nm, respectively, which are larger than the ionic radius of \\mathrm{Na}^{+}(0.102nm). In fact, soda-lime glasses are tempered by a K^{+}-\\mathrm{Na}^{+}ion exchange.",
        "type": 4,
        "wrong_answers_1": "Inasmuch as \\left[\\mathrm{M}_{\\mathrm{L}}^{2+}\\right] \\prec\\left[\\mathrm{M}_{H}^{2+}\\right] then the natural logarithm of the \\left[\\mathrm{M}^{2+}\\right] ratio is negative, which yields a positive value for \\Delta V. This means that the electrochemical reaction is spontaneous as written, or that oxidation occurs at the electrode having the lower \\mathrm{M}^{2+} concentration.",
        "wrong_answers_2": "The Na+ ion has an electron configuration the same as neon.",
        "wrong_answers_3": "The alloying elements in tool steels (e.g., \\mathrm{Cr}, V, \\mathrm{W}, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds."
    },
    {
        "idx": 897,
        "question": "Cite the two desirable characteristics of clay minerals relative to fabrication processes.",
        "answer": "Two desirable characteristics of clay minerals relative to fabrication processes are (1) they become hydroplastic (and therefore formable) when mixed with water; and (2) during firing, clays melt over a range of temperatures, which allows some fusion and bonding of the ware without complete melting and a loss of mechanical integrity and shape.",
        "type": 4,
        "wrong_answers_1": "Quartz acts as a filler material. Clay facilitates the forming operation since, when mixed with water, the mass may be made to become either hydroplastic or form a slip. Also, since clays melt over a range of temperatures, the shape of the piece being fired will be maintained. The flux facilitates the formation of a glass having a relatively low melting temperature.",
        "wrong_answers_2": "Two desirable characteristics of glasses are optical transparency and ease of fabrication.",
        "wrong_answers_3": "Clays become hydroplastic when water is added because the water molecules occupy regions between the layered molecular sheets; these water molecules essentially eliminate the secondary molecular bonds between adjacent sheets, and also form a thin film around the clay particles. The net result is that the clay particles are relatively free to move past one another, which is manifested as the hydroplasticity phenomenon."
    },
    {
        "idx": 898,
        "question": "From a molecular perspective, briefly explain the mechanism by which clay minerals become hydroplastic when water is added.",
        "answer": "Clays become hydroplastic when water is added because the water molecules occupy regions between the layered molecular sheets; these water molecules essentially eliminate the secondary molecular bonds between adjacent sheets, and also form a thin film around the clay particles. The net result is that the clay particles are relatively free to move past one another, which is manifested as the hydroplasticity phenomenon.",
        "type": 4,
        "wrong_answers_1": "The reason that drying shrinkage is greater for products having smaller clay particles is because there is more particle surface area, and, consequently, more water will surround a given volume of particles. The drying shrinkage will thus be greater as this water is removed, and as the interparticle separation decreases.",
        "wrong_answers_2": "Quartz acts as a filler material. Clay facilitates the forming operation since, when mixed with water, the mass may be made to become either hydroplastic or form a slip. Also, since clays melt over a range of temperatures, the shape of the piece being fired will be maintained. The flux facilitates the formation of a glass having a relatively low melting temperature.",
        "wrong_answers_3": "With cements, the bonding process is a chemical, hydration reaction between the water that has been added and the various cement constituents. The cement particles are bonded together by reactions that occur at the particle surfaces."
    },
    {
        "idx": 899,
        "question": "What are the three main components of a whiteware ceramic such as porcelain?",
        "answer": "clay, quartz, and a flux.",
        "type": 4,
        "wrong_answers_1": "Quartz will have a larger conductivity than fused silica because fused silica is noncrystalline (whereas quartz is crystalline) and lattice vibrations are more effectively scattered in noncrystalline materials.",
        "wrong_answers_2": "Quartz acts as a filler material. Clay facilitates the forming operation since, when mixed with water, the mass may be made to become either hydroplastic or form a slip. Also, since clays melt over a range of temperatures, the shape of the piece being fired will be maintained. The flux facilitates the formation of a glass having a relatively low melting temperature.",
        "wrong_answers_3": "Increasing the temperature should increase the diffusivity and therefore increase the diffusion flux and flow rate."
    },
    {
        "idx": 900,
        "question": "What role does each component play in the forming and firing procedures?",
        "answer": "Quartz acts as a filler material. Clay facilitates the forming operation since, when mixed with water, the mass may be made to become either hydroplastic or form a slip. Also, since clays melt over a range of temperatures, the shape of the piece being fired will be maintained. The flux facilitates the formation of a glass having a relatively low melting temperature.",
        "type": 4,
        "wrong_answers_1": "Two desirable characteristics of clay minerals relative to fabrication processes are (1) they become hydroplastic (and therefore formable) when mixed with water; and (2) during firing, clays melt over a range of temperatures, which allows some fusion and bonding of the ware without complete melting and a loss of mechanical integrity and shape.",
        "wrong_answers_2": "clay, quartz, and a flux.",
        "wrong_answers_3": "Clays become hydroplastic when water is added because the water molecules occupy regions between the layered molecular sheets; these water molecules essentially eliminate the secondary molecular bonds between adjacent sheets, and also form a thin film around the clay particles. The net result is that the clay particles are relatively free to move past one another, which is manifested as the hydroplasticity phenomenon."
    },
    {
        "idx": 901,
        "question": "Why is it so important to control the rate of drying of a ceramic body that has been hydroplastically formed or slip cast?",
        "answer": "It is important to control the rate of drying inasmuch as if the rate of drying is too rapid, there will be nonuniform shrinkage between surface and interior regions, such that warping and/or cracking of the ceramic ware may result.",
        "type": 4,
        "wrong_answers_1": "Three factors that affect the rate of drying are temperature, humidity, and rate of air flow. The rate of drying is enhanced by increasing both the temperature and rate of air flow, and by decreasing the humidity of the air.",
        "wrong_answers_2": "The reason that drying shrinkage is greater for products having smaller clay particles is because there is more particle surface area, and, consequently, more water will surround a given volume of particles. The drying shrinkage will thus be greater as this water is removed, and as the interparticle separation decreases.",
        "wrong_answers_3": "Thermal stresses may be introduced into a structure by rapid heating or cooling because temperature gradients are established across the cross section due to more rapid temperature changes at the surface than within the interior; thus, the surface will expand or contract at a different rate than the interior and since this surface expansion or contraction will be restrained by the interior, stresses will be introduced."
    },
    {
        "idx": 902,
        "question": "Cite three factors that influence the rate of drying, and explain how each affects the rate.",
        "answer": "Three factors that affect the rate of drying are temperature, humidity, and rate of air flow. The rate of drying is enhanced by increasing both the temperature and rate of air flow, and by decreasing the humidity of the air.",
        "type": 4,
        "wrong_answers_1": "It is important to control the rate of drying inasmuch as if the rate of drying is too rapid, there will be nonuniform shrinkage between surface and interior regions, such that warping and/or cracking of the ceramic ware may result.",
        "wrong_answers_2": "Increasing the temperature should increase the diffusivity and therefore increase the diffusion flux and flow rate.",
        "wrong_answers_3": "At a rate of 100°C/s, both martensite and pearlite form."
    },
    {
        "idx": 903,
        "question": "Cite one reason why drying shrinkage is greater for slip cast or hydroplastic products that have smaller clay particles.",
        "answer": "The reason that drying shrinkage is greater for products having smaller clay particles is because there is more particle surface area, and, consequently, more water will surround a given volume of particles. The drying shrinkage will thus be greater as this water is removed, and as the interparticle separation decreases.",
        "type": 4,
        "wrong_answers_1": "It is important to control the rate of drying inasmuch as if the rate of drying is too rapid, there will be nonuniform shrinkage between surface and interior regions, such that warping and/or cracking of the ceramic ware may result.",
        "wrong_answers_2": "Tempered martensite is harder and stronger inasmuch as there is much more ferrite-cementite phase boundary area for the smaller particles; thus, there is greater reinforcement of the ferrite phase, and more phase boundary barriers to dislocation motion.",
        "wrong_answers_3": "Clays become hydroplastic when water is added because the water molecules occupy regions between the layered molecular sheets; these water molecules essentially eliminate the secondary molecular bonds between adjacent sheets, and also form a thin film around the clay particles. The net result is that the clay particles are relatively free to move past one another, which is manifested as the hydroplasticity phenomenon."
    },
    {
        "idx": 904,
        "question": "Name three factors that influence the degree to which vitrification occurs in clay-based ceramic wares.",
        "answer": "(1) composition (2) the temperature of firing(3) the time at the firing temperature.",
        "type": 4,
        "wrong_answers_1": "The glass transition temperature is, for a noncrystalline ceramic, that temperature at which there is a change of slope for the specific volume versus temperature curve.",
        "wrong_answers_2": "The melting temperature is, for a crystalline material and upon cooling, that temperature at which there is a sudden and discontinuous decrease in the specific-volume-versus-temperature curve.",
        "wrong_answers_3": "For clay-based aggregates, a liquid phase forms during firing, which infiltrates the pores between the unmelted particles; upon cooling, this liquid becomes a glass, that serves as the bonding phase."
    },
    {
        "idx": 905,
        "question": "Explain how density, firing distortion, strength, corrosion resistance, and thermal conductivity are affected by the extent of vitrification.",
        "answer": "Density will increase with degree of vitrification since the total remaining pore volume decreases. Strength will also increase with degree of vitrification inasmuch as more of the liquid phase forms, which fills in a greater fraction of pore volume. Upon cooling, the liquid forms a glass matrix of relatively high strength. Corrosion resistance normally increases also, especially at service temperatures below that at which the glass phase begins to soften. The rate of corrosion is dependent on the amount of surface area exposed to the corrosive medium; hence, decreasing the total surface area by filling in some of the surface pores, diminishes the corrosion rate. Thermal conductivity will increase with degree of vitrification. The glass phase has a higher conductivity than the pores that it has filled.",
        "type": 4,
        "wrong_answers_1": "For clay-based aggregates, a liquid phase forms during firing, which infiltrates the pores between the unmelted particles; upon cooling, this liquid becomes a glass, that serves as the bonding phase.",
        "wrong_answers_2": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "wrong_answers_3": "For some ceramic materials, the thermal conductivity first decreases with rising temperature because the scattering of lattice vibrations increases with temperature. At higher temperatures, the thermal conductivity will increase for some ceramics that are porous because radiant heat transfer across pores may become important, which process increases with rising temperature."
    },
    {
        "idx": 906,
        "question": "Some ceramic materials are fabricated by hot isostatic pressing. Cite some of the limitations and difficulties associated with this technique.",
        "answer": "The principal disadvantage of hot-isostatic pressing is that it is expensive. The pressure is applied on a pre-formed green piece by a gas. Thus, the process is slow, and the equipment required to supply the gas and withstand the elevated temperature and pressure is costly.",
        "type": 4,
        "wrong_answers_1": "The head for a carpenter's hammer is produced by forging, a metalworking process; a simple steel shape is heated and formed in several steps while hot into the required shape. The head is then heat treated to produce the required mechanical and physical properties.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}) electron configuration is that of an inert gas because of filled (3 s) and (3 p) subshells.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}) electron configuration is that of an inert gas because of filled (4 s) and (4 p) subshells."
    },
    {
        "idx": 907,
        "question": "Compute repeat unit molecular weight for poly(vinyl chloride)",
        "answer": "62.49g/mol",
        "type": 1,
        "wrong_answers_1": "67.87g/mol",
        "wrong_answers_2": "66.81g/mol",
        "wrong_answers_3": "60.59g/mol"
    },
    {
        "idx": 908,
        "question": "Compute repeat unit molecular weight for poly(ethylene terephthalate)",
        "answer": "192.16g/mol",
        "type": 1,
        "wrong_answers_1": "215.03g/mol",
        "wrong_answers_2": "219.16g/mol",
        "wrong_answers_3": "166.61g/mol"
    },
    {
        "idx": 909,
        "question": "Compute repeat unit molecular weight for polycarbonate",
        "answer": "254.27g/mol",
        "type": 1,
        "wrong_answers_1": "226.25g/mol",
        "wrong_answers_2": "284.73g/mol",
        "wrong_answers_3": "263.43g/mol"
    },
    {
        "idx": 910,
        "question": "Compute repeat unit molecular weight for polydimethylsiloxane",
        "answer": "74.16g/mol",
        "type": 1,
        "wrong_answers_1": "64.49g/mol",
        "wrong_answers_2": "71.58g/mol",
        "wrong_answers_3": "65.85g/mol"
    },
    {
        "idx": 911,
        "question": "The number-average molecular weight of a polypropylene is 1,000,000g / mol. Compute the degree of polymerization.",
        "answer": "23760",
        "type": 1,
        "wrong_answers_1": "22566",
        "wrong_answers_2": "25767",
        "wrong_answers_3": "20862"
    },
    {
        "idx": 912,
        "question": "Compute the repeat unit molecular weight of polystyrene.",
        "answer": "the repeat unit molecular weight of polystyrene is 104.14 g/mol.",
        "type": 1,
        "wrong_answers_1": "the repeat unit molecular weight of polystyrene is 114.69 g/mol.",
        "wrong_answers_2": "the repeat unit molecular weight of polystyrene is 100.65 g/mol.",
        "wrong_answers_3": "the repeat unit molecular weight of polystyrene is 97.69 g/mol."
    },
    {
        "idx": 913,
        "question": "Compute the number-average molecular weight for a polystyrene for which the degree of polymerization is 25,000.",
        "answer": "2.60 x 10^6 g/mol.",
        "type": 1,
        "wrong_answers_1": "2.86 x 10^6 g/mol.",
        "wrong_answers_2": "2.73 x 10^6 g/mol.",
        "wrong_answers_3": "2.77 x 10^6 g/mol."
    },
    {
        "idx": 914,
        "question": "High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen. Determine the concentration of Cl (in wt%) that must be added if this substitution occurs for 5% of all the original hydrogen atoms.",
        "answer": "the concentration of Cl that must be added is 20.3 wt%.",
        "type": 1,
        "wrong_answers_1": "the concentration of Cl that must be added is 19.4 wt%.",
        "wrong_answers_2": "the concentration of Cl that must be added is 23.3 wt%.",
        "wrong_answers_3": "the concentration of Cl that must be added is 23.0 wt%."
    },
    {
        "idx": 915,
        "question": "In what ways does this chlorinated polyethylene differ from poly(vinyl chloride)?",
        "answer": "chlorinated polyethylene differs from poly(vinyl chloride) in that, for pvc, (1) 25% of the side-bonding sites are substituted with Cl, and (2) the substitution is probably much less random.",
        "type": 4,
        "wrong_answers_1": "chlorinated polyethylene differs from poly(vinyl chloride) in that, for pvc, (1) 25% of the side-bonding sites are substituted with Cl, and (2) the substitution is probably much less random.",
        "wrong_answers_2": "Yes, for these two polymers it is possible to decide. The linear and syndiotactic poly(vinyl chloride) is more likely to crystallize; the phenyl side-group for polystyrene is bulkier than the Cl side-group for poly(vinyl chloride). Syndiotactic and isotactic isomers are equally likely to crystallize.",
        "wrong_answers_3": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity."
    },
    {
        "idx": 916,
        "question": "Make comparisons of thermoplastic and thermosetting polymers on the basis of mechanical characteristics upon heating.",
        "answer": "Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers harden upon heating, while further heating will not lead to softening.",
        "type": 4,
        "wrong_answers_1": "Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers harden upon heating, while further heating will not lead to softening.",
        "wrong_answers_2": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked.",
        "wrong_answers_3": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked."
    },
    {
        "idx": 917,
        "question": "Make comparisons of thermoplastic and thermosetting polymers according to possible molecular structures.",
        "answer": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked.",
        "type": 4,
        "wrong_answers_1": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked.",
        "wrong_answers_2": "No, it is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize.",
        "wrong_answers_3": "Yes, it is possible. The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's crystallinity leads to an increase in its tensile modulus. In addition, tensile modulus is independent of molecular weight - the atactic/branched material has the higher molecular weight."
    },
    {
        "idx": 918,
        "question": "Is it possible to grind up and reuse phenol-formaldehyde? Why or why not?",
        "answer": "It is not possible to grind up and reuse phenol-formaldehyde because it is a network thermoset polymer and, therefore, is not amenable to remolding.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to grind up and reuse polypropylene since it is a thermoplastic polymer, will soften when reheated, and, thus, may be remolded.",
        "wrong_answers_2": "Yes, it is possible to decide for these two polymers. The linear and isotactic polystyrene is more likely to crystallize than network phenol-formaldehyde; network polymers rarely crystallize, whereas isotactic ones crystallize relatively easily.",
        "wrong_answers_3": "No, it is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize."
    },
    {
        "idx": 919,
        "question": "Is it possible to grind up and reuse polypropylene? Why or why not?",
        "answer": "Yes, it is possible to grind up and reuse polypropylene since it is a thermoplastic polymer, will soften when reheated, and, thus, may be remolded.",
        "type": 4,
        "wrong_answers_1": "It is not possible to grind up and reuse phenol-formaldehyde because it is a network thermoset polymer and, therefore, is not amenable to remolding.",
        "wrong_answers_2": "Yes, it is possible to determine which polymer has the higher melting temperature. The polypropylene will have the higher Tm because it has a bulky phenyl side group in its repeat unit structure, which is absent in the polyethylene. Furthermore, the polypropylene has a higher degree of polymerization.",
        "wrong_answers_3": "Yes, it is possible. The linear and isotactic polypropylene will have a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. "
    },
    {
        "idx": 920,
        "question": "The number-average molecular weight of a poly(styrene-butadiene) alternating copolymer is 1,350,000g / mol; determine the average number of styrene and butadiene repeat units per molecule.",
        "answer": "the average number of styrene and butadiene repeat units per molecule is 8530.",
        "type": 1,
        "wrong_answers_1": "the average number of styrene and butadiene repeat units per molecule is 7440.",
        "wrong_answers_2": "the average number of styrene and butadiene repeat units per molecule is 9000.",
        "wrong_answers_3": "the average number of styrene and butadiene repeat units per molecule is 9120."
    },
    {
        "idx": 921,
        "question": "Calculate the number-average molecular weight of a random nitrile rubber [poly(acrylonitrilebutadiene) copolymer] in which the fraction of butadiene repeat units is 0.30 ; assume that this concentration corresponds to a degree of polymerization of 2000 .",
        "answer": "the number-average molecular weight of the nitrile rubber copolymer is \\bar{m}_{n} = 106740, \\text{g/mol}.",
        "type": 1,
        "wrong_answers_1": "the number-average molecular weight of the nitrile rubber copolymer is \\bar{m}_{n} = 102811, \\text{g/mol}.",
        "wrong_answers_2": "the number-average molecular weight of the nitrile rubber copolymer is \\bar{m}_{n} = 120805, \\text{g/mol}.",
        "wrong_answers_3": "the number-average molecular weight of the nitrile rubber copolymer is \\bar{m}_{n} = 111630, \\text{g/mol}."
    },
    {
        "idx": 922,
        "question": "An alternating copolymer is known to have a number-average molecular weight of 250,000g / mol and a degree of polymerization of 3420 . If one of the repeat units is styrene, which of ethylene, propylene, tetrafluoroethylene, and vinyl chloride is the other repeat unit? ",
        "answer": "propylene",
        "type": 2
    },
    {
        "idx": 923,
        "question": "(a) Determine the ratio of butadiene to styrene repeat units in a copolymer having a number-average molecular weight of 350,000 g/mol and degree of polymerization of 4425.",
        "answer": "1.0.",
        "type": 1,
        "wrong_answers_1": "0.9.",
        "wrong_answers_2": "1.1.",
        "wrong_answers_3": "1.1."
    },
    {
        "idx": 924,
        "question": "(b) Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block?",
        "answer": "random, alternating, graft, and block",
        "type": 2
    },
    {
        "idx": 925,
        "question": "Crosslinked copolymers consisting of 60 wt% ethylene and 40 wt% propylene may have elastic properties similar to those for natural rubber. For a copolymer of this composition, determine the fraction of both repeat unit types.",
        "answer": "the fraction of the ethylene repeat unit is  f(\\text{ethylene}) = 0.69 . the fraction of the propylene repeat unit is  f(\\text{propylene}) = 0.31 .",
        "type": 1,
        "wrong_answers_1": "the fraction of the ethylene repeat unit is  f(\\text{ethylene}) = 0.62 . the fraction of the propylene repeat unit is  f(\\text{propylene}) = 0.33 .",
        "wrong_answers_2": "the fraction of the ethylene repeat unit is  f(\\text{ethylene}) = 0.64 . the fraction of the propylene repeat unit is  f(\\text{propylene}) = 0.34 .",
        "wrong_answers_3": "the fraction of the ethylene repeat unit is  f(\\text{ethylene}) = 0.79 . the fraction of the propylene repeat unit is  f(\\text{propylene}) = 0.34 ."
    },
    {
        "idx": 926,
        "question": "Explain briefly why the tendency of a polymer to crystallize decreases with increasing molecular weight.",
        "answer": "The tendency of a polymer to crystallize decreases with increasing molecular weight because as the chains become longer it is more difficult for all regions along adjacent chains to align so as to produce the ordered atomic array.",
        "type": 4,
        "wrong_answers_1": "Increasing the degree of crystallinity of a semicrystalline polymer enhances its thermal conductivity; the vibrations, rotations, etc. of the molecular chains are more effective modes of thermal transport when a crystalline structure prevails.",
        "wrong_answers_2": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration.",
        "wrong_answers_3": "No, it is not possible to decide for these two polymers. On the basis of tacticity, the isotactic PP is more likely to crystallize than the atactic PVC. On the other hand, with regard to side-group bulkiness, the PVC is more likely to crystallize."
    },
    {
        "idx": 927,
        "question": "For the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (a) Linear and syndiotactic poly(vinyl chloride); linear and isotactic polystyrene.",
        "answer": "Yes, for these two polymers it is possible to decide. The linear and syndiotactic poly(vinyl chloride) is more likely to crystallize; the phenyl side-group for polystyrene is bulkier than the Cl side-group for poly(vinyl chloride). Syndiotactic and isotactic isomers are equally likely to crystallize.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is more likely to crystallize than crosslinked cis-isoprene because linear polymers are more likely to crystallize than crosslinked ones.",
        "wrong_answers_2": "No, it is not possible to decide for these two polymers. On the basis of tacticity, the isotactic PP is more likely to crystallize than the atactic PVC. On the other hand, with regard to side-group bulkiness, the PVC is more likely to crystallize.",
        "wrong_answers_3": "Yes, it is possible to decide for these two polymers. The linear and isotactic polystyrene is more likely to crystallize than network phenol-formaldehyde; network polymers rarely crystallize, whereas isotactic ones crystallize relatively easily."
    },
    {
        "idx": 928,
        "question": "For the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (b) Network phenol-formaldehyde; linear and heavily crosslinked cis-isoprene.",
        "answer": "No, it is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to decide for these two polymers. The linear and isotactic polystyrene is more likely to crystallize than network phenol-formaldehyde; network polymers rarely crystallize, whereas isotactic ones crystallize relatively easily.",
        "wrong_answers_2": "Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is more likely to crystallize than crosslinked cis-isoprene because linear polymers are more likely to crystallize than crosslinked ones.",
        "wrong_answers_3": "No, it is not possible to decide for these two polymers. On the basis of tacticity, the isotactic PP is more likely to crystallize than the atactic PVC. On the other hand, with regard to side-group bulkiness, the PVC is more likely to crystallize."
    },
    {
        "idx": 929,
        "question": "For the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (c) Linear polyethylene; lightly branched isotactic polypropylene.",
        "answer": "Yes, it is possible to decide for these two polymers. The linear polyethylene is more likely to crystallize. The repeat unit structure for polypropylene is chemically more complicated than is the repeat unit structure for polyethylene. Furthermore, branched structures are less likely to crystallize than are linear structures.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is more likely to crystallize than crosslinked cis-isoprene because linear polymers are more likely to crystallize than crosslinked ones.",
        "wrong_answers_2": "Yes, it is possible. The linear and isotactic polypropylene will have a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. ",
        "wrong_answers_3": "No, it is not possible. Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don't normally crystallize. The atactic polypropylene probably also has a relatively low degree of crystallinity; atactic structures also don't tend to crystallize, and polypropylene has a more complex repeat unit structure than does polyethylene.Tensile modulus increases with degree of crystallinity, and it is not possible to determine which polymer is more crystalline. Furthermore, tensile modulus is independent of molecular weight."
    },
    {
        "idx": 930,
        "question": "For the following pairs of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (d) Alternating poly(styrene-ethylene) copolymer; random poly(vinyl chloride-tetrafluoroethylene) copolymer.",
        "answer": "Yes, it is possible to decide for these two copolymers. The alternating poly(styrene-ethylene) copolymer is more likely to crystallize. Alternating copolymers crystallize more easily than do random copolymers.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to decide for these two copolymers. The block poly(acrylonitrileisoprene) copolymer is more likely to crystallize than the graft poly(chloroprene-isobutylene) copolymer. Block copolymers crystallize more easily than graft ones.",
        "wrong_answers_2": "The alternating poly(styrene-butadiene) copolymer will have a higher crystallinity than the random copolymer; alternating copolymers crystallize more easily than random ones. The influence of crystallinity on conductivity is explained in part (c).",
        "wrong_answers_3": "Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is more likely to crystallize than crosslinked cis-isoprene because linear polymers are more likely to crystallize than crosslinked ones."
    },
    {
        "idx": 931,
        "question": "Consider the diffusion of water vapor through a polypropylene (PP) sheet 2mm thick. The pressures of H_{2} \\mathrm{O} at the two faces are 1 kPa and 10 kPa, which are maintained constant. Assuming conditions of steady state, what is the diffusion flux [in \\left[{cm}^{3} STP\\right] / {cm}^{2}-s ] at 298 K ?",
        "answer": "the diffusion flux is 1.71 × 10^{-7} \\ \\text{cm}^{3}\\text{stp}/\\text{cm}^{2}-\\text{s}.",
        "type": 1,
        "wrong_answers_1": "the diffusion flux is 1.59 × 10^{-7} \\ \\text{cm}^{3}\\text{stp}/\\text{cm}^{2}-\\text{s}.",
        "wrong_answers_2": "the diffusion flux is 1.58 × 10^{-7} \\ \\text{cm}^{3}\\text{stp}/\\text{cm}^{2}-\\text{s}.",
        "wrong_answers_3": "the diffusion flux is 1.58 × 10^{-7} \\ \\text{cm}^{3}\\text{stp}/\\text{cm}^{2}-\\text{s}."
    },
    {
        "idx": 932,
        "question": "Argon diffuses through a high density polyethylene (HDPE) sheet 40mm thick at a rate of 4.0 × 10^{-7} \\left({cm}^{3} STP / {cm}^{2}-s\\right. at 325 K. The pressures of argon at the two faces are 5000 kPa and 1500 kPa, which are maintained constant. Assuming conditions of steady state, what is the permeability coefficient at 325 K ?",
        "answer": "the permeability coefficient of ar through hdpe at 325k is 4.57 × 10^{-13} \\frac{{cm}^{3} \\mathrm{stp}·{cm}}{{cm}^{2}-s-\\mathrm{pa}}.",
        "type": 1,
        "wrong_answers_1": "the permeability coefficient of ar through hdpe at 292k is 4.29 × 10^{-13} \\frac{{cm}^{3} \\mathrm{stp}·{cm}}{{cm}^{2}-s-\\mathrm{pa}}.",
        "wrong_answers_2": "the permeability coefficient of ar through hdpe at 278k is 4.71 × 10^{-13} \\frac{{cm}^{3} \\mathrm{stp}·{cm}}{{cm}^{2}-s-\\mathrm{pa}}.",
        "wrong_answers_3": "the permeability coefficient of ar through hdpe at 313k is 5.10 × 10^{-13} \\frac{{cm}^{3} \\mathrm{stp}·{cm}}{{cm}^{2}-s-\\mathrm{pa}}."
    },
    {
        "idx": 933,
        "question": "The permeability coefficient of a type of small gas molecule in a polymer is dependent on absolute temperature according to the following equation:\n\\[\nP_{M}=P_{M_{0}} \\exp \\left(-\\frac{Q_{p}}{R T}\\right)\n\\]\nwhere P_{M_{0}} and Q_{p} are constants for a given gas-polymer pair. Consider the diffusion of hydrogen through a poly(dimethyl siloxane) (PDMSO) sheet 20mm thick. The hydrogen pressures at the two faces are 10 kPa and 1 kPa, which are maintained constant. Compute the diffusion flux [in \\left({cm}^{3} STP\\right) / {cm}^{2}-s ] at 350 K. For this diffusion system\n\\[\n\\begin{array}{l}\nP_{M_{0}}=1.45 × 10^{-8}\\left(cm^{3} STP\\right)\\left({cm}^{3} / {cm}^{2}-s-\\mathrm{Pa}\\right. \\\\\nQ_{p}=13.7kJ / mol\n\\end{array}\n\\]\nAlso, assume a condition of steady state diffusion",
        "answer": "the diffusion flux is 5.90 × 10^{-7} \\frac{\\text{cm}^{3} \\text{stp}}{\\text{cm}^{2}·\\text{s}}.",
        "type": 1,
        "wrong_answers_1": "the diffusion flux is 5.44 × 10^{-7} \\frac{\\text{cm}^{3} \\text{stp}}{\\text{cm}^{2}·\\text{s}}.",
        "wrong_answers_2": "the diffusion flux is 6.36 × 10^{-7} \\frac{\\text{cm}^{3} \\text{stp}}{\\text{cm}^{2}·\\text{s}}.",
        "wrong_answers_3": "the diffusion flux is 6.66 × 10^{-7} \\frac{\\text{cm}^{3} \\text{stp}}{\\text{cm}^{2}·\\text{s}}."
    },
    {
        "idx": 934,
        "question": "For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to\n\\[\n\\sigma(t)=\\sigma(0) \\exp \\left(-\\frac{t}{\\tau}\\right)\n\\]\nwhere \\sigma(t) and \\sigma(t) represent the time-dependent and initial (i.e., time =0 ) stresses, respectively, and t and \\tau denote elapsed time and the relaxation time; \\tau is a time-independent constant characteristic of the material. A specimen of some viscoelastic polymer the stress relaxation of which obeys Equation 15.10 was suddenly pulled in tension to a measured strain of 0.6 ; the stress necessary to maintain this constant strain was measured as a function of time. Determine E_{\\tau}(10) for this material if the initial stress level was 2.76 MPa(400 psi), which dropped to 1.72 MPa (250 psi) after 60s.",
        "answer": "e_{\\tau}(10) = 4.25 \\text{ mpa (616 psi)}",
        "type": 1,
        "wrong_answers_1": "e_{\\tau}(11) = 4.40 \\text{ mpa (596 psi)}",
        "wrong_answers_2": "e_{\\tau}(9) = 4.06 \\text{ mpa (570 psi)}",
        "wrong_answers_3": "e_{\\tau}(9) = 4.08 \\text{ mpa (538 psi)}"
    },
    {
        "idx": 935,
        "question": "(a) Contrast the manner in which stress relaxation and viscoelastic creep tests are conducted.",
        "answer": "Stress relaxation tests are conducted by rapidly straining the material elastically in tension, holding the strain level constant, and then measuring the stress as a function of time. For viscoelastic creep tests, a stress (usually tensile) is applied instantaneously and maintained constant while strain is measured as a function of time.",
        "type": 4,
        "wrong_answers_1": "The experimental parameters of interest from the stress relaxation and viscoelastic creep tests are the relaxation modulus and creep modulus (or creep compliance), respectively. The relaxation modulus is the ratio of stress measured after 10 s and strain; creep modulus is the ratio of stress and strain taken at a specific time.",
        "wrong_answers_2": "A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time.",
        "wrong_answers_3": "Gray iron is weak and brittle in tension because the tips of the graphite flakes act as points of stress concentration."
    },
    {
        "idx": 936,
        "question": "(b) For each of these tests, cite the experimental parameter of interest and how it is determined.",
        "answer": "The experimental parameters of interest from the stress relaxation and viscoelastic creep tests are the relaxation modulus and creep modulus (or creep compliance), respectively. The relaxation modulus is the ratio of stress measured after 10 s and strain; creep modulus is the ratio of stress and strain taken at a specific time.",
        "type": 4,
        "wrong_answers_1": "Stress relaxation tests are conducted by rapidly straining the material elastically in tension, holding the strain level constant, and then measuring the stress as a function of time. For viscoelastic creep tests, a stress (usually tensile) is applied instantaneously and maintained constant while strain is measured as a function of time.",
        "wrong_answers_2": "of the possible copolymers, the only one for which there is a restriction on the ratio of repeat unit types is alternating; the ratio must be 1:1. therefore, on the basis of the result in part (a), the possibilities for this copolymer are random, graft, and block.",
        "wrong_answers_3": "Yes, it is possible. The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's crystallinity leads to an increase in its tensile modulus. In addition, tensile modulus is independent of molecular weight - the atactic/branched material has the higher molecular weight."
    },
    {
        "idx": 937,
        "question": "For thermoplastic polymers, cite five factors that favor brittle fracture.",
        "answer": "For thermoplastic polymers, five factors that favor brittle fracture are as follows: (1) a reduction in temperature, (2) an increase in strain rate, (3) the presence of a sharp notch, (4) increased specimen thickness, and (5) modifications of the polymer structure.",
        "type": 4,
        "wrong_answers_1": "Increasing the temperature should increase the diffusivity and therefore increase the diffusion flux and flow rate.",
        "wrong_answers_2": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material.",
        "wrong_answers_3": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material."
    },
    {
        "idx": 938,
        "question": "For the pair of polymers: random acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked; alternating acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
        "answer": "No, it is not possible. The random acrylonitrile-butadiene copolymer will tend to a lower degree of crystallinity than the alternating acrylonitrile-butadiene copolymer inasmuch as random copolymers don't normally crystallize. On this basis only, the alternating material would have a higher modulus inasmuch as tensile modulus increases with degree of crystallinity. On the other hand, the random copolymer has a higher degree of crosslinking (10% versus 5% for the alternating copolymer), and, on this basis only would have the higher tensile modulus-an increase in crosslinking leads to an increase in E. Thus, this determination is not possible; with regard to degree of crystallinity the alternating material has the higher E, whereas the random copolymer would have a higher E value on the basis of degree of crosslinking.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible. The block styrene-butadiene copolymer with 10% of possible sites crosslinked will have the higher modulus. Block copolymers normally have higher degrees of crystallinity than random copolymers of the same material. A higher degree of crystallinity favors larger moduli. In addition, the block copolymer also has a higher degree of crosslinking; increasing the amount of crosslinking also enhances the tensile modulus.",
        "wrong_answers_2": "The alternating poly(styrene-butadiene) copolymer will have a higher crystallinity than the random copolymer; alternating copolymers crystallize more easily than random ones. The influence of crystallinity on conductivity is explained in part (c).",
        "wrong_answers_3": "No it is not possible. The random acrylonitrile-butadiene copolymer has more crosslinking; increased crosslinking leads to an increase in strength. However, the block copolymeric material will most likely have a higher degree of crystallinity; and increasing crystallinity improves the strength."
    },
    {
        "idx": 939,
        "question": "For the pair of polymers: branched and syndiotactic polypropylene with a degree of polymerization of 5000; linear and isotactic polypropylene with a degree of polymerization of 3000, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
        "answer": "Yes, it is possible. The linear and isotactic polypropylene will have a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. ",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible. The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's crystallinity leads to an increase in its tensile modulus. In addition, tensile modulus is independent of molecular weight - the atactic/branched material has the higher molecular weight.",
        "wrong_answers_2": "Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is more likely to crystallize than crosslinked cis-isoprene because linear polymers are more likely to crystallize than crosslinked ones.",
        "wrong_answers_3": "Yes, it is possible to decide for these two polymers. The linear and isotactic polystyrene is more likely to crystallize than network phenol-formaldehyde; network polymers rarely crystallize, whereas isotactic ones crystallize relatively easily."
    },
    {
        "idx": 940,
        "question": "For the pair of polymers: branched polyethylene with a number-average molecular weight of 250000 g/mol; linear and isotactic poly(vinyl chloride) with a number-average molecular weight of 200000 g/mol, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why.",
        "answer": "No, it is not possible. Linear polymers have higher degrees of crystallization (and higher tensile moduli) than branched polymers - on this basis, the PVC material should have the higher value of E. On the other hand, PVC has a more complex repeat unit structure than does polyethylene, which means that, on this basis, the PE would have a higher degree of crystallinity and also a greater tensile modulus. Also, tensile modulus is relatively independent of number-average molecular weight. Therefore, this determination is not possible since it is not possible to determine which of the two materials has the greater degree of crystallinity.",
        "type": 4,
        "wrong_answers_1": "No, it is not possible. Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don't normally crystallize. The atactic polypropylene probably also has a relatively low degree of crystallinity; atactic structures also don't tend to crystallize, and polypropylene has a more complex repeat unit structure than does polyethylene.Tensile modulus increases with degree of crystallinity, and it is not possible to determine which polymer is more crystalline. Furthermore, tensile modulus is independent of molecular weight.",
        "wrong_answers_2": "No, it is not possible. The random acrylonitrile-butadiene copolymer will tend to a lower degree of crystallinity than the alternating acrylonitrile-butadiene copolymer inasmuch as random copolymers don't normally crystallize. On this basis only, the alternating material would have a higher modulus inasmuch as tensile modulus increases with degree of crystallinity. On the other hand, the random copolymer has a higher degree of crosslinking (10% versus 5% for the alternating copolymer), and, on this basis only would have the higher tensile modulus-an increase in crosslinking leads to an increase in E. Thus, this determination is not possible; with regard to degree of crystallinity the alternating material has the higher E, whereas the random copolymer would have a higher E value on the basis of degree of crosslinking.",
        "wrong_answers_3": "Yes, it is possible. The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's crystallinity leads to an increase in its tensile modulus. In addition, tensile modulus is independent of molecular weight - the atactic/branched material has the higher molecular weight."
    },
    {
        "idx": 941,
        "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why. Syndiotactic polystyrene having a number-average molecular weight of 600,000 g/mol; atactic polystyrene having a number-average molecular weight of 500,000 g/mol",
        "answer": "Yes it is possible. The syndiotactic polystyrene has the higher tensile strength. Syndiotactic polymers are more likely to crystallize than atactic ones; the greater the crystallinity, the higher the tensile strength. Furthermore, the syndiotactic also has a higher molecular weight; increasing molecular weight also enhances the strength.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to determine which of the two polystyrenes has the higher Tm. The isotactic polystyrene will have the higher melting temperature because it has a higher density (i.e., less branching) and also the greater weight-average molecular weight.",
        "wrong_answers_2": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The syndiotactic polypropylene will have a higher degree of crystallinity than the atactic material. On the basis of this effect alone, the syndiotactic PP should have the greater Tm, since melting temperature increases with degree of crystallinity. However, the molecular weight for the syndiotactic polypropylene (500,000 g/mol) is less than for the atactic material (750,000 g/mol); a lowering of molecular weight generally results in a reduction of melting temperature.",
        "wrong_answers_3": "Yes, it is possible. The linear and isotactic material will have the higher tensile strength. Both linearity and isotacticity favor a higher degree of crystallinity than do branching and atacticity; and tensile strength increases with increasing degree of crystallinity. Furthermore, the molecular weight of the linear/isotactic material is higher (100,000 g/mol versus 75,000 g/mol), and tensile strength increases with increasing molecular weight."
    },
    {
        "idx": 942,
        "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why. Random acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked; block acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked",
        "answer": "No it is not possible. The random acrylonitrile-butadiene copolymer has more crosslinking; increased crosslinking leads to an increase in strength. However, the block copolymeric material will most likely have a higher degree of crystallinity; and increasing crystallinity improves the strength.",
        "type": 4,
        "wrong_answers_1": "No, it is not possible. The random acrylonitrile-butadiene copolymer will tend to a lower degree of crystallinity than the alternating acrylonitrile-butadiene copolymer inasmuch as random copolymers don't normally crystallize. On this basis only, the alternating material would have a higher modulus inasmuch as tensile modulus increases with degree of crystallinity. On the other hand, the random copolymer has a higher degree of crosslinking (10% versus 5% for the alternating copolymer), and, on this basis only would have the higher tensile modulus-an increase in crosslinking leads to an increase in E. Thus, this determination is not possible; with regard to degree of crystallinity the alternating material has the higher E, whereas the random copolymer would have a higher E value on the basis of degree of crosslinking.",
        "wrong_answers_2": "Yes, it is possible. The block styrene-butadiene copolymer with 10% of possible sites crosslinked will have the higher modulus. Block copolymers normally have higher degrees of crystallinity than random copolymers of the same material. A higher degree of crystallinity favors larger moduli. In addition, the block copolymer also has a higher degree of crosslinking; increasing the amount of crosslinking also enhances the tensile modulus.",
        "wrong_answers_3": "No, it is not possible. Alternating copolymers tend to be more crystalline than graft copolymers, and tensile strength increases with degree of crystallinity. However, the graft material has a higher degree of crosslinking, and tensile strength increases with the percentage of crosslinks."
    },
    {
        "idx": 943,
        "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and then cite the reason(s) for your choice; and (3) if it is not possible to decide, then state why. Network polyester; lightly branched polypropylene",
        "answer": "Yes it is possible. The network polyester will display a greater tensile strength. Relative chain motion is much more restricted than for the lightly branched polypropylene since there are many more of the strong covalent bonds for the network structure.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible. The network polyester will display a greater tensile strength. Relative chain motion is much more restricted than for the lightly branched polytetrafluoroethylene since there are many more of the strong covalent bonds for the network structure.",
        "wrong_answers_2": "Yes, it is possible. The linear and isotactic polypropylene will have a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. ",
        "wrong_answers_3": "Yes, it is possible to decide for these two polymers. The linear polyethylene is more likely to crystallize. The repeat unit structure for polypropylene is chemically more complicated than is the repeat unit structure for polyethylene. Furthermore, branched structures are less likely to crystallize than are linear structures."
    },
    {
        "idx": 944,
        "question": "List the two molecular characteristics that are essential for elastomers.",
        "answer": "Two molecular characteristics essential for elastomers are: (1) they must be amorphous, having chains that are extensively coiled and kinked in the unstressed state; and (2) there must be some crosslinking.",
        "type": 4,
        "wrong_answers_1": "Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of crystallinity.",
        "wrong_answers_2": "Fiber materials that are drawn must be thermoplastic because during drawing, mechanical elongation must be possible; inasmuch as thermosetting materials are, in general, hard and relatively brittle, they are not easily elongated.",
        "wrong_answers_3": "Fiber materials that are melt spun must be thermoplastic because they must be capable of forming a viscous liquid when heated, which is not possible for thermosets."
    },
    {
        "idx": 945,
        "question": "Name the following polymer(s) that would be suitable for the fabrication of cups to contain hot coffee: polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate. Why?",
        "answer": "This question asks us to name which, of several polymers, would be suitable for the fabrication of cups to contain hot coffee. At its glass transition temperature, an amorphous polymer begins to soften. The maximum temperature of hot coffee is probably slightly below 100^{\\circ} C\\left(212^{\\circ} F\\right). Of the polymers listed, only polystyrene and polycarbonate have glass transition temperatures of 100^{\\circ} C or above, and would be suitable for this application.",
        "type": 4,
        "wrong_answers_1": "The softening point of a glass is that temperature at which the viscosity is 4 × 10^{6} \\mathrm{~Pa}·s; from Figure 13.7, these temperatures for the 96% silica, borosilicate, and soda-lime glasses are 1540^{\\circ} C\\left(2800^{\\circ} F\\right), 830^{\\circ} C\\left(1525^{\\circ} F\\right), and 700^{\\circ} C\\left(1290^{\\circ} F\\right), respectively.",
        "wrong_answers_2": "Austenitize at approximately 750^{\\circ} C,\nQuench to below 130^{\\circ} C (the M_{\\gamma} temperature)\nTemper at 620^{\\circ} C or less.",
        "wrong_answers_3": "The glass transition temperature is, for a noncrystalline ceramic, that temperature at which there is a change of slope for the specific volume versus temperature curve."
    },
    {
        "idx": 946,
        "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Isotactic polystyrene that has a density of 1.12 g/cm^3 and a weight-average molecular weight of 150,000 g/mol; syndiotactic polystyrene that has a density of 1.10 g/cm^3 and a weight-average molecular weight of 125,000 g/mol",
        "answer": "Yes, it is possible to determine which of the two polystyrenes has the higher Tm. The isotactic polystyrene will have the higher melting temperature because it has a higher density (i.e., less branching) and also the greater weight-average molecular weight.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to determine which polymer has the higher melting temperature. Of these two polytetrafluoroethylene polymers, the PTFE with the higher density (2.20 g/cm^3) will have the higher percent crystallinity, and, therefore, a higher melting temperature than the lower density PTFE. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration.",
        "wrong_answers_2": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will most likely have a higher percent crystallinity, and, therefore, a higher melting temperature than the branched polyethylene. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration.",
        "wrong_answers_3": "Yes, it is possible to determine which polymer has the higher melting temperature. The polypropylene will have the higher Tm because it has a bulky phenyl side group in its repeat unit structure, which is absent in the polyethylene. Furthermore, the polypropylene has a higher degree of polymerization."
    },
    {
        "idx": 947,
        "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Linear polyethylene that has a degree of polymerization of 5,000; linear and isotactic polypropylene that has a degree of polymerization of 6,500",
        "answer": "Yes, it is possible to determine which polymer has the higher melting temperature. The polypropylene will have the higher Tm because it has a bulky phenyl side group in its repeat unit structure, which is absent in the polyethylene. Furthermore, the polypropylene has a higher degree of polymerization.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to determine which of the two polystyrenes has the higher Tm. The isotactic polystyrene will have the higher melting temperature because it has a higher density (i.e., less branching) and also the greater weight-average molecular weight.",
        "wrong_answers_2": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The polystyrene has a bulkier side group than the polypropylene; on the basis of this effect alone, the polystyrene should have the greater Tm. However, the polystyrene has more branching and a lower degree of polymerization; both of these factors lead to a lowering of the melting temperature.",
        "wrong_answers_3": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will most likely have a higher percent crystallinity, and, therefore, a higher melting temperature than the branched polyethylene. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration."
    },
    {
        "idx": 948,
        "question": "For the following pair of polymers, do the following: (1) state whether or not it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Branched and isotactic polystyrene that has a degree of polymerization of 4,000; linear and isotactic polypropylene that has a degree of polymerization of 7,500",
        "answer": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The polystyrene has a bulkier side group than the polypropylene; on the basis of this effect alone, the polystyrene should have the greater Tm. However, the polystyrene has more branching and a lower degree of polymerization; both of these factors lead to a lowering of the melting temperature.",
        "type": 4,
        "wrong_answers_1": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The syndiotactic polypropylene will have a higher degree of crystallinity than the atactic material. On the basis of this effect alone, the syndiotactic PP should have the greater Tm, since melting temperature increases with degree of crystallinity. However, the molecular weight for the syndiotactic polypropylene (500,000 g/mol) is less than for the atactic material (750,000 g/mol); a lowering of molecular weight generally results in a reduction of melting temperature.",
        "wrong_answers_2": "Yes, it is possible to determine which of the two polystyrenes has the higher Tm. The isotactic polystyrene will have the higher melting temperature because it has a higher density (i.e., less branching) and also the greater weight-average molecular weight.",
        "wrong_answers_3": "Yes, it is possible to determine which polymer has the higher melting temperature. The polypropylene will have the higher Tm because it has a bulky phenyl side group in its repeat unit structure, which is absent in the polyethylene. Furthermore, the polypropylene has a higher degree of polymerization."
    },
    {
        "idx": 949,
        "question": "Briefly explain the difference in molecular chemistry between silicone polymers and other polymeric materials.",
        "answer": "The backbone chain of most polymers consists of carbon atoms that are linked together. For the silicone polymers, this backbone chain is composed of silicon and oxygen atoms that alternate positions.",
        "type": 4,
        "wrong_answers_1": "Concrete consists of an aggregate of particles that are bonded together by a cement.",
        "wrong_answers_2": "Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of crystallinity.",
        "wrong_answers_3": "diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen."
    },
    {
        "idx": 950,
        "question": "List two important characteristics for polymers that are to be used in fiber applications.",
        "answer": "Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of crystallinity.",
        "type": 4,
        "wrong_answers_1": "Disadvantages of ferrous alloys are: (1) They are susceptible to corrosion. (2) They have a relatively high density. (3) They have relatively low electrical conductivities.",
        "wrong_answers_2": "Two molecular characteristics essential for elastomers are: (1) they must be amorphous, having chains that are extensively coiled and kinked in the unstressed state; and (2) there must be some crosslinking.",
        "wrong_answers_3": "Five important characteristics for polymers that are to be used in thin-film applications are: (1) low density; (2) high flexibility; (3) high tensile and tear strengths; (4) resistance to moisture/chemical attack; and (5) low gas permeability.\n}"
    },
    {
        "idx": 951,
        "question": "Cite five important characteristics for polymers that are to be used in thin-film applications.",
        "answer": "Five important characteristics for polymers that are to be used in thin-film applications are: (1) low density; (2) high flexibility; (3) high tensile and tear strengths; (4) resistance to moisture/chemical attack; and (5) low gas permeability.\n}",
        "type": 4,
        "wrong_answers_1": "Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of crystallinity.",
        "wrong_answers_2": "Distinctive features: relatively low density, high melting temperatures, and high strengths are possible. Limitation: because of chemical reactivity with other materials at elevated temperatures, these alloys are expensive to refine. Applications: aircraft structures, space vehicles, and in chemical and petroleum industries.",
        "wrong_answers_3": "Chemical resistance decreases."
    },
    {
        "idx": 952,
        "question": "Cite the primary differences between addition and condensation polymerization techniques.",
        "answer": "For addition polymerization, the reactant species have the same chemical composition as the monomer species in the molecular chain. This is not the case for condensation polymerization, wherein there is a chemical reaction between two or more monomer species, producing the repeating unit. There is often a low molecular weight by-product for condensation polymerization; such is not found for addition polymerization.",
        "type": 4,
        "wrong_answers_1": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "wrong_answers_2": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration.",
        "wrong_answers_3": "Activation polarization is rate controlling when the reaction rate is low and/or the concentration of active species in the liquid solution is high."
    },
    {
        "idx": 953,
        "question": "Estimate the maximum thermal conductivity value for a cermet that contains 85 vol % titanium carbide (TiC) particles in a cobalt matrix. Assume thermal conductivities of 27 and 69 W/m-K for TiC and Co, respectively.",
        "answer": "the maximum thermal conductivity k_max is 33.3 W/m-K.",
        "type": 1,
        "wrong_answers_1": "the maximum thermal conductivity k_max is 37.0 W/m-K.",
        "wrong_answers_2": "the maximum thermal conductivity k_max is 36.8 W/m-K.",
        "wrong_answers_3": "the maximum thermal conductivity k_max is 36.0 W/m-K."
    },
    {
        "idx": 954,
        "question": "Estimate the minimum thermal conductivity value for a cermet that contains 85 vol % titanium carbide (TiC) particles in a cobalt matrix. Assume thermal conductivities of 27 and 69 W/m-K for TiC and Co, respectively.",
        "answer": "the minimum thermal conductivity k_min is 29.7 W/m-K.",
        "type": 1,
        "wrong_answers_1": "the minimum thermal conductivity k_min is 28.5 W/m-K.",
        "wrong_answers_2": "the minimum thermal conductivity k_min is 34.0 W/m-K.",
        "wrong_answers_3": "the minimum thermal conductivity k_min is 28.4 W/m-K."
    },
    {
        "idx": 955,
        "question": "A large-particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of tungsten and copper are 0.60 and 0.40 , respectively, estimate the upper limit for the specific stiffness of this composite given the data that follow.\n\\begin{tabular}{lcc}\n\\hline & Specific Gravity & \\begin{tabular}{c} \nModulus of Elasticity \\\\\n(GPa)\n\\end{tabular} \\\\\n\\hline Copper & 8.9 & 110 \\\\\nTungsten & 19.3 & 407 \\\\\n\\hline\n\\end{tabular}",
        "answer": "the upper limit for the specific stiffness of this composite is 19.0 gpa. an alternate approach yields a specific stiffness of 17.6 gpa.",
        "type": 1,
        "wrong_answers_1": "the upper limit for the specific stiffness of this composite is 20.8 gpa. an alternate approach yields a specific stiffness of 15.4 gpa.",
        "wrong_answers_2": "the upper limit for the specific stiffness of this composite is 17.6 gpa. an alternate approach yields a specific stiffness of 16.1 gpa.",
        "wrong_answers_3": "the upper limit for the specific stiffness of this composite is 20.6 gpa. an alternate approach yields a specific stiffness of 19.0 gpa."
    },
    {
        "idx": 956,
        "question": "(a) What is the distinction between cement and concrete?",
        "answer": "Concrete consists of an aggregate of particles that are bonded together by a cement.",
        "type": 4,
        "wrong_answers_1": "With cements, the bonding process is a chemical, hydration reaction between the water that has been added and the various cement constituents. The cement particles are bonded together by reactions that occur at the particle surfaces.",
        "wrong_answers_2": "The temperature indicator is made from bimetallic materials, which consist of two materials with different coefficients of thermal expansion bonded together.",
        "wrong_answers_3": "The backbone chain of most polymers consists of carbon atoms that are linked together. For the silicone polymers, this backbone chain is composed of silicon and oxygen atoms that alternate positions."
    },
    {
        "idx": 957,
        "question": "(b) Cite three important limitations that restrict the use of concrete as a structural material.",
        "answer": "Three limitations of concrete are: (1) it is a relatively weak and brittle material; (2) it experiences relatively large thermal expansions (contractions) with changes in temperature; and (3) it may crack when exposed to freeze-thaw cycles.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to grind up and reuse polypropylene since it is a thermoplastic polymer, will soften when reheated, and, thus, may be remolded.",
        "wrong_answers_2": "Gray iron--Relatively weak and brittle in tension; good capacity for damping vibrations. Malleable iron--Moderate strength and ductility.",
        "wrong_answers_3": "For refractory ceramic materials, three characteristics that improve with increasing porosity are (1) decreased thermal expansion and contraction upon thermal cycling, (2) improved thermal insulation, and (3) improved resistance to thermal shock."
    },
    {
        "idx": 958,
        "question": "(c) Briefly explain three techniques that are utilized to strengthen concrete by reinforcement.",
        "answer": "Three reinforcement strengthening techniques are: (1) reinforcement with steel wires, rods, etc.; (2) reinforcement with fine fibers of a high modulus material; and (3) introduction of residual compressive stresses by prestressing or postensioning.",
        "type": 4,
        "wrong_answers_1": "Fibers are polycrystalline or amorphous materials with small diameters.",
        "wrong_answers_2": "For polymer-matrix fiber-reinforced composites, three functions of the polymer-matrix phase are: (1) to bind the fibers together so that the applied stress is distributed among the fibers; (2) to protect the surface of the fibers from being damaged; and (3) to separate the fibers and inhibit crack propagation.",
        "wrong_answers_3": "Tempered martensite is harder and stronger inasmuch as there is much more ferrite-cementite phase boundary area for the smaller particles; thus, there is greater reinforcement of the ferrite phase, and more phase boundary barriers to dislocation motion."
    },
    {
        "idx": 959,
        "question": "Cite one similarity between precipitation hardening and dispersion strengthening.",
        "answer": "The similarity between precipitation hardening and dispersion strengthening is the strengthening mechanism--i.e., the precipitates/particles effectively hinder dislocation motion.",
        "type": 4,
        "wrong_answers_1": "The two differences are: (1) the hardening/strengthening effect is not retained at elevated temperatures for precipitation hardening--however, it is retained for dispersion strengthening; and (2) the strength is developed by a heat treatment for precipitation hardening--such is not the case for dispersion strengthening.",
        "wrong_answers_2": "For precipitation hardening, natural aging is allowing the precipitation process to occur at the ambient temperature; artificial aging is carried out at an elevated temperature.",
        "wrong_answers_3": "The chief difference between heat-treatable and nonheat-treatable alloys is that heat-treatable alloys may be strengthened by a heat treatment wherein a precipitate phase is formed (precipitation hardening) or a martensitic transformation occurs. Nonheat-treatable alloys are not amenable to strengthening by such treatments."
    },
    {
        "idx": 960,
        "question": "Cite two differences between precipitation hardening and dispersion strengthening.",
        "answer": "The two differences are: (1) the hardening/strengthening effect is not retained at elevated temperatures for precipitation hardening--however, it is retained for dispersion strengthening; and (2) the strength is developed by a heat treatment for precipitation hardening--such is not the case for dispersion strengthening.",
        "type": 4,
        "wrong_answers_1": "The similarity between precipitation hardening and dispersion strengthening is the strengthening mechanism--i.e., the precipitates/particles effectively hinder dislocation motion.",
        "wrong_answers_2": "For precipitation hardening, natural aging is allowing the precipitation process to occur at the ambient temperature; artificial aging is carried out at an elevated temperature.",
        "wrong_answers_3": "The chief difference between heat-treatable and nonheat-treatable alloys is that heat-treatable alloys may be strengthened by a heat treatment wherein a precipitate phase is formed (precipitation hardening) or a martensitic transformation occurs. Nonheat-treatable alloys are not amenable to strengthening by such treatments."
    },
    {
        "idx": 961,
        "question": "A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibers and 70 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows: Modulus of Elasticity for Aramid fiber is 131 GPa (19x10^6 psi) and Tensile Strength is 3600 MPa (520,000 psi); Modulus of Elasticity for Polycarbonate is 2.4 GPa (3.5x10^5 psi) and Tensile Strength is 65 MPa (9425 psi). Also, the stress on the polycarbonate matrix when the aramid fibers fail is 45 MPa (6500 psi). For this composite, compute the longitudinal tensile strength.",
        "answer": "the longitudinal tensile strength is 1100 mpa (160,000 psi).",
        "type": 1,
        "wrong_answers_1": "the longitudinal tensile strength is 1060 mpa (140992 psi).",
        "wrong_answers_2": "the longitudinal tensile strength is 1260 mpa (179668 psi).",
        "wrong_answers_3": "the longitudinal tensile strength is 1150 mpa (179497 psi)."
    },
    {
        "idx": 962,
        "question": "A continuous and aligned fiber-reinforced composite is to be produced consisting of 30 vol% aramid fibers and 70 vol% of a polycarbonate matrix; mechanical characteristics of these two materials are as follows: Modulus of Elasticity for Aramid fiber is 131 GPa (19x10^6 psi) and Tensile Strength is 3600 MPa (520,000 psi); Modulus of Elasticity for Polycarbonate is 2.4 GPa (3.5x10^5 psi) and Tensile Strength is 65 MPa (9425 psi). Also, the stress on the polycarbonate matrix when the aramid fibers fail is 45 MPa (6500 psi). For this composite, compute the longitudinal modulus of elasticity.",
        "answer": "the longitudinal modulus of elasticity is 41 gpa (5.95x10^6 psi).",
        "type": 1,
        "wrong_answers_1": "the longitudinal modulus of elasticity is 44 gpa (5.19x10^6 psi).",
        "wrong_answers_2": "the longitudinal modulus of elasticity is 39 gpa (6.50x10^6 psi).",
        "wrong_answers_3": "the longitudinal modulus of elasticity is 46 gpa (5.33x10^6 psi)."
    },
    {
        "idx": 963,
        "question": "For a continuous and oriented fiber-reinforced composite, the modulus of elasticity in the longitudinal direction is 19.7 GPa (2.8 x 10^5 psi). If the volume fraction of fibers is 0.25, determine the modulus of elasticity of the fiber phase.",
        "answer": "the modulus of elasticity of the fiber phase e_f is 70.4 GPa (10.2 x 10^6 psi).",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity of the fiber phase e_f is 75.9 GPa (11.7 x 10^6 psi).",
        "wrong_answers_2": "the modulus of elasticity of the fiber phase e_f is 60.8 GPa (10.8 x 10^6 psi).",
        "wrong_answers_3": "the modulus of elasticity of the fiber phase e_f is 66.1 GPa (11.0 x 10^6 psi)."
    },
    {
        "idx": 964,
        "question": "For a continuous and oriented fiber-reinforced composite, the modulus of elasticity in the transverse direction is 3.66 GPa (5.3 x 10^5 psi). If the volume fraction of fibers is 0.25, determine the modulus of elasticity of the matrix phase.",
        "answer": "the modulus of elasticity of the matrix phase e_m is 2.79 GPa (4.04 x 10^5 psi).",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity of the matrix phase e_m is 2.39 GPa (3.67 x 10^5 psi).",
        "wrong_answers_2": "the modulus of elasticity of the matrix phase e_m is 3.10 GPa (3.62 x 10^5 psi).",
        "wrong_answers_3": "the modulus of elasticity of the matrix phase e_m is 2.52 GPa (3.48 x 10^5 psi)."
    },
    {
        "idx": 965,
        "question": "In an aligned and continuous glass fiber-reinforced nylon 6,6 composite, the fibers are to carry 94% of a load applied in the longitudinal direction. Using the data provided, determine the volume fraction of fibers that will be required.",
        "answer": "the required volume fraction of fibers is 0.393.",
        "type": 1,
        "wrong_answers_1": "the required volume fraction of fibers is 0.414.",
        "wrong_answers_2": "the required volume fraction of fibers is 0.356.",
        "wrong_answers_3": "the required volume fraction of fibers is 0.335."
    },
    {
        "idx": 966,
        "question": "In an aligned and continuous glass fiber-reinforced nylon 6,6 composite, the fibers are to carry 94% of a load applied in the longitudinal direction. What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 30 MPa (4350 psi).",
        "answer": "the tensile strength of this composite is 1354 mpa (196,400 psi).",
        "type": 1,
        "wrong_answers_1": "the tensile strength of this composite is 1524 mpa (181995 psi).",
        "wrong_answers_2": "the tensile strength of this composite is 1204 mpa (210861 psi).",
        "wrong_answers_3": "the tensile strength of this composite is 1405 mpa (178990 psi)."
    },
    {
        "idx": 967,
        "question": "For a polymer-matrix fiber-reinforced composite, list three functions of the matrix phase.",
        "answer": "For polymer-matrix fiber-reinforced composites, three functions of the polymer-matrix phase are: (1) to bind the fibers together so that the applied stress is distributed among the fibers; (2) to protect the surface of the fibers from being damaged; and (3) to separate the fibers and inhibit crack propagation.",
        "type": 4,
        "wrong_answers_1": "Fibers are polycrystalline or amorphous materials with small diameters.",
        "wrong_answers_2": "A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time.",
        "wrong_answers_3": "Reasons why fiberglass-reinforced composites are utilized extensively are: (1) glass fibers are very inexpensive to produce; (2) these composites have relatively high specific strengths; and (3) they are chemically inert in a wide variety of environments."
    },
    {
        "idx": 968,
        "question": "For a polymer-matrix fiber-reinforced composite, compare the desired mechanical characteristics of matrix and fiber phases.",
        "answer": "The matrix phase must be ductile and is usually relatively soft, whereas the fiber phase must be stiff and strong.",
        "type": 4,
        "wrong_answers_1": "In general, the matrix phase is relatively weak, has a low elastic modulus, but is quite ductile. On the other hand, the fiber phase is normally quite strong, stiff, and brittle.",
        "wrong_answers_2": "There must be a strong interfacial bond between fiber and matrix in order to: (1) maximize the stress transmittance between matrix and fiber phases; and (2) minimize fiber pull-out, and the probability of failure.",
        "wrong_answers_3": "The matrix phase is a continuous phase that surrounds the noncontinuous dispersed phase."
    },
    {
        "idx": 969,
        "question": "For a polymer-matrix fiber-reinforced composite, cite two reasons why there must be a strong bond between fiber and matrix at their interface.",
        "answer": "There must be a strong interfacial bond between fiber and matrix in order to: (1) maximize the stress transmittance between matrix and fiber phases; and (2) minimize fiber pull-out, and the probability of failure.",
        "type": 4,
        "wrong_answers_1": "The matrix phase must be ductile and is usually relatively soft, whereas the fiber phase must be stiff and strong.",
        "wrong_answers_2": "In general, the matrix phase is relatively weak, has a low elastic modulus, but is quite ductile. On the other hand, the fiber phase is normally quite strong, stiff, and brittle.",
        "wrong_answers_3": "For polymer-matrix fiber-reinforced composites, three functions of the polymer-matrix phase are: (1) to bind the fibers together so that the applied stress is distributed among the fibers; (2) to protect the surface of the fibers from being damaged; and (3) to separate the fibers and inhibit crack propagation."
    },
    {
        "idx": 970,
        "question": "What is the distinction between matrix and dispersed phases in a composite material?",
        "answer": "The matrix phase is a continuous phase that surrounds the noncontinuous dispersed phase.",
        "type": 4,
        "wrong_answers_1": "ferrite is the proeutectoid phase.",
        "wrong_answers_2": "The matrix phase must be ductile and is usually relatively soft, whereas the fiber phase must be stiff and strong.",
        "wrong_answers_3": "the proeutectoid phase is fe3c."
    },
    {
        "idx": 971,
        "question": "Contrast the mechanical characteristics of matrix and dispersed phases for fiber-reinforced composites.",
        "answer": "In general, the matrix phase is relatively weak, has a low elastic modulus, but is quite ductile. On the other hand, the fiber phase is normally quite strong, stiff, and brittle.",
        "type": 4,
        "wrong_answers_1": "The matrix phase must be ductile and is usually relatively soft, whereas the fiber phase must be stiff and strong.",
        "wrong_answers_2": "The matrix phase is a continuous phase that surrounds the noncontinuous dispersed phase.",
        "wrong_answers_3": "There must be a strong interfacial bond between fiber and matrix in order to: (1) maximize the stress transmittance between matrix and fiber phases; and (2) minimize fiber pull-out, and the probability of failure."
    },
    {
        "idx": 972,
        "question": "Briefly explain the difference between oxidation and reduction electrochemical reactions.",
        "answer": "Oxidation is the process by which an atom gives up an electron (or electrons) to become a cation. Reduction is the process by which an atom acquires an extra electron (or electrons) and becomes an anion.",
        "type": 4,
        "wrong_answers_1": "The I- ion is an iodine atom that has acquired one extra electron; therefore, it has an electron configuration the same as xenon.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) electron configuration is that of an alkaline earth metal because of two (s) electrons.",
        "wrong_answers_3": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom."
    },
    {
        "idx": 973,
        "question": "Which reaction occurs at the anode and which at the cathode?",
        "answer": "Oxidation occurs at the anode; reduction at the cathode.",
        "type": 4,
        "wrong_answers_1": "A sacrificial anode is electrically coupled to the metal piece to be protected, which anode is also situated in the corrosion environment. The sacrificial anode is a metal or alloy that is chemically more reactive in the particular environment. It (the anode) preferentially oxidizes, and, upon giving up electrons to the other metal, protects it from electrochemical corrosion.",
        "wrong_answers_2": "Oxidation occurs in the cell half with the lower Fe2+ concentration (2 × 10^-2 M).",
        "wrong_answers_3": "For a small anode-to-cathode area ratio, the corrosion rate will be higher than for A large ratio. The reason for this is that for some given current flow associated with the corrosion reaction, for a small area ratio the current density at the anode will be greater than for a large ratio. The corrosion rate is proportional to the current density (i) according to Equation."
    },
    {
        "idx": 974,
        "question": "(a) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in each of the following solutions: (i) HCl",
        "answer": "In HCl, possible reactions are Mg -> Mg2+ + 2 e- (oxidation) and 2 H+ + 2 e- -> H2 (reduction).",
        "type": 4,
        "wrong_answers_1": "In an HCl solution containing dissolved oxygen, possible reactions are Mg -> Mg2+ + 2 e- (oxidation) and 4 H+ + O2 + 4 e- -> 2 H2O (reduction).",
        "wrong_answers_2": "In an HCl solution containing dissolved oxygen and Fe2+ ions, possible reactions are Mg -> Mg2+ + 2 e- (oxidation), 4 H+ + O2 + 4 e- -> 2 H2O (reduction), and Fe2+ + 2 e- -> Fe (reduction).",
        "wrong_answers_3": "Activation polarization is the condition wherein a reaction rate is controlled by one step in a series of steps that takes place at the slowest rate. For corrosion, activation polarization is possible for both oxidation and reduction reactions. Concentration polarization occurs when a reaction rate is limited by diffusion in a solution. For corrosion, concentration polarization is possible only for reduction reactions."
    },
    {
        "idx": 975,
        "question": "(a) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in each of the following solutions: (ii) an HCl solution containing dissolved oxygen",
        "answer": "In an HCl solution containing dissolved oxygen, possible reactions are Mg -> Mg2+ + 2 e- (oxidation) and 4 H+ + O2 + 4 e- -> 2 H2O (reduction).",
        "type": 4,
        "wrong_answers_1": "In an HCl solution containing dissolved oxygen and Fe2+ ions, possible reactions are Mg -> Mg2+ + 2 e- (oxidation), 4 H+ + O2 + 4 e- -> 2 H2O (reduction), and Fe2+ + 2 e- -> Fe (reduction).",
        "wrong_answers_2": "In HCl, possible reactions are Mg -> Mg2+ + 2 e- (oxidation) and 2 H+ + 2 e- -> H2 (reduction).",
        "wrong_answers_3": "Activation polarization is the condition wherein a reaction rate is controlled by one step in a series of steps that takes place at the slowest rate. For corrosion, activation polarization is possible for both oxidation and reduction reactions. Concentration polarization occurs when a reaction rate is limited by diffusion in a solution. For corrosion, concentration polarization is possible only for reduction reactions."
    },
    {
        "idx": 976,
        "question": "(a) Write the possible oxidation and reduction half-reactions that occur when magnesium is immersed in each of the following solutions: (iii) an HCl solution containing dissolved oxygen and, in addition, Fe2+ ions",
        "answer": "In an HCl solution containing dissolved oxygen and Fe2+ ions, possible reactions are Mg -> Mg2+ + 2 e- (oxidation), 4 H+ + O2 + 4 e- -> 2 H2O (reduction), and Fe2+ + 2 e- -> Fe (reduction).",
        "type": 4,
        "wrong_answers_1": "In an HCl solution containing dissolved oxygen, possible reactions are Mg -> Mg2+ + 2 e- (oxidation) and 4 H+ + O2 + 4 e- -> 2 H2O (reduction).",
        "wrong_answers_2": "In HCl, possible reactions are Mg -> Mg2+ + 2 e- (oxidation) and 2 H+ + 2 e- -> H2 (reduction).",
        "wrong_answers_3": "Oxidation occurs in the cell half with the lower Fe2+ concentration (2 × 10^-2 M)."
    },
    {
        "idx": 977,
        "question": "(b) In which of these solutions would you expect the magnesium to oxidize most rapidly? Why?",
        "answer": "The magnesium would probably oxidize most rapidly in the HCl solution containing dissolved oxygen and Fe2+ ions because there are two reduction reactions that will consume electrons from the oxidation of magnesium.",
        "type": 2
    },
    {
        "idx": 978,
        "question": "Demonstrate that the value of F in Equation 17.19 is 96,500 C/mol.",
        "answer": "The faraday constant F is 96,500 C/mol.",
        "type": 1,
        "wrong_answers_1": "The faraday constant F is 105302 C/mol.",
        "wrong_answers_2": "The faraday constant F is 109069 C/mol.",
        "wrong_answers_3": "The faraday constant F is 110327 C/mol."
    },
    {
        "idx": 979,
        "question": "Demonstrate that at 25 C (298 K), (RT)/(nF) ln x = (0.0592)/(n) log x.",
        "answer": "At 25 C (298 K), (RT)/(nF) ln x = (0.0592)/(n) log x. This gives units in volts since a volt is a J/C.",
        "type": 1,
        "wrong_answers_1": "At 27 C (285 K), (RT)/(nF) ln x = (0.0543)/(n) log x. This gives units in volts since a volt is a J/C.",
        "wrong_answers_2": "At 28 C (280 K), (RT)/(nF) ln x = (0.0526)/(n) log x. This gives units in volts since a volt is a J/C.",
        "wrong_answers_3": "At 27 C (273 K), (RT)/(nF) ln x = (0.0555)/(n) log x. This gives units in volts since a volt is a J/C."
    },
    {
        "idx": 980,
        "question": "Is a voltage generated between the two cell halves of a Zn/Zn2+ concentration cell where both electrodes are pure zinc, with Zn2+ concentrations of 1.0 M and 10^-2 M?",
        "answer": "Yes, a voltage is generated.",
        "type": 3
    },
    {
        "idx": 981,
        "question": "What is the magnitude of the voltage generated in the Zn/Zn2+ concentration cell with Zn2+ concentrations of 1.0 M and 10^-2 M?",
        "answer": "The magnitude of the voltage is 0.0592 v.",
        "type": 1,
        "wrong_answers_1": "The magnitude of the voltage is 0.0635 v.",
        "wrong_answers_2": "The magnitude of the voltage is 0.0511 v.",
        "wrong_answers_3": "The magnitude of the voltage is 0.0528 v."
    },
    {
        "idx": 982,
        "question": "Which electrode will be oxidized in the Zn/Zn2+ concentration cell with Zn2+ concentrations of 1.0 M and 10^-2 M?",
        "answer": "The electrode in the cell half with the Zn2+ concentration of 10^-2 M will be oxidized.",
        "type": 4,
        "wrong_answers_1": "Oxidation occurs in the cell half with the lower Fe2+ concentration (2 × 10^-2 M).",
        "wrong_answers_2": "Increasing the concentration of oxygen in the ambient environment will increase the concentration gradient, thereby increasing the diffusion flux of oxygen.",
        "wrong_answers_3": "Concentration polarization is rate controlling when the reaction rate is high and/or the concentration of active species in the liquid solution is low."
    },
    {
        "idx": 983,
        "question": "A piece of corroded steel plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was 10 in^{2} and that approximately 2.6kg had corroded away during the submersion. Assuming a corrosion penetration rate of 200 \\mathrm{mpy} for this alloy in seawater, estimate the time of submersion in years. The density of steel is 7.9g / {cm}^{3}.",
        "answer": "the time of submersion is 10 \\mathrm{yr}.",
        "type": 1,
        "wrong_answers_1": "the time of submersion is 9 \\mathrm{yr}.",
        "wrong_answers_2": "the time of submersion is 11 \\mathrm{yr}.",
        "wrong_answers_3": "the time of submersion is 11 \\mathrm{yr}."
    },
    {
        "idx": 984,
        "question": "A thick steel sheet of area 400 cm^{2} is exposed to air near the ocean. After a one-year period it was found to experience a weight loss of 375g due to corrosion. To what rate of corrosion, in both mpy and mm / \\mathrm{yr}, does this correspond?",
        "answer": "the corrosion rate is 1.2mm / \\mathrm{yr}. the corrosion rate is 46.7 \\mathrm{mpy}.",
        "type": 1,
        "wrong_answers_1": "the corrosion rate is 1.1mm / \\mathrm{yr}. the corrosion rate is 44.0 \\mathrm{mpy}.",
        "wrong_answers_2": "the corrosion rate is 1.3mm / \\mathrm{yr}. the corrosion rate is 42.7 \\mathrm{mpy}.",
        "wrong_answers_3": "the corrosion rate is 1.3mm / \\mathrm{yr}. the corrosion rate is 50.1 \\mathrm{mpy}."
    },
    {
        "idx": 985,
        "question": "Cite the major differences between activation and concentration polarizations.",
        "answer": "Activation polarization is the condition wherein a reaction rate is controlled by one step in a series of steps that takes place at the slowest rate. For corrosion, activation polarization is possible for both oxidation and reduction reactions. Concentration polarization occurs when a reaction rate is limited by diffusion in a solution. For corrosion, concentration polarization is possible only for reduction reactions.",
        "type": 4,
        "wrong_answers_1": "Activation polarization is rate controlling when the reaction rate is low and/or the concentration of active species in the liquid solution is high.",
        "wrong_answers_2": "Concentration polarization is rate controlling when the reaction rate is high and/or the concentration of active species in the liquid solution is low.",
        "wrong_answers_3": "In HCl, possible reactions are Mg -> Mg2+ + 2 e- (oxidation) and 2 H+ + 2 e- -> H2 (reduction)."
    },
    {
        "idx": 986,
        "question": "Under what conditions is activation polarization rate controlling?",
        "answer": "Activation polarization is rate controlling when the reaction rate is low and/or the concentration of active species in the liquid solution is high.",
        "type": 4,
        "wrong_answers_1": "Concentration polarization is rate controlling when the reaction rate is high and/or the concentration of active species in the liquid solution is low.",
        "wrong_answers_2": "Activation polarization is the condition wherein a reaction rate is controlled by one step in a series of steps that takes place at the slowest rate. For corrosion, activation polarization is possible for both oxidation and reduction reactions. Concentration polarization occurs when a reaction rate is limited by diffusion in a solution. For corrosion, concentration polarization is possible only for reduction reactions.",
        "wrong_answers_3": "Three factors that affect the rate of drying are temperature, humidity, and rate of air flow. The rate of drying is enhanced by increasing both the temperature and rate of air flow, and by decreasing the humidity of the air."
    },
    {
        "idx": 987,
        "question": "Under what conditions is concentration polarization rate controlling?",
        "answer": "Concentration polarization is rate controlling when the reaction rate is high and/or the concentration of active species in the liquid solution is low.",
        "type": 4,
        "wrong_answers_1": "Activation polarization is rate controlling when the reaction rate is low and/or the concentration of active species in the liquid solution is high.",
        "wrong_answers_2": "Activation polarization is the condition wherein a reaction rate is controlled by one step in a series of steps that takes place at the slowest rate. For corrosion, activation polarization is possible for both oxidation and reduction reactions. Concentration polarization occurs when a reaction rate is limited by diffusion in a solution. For corrosion, concentration polarization is possible only for reduction reactions.",
        "wrong_answers_3": "Three factors that affect the rate of drying are temperature, humidity, and rate of air flow. The rate of drying is enhanced by increasing both the temperature and rate of air flow, and by decreasing the humidity of the air."
    },
    {
        "idx": 988,
        "question": "Assuming that activation polarization controls both oxidation and reduction reactions, determine the rate of corrosion of metal M (in mol/cm2-s) given the following corrosion data: For Metal M: V(M/M2+)=-0.47 V, i0=5x10-10 A/cm2, β=+0.15; For Hydrogen: V(H+/H2)=0 V, i0=2x10-9 A/cm2, β=-0.12.",
        "answer": "the rate of corrosion of metal m is 2.64x10-13 mol/cm2-s.",
        "type": 1,
        "wrong_answers_1": "the rate of corrosion of metal m is 2.75x10-13 mol/cm2-s.",
        "wrong_answers_2": "the rate of corrosion of metal m is 2.85x10-13 mol/cm2-s.",
        "wrong_answers_3": "the rate of corrosion of metal m is 2.98x10-13 mol/cm2-s."
    },
    {
        "idx": 989,
        "question": "Compute the corrosion potential for this reaction given the following corrosion data: For Metal M: V(M/M2+)=-0.47 V, i0=5x10-10 A/cm2, β=+0.15; For Hydrogen: V(H+/H2)=0 V, i0=2x10-9 A/cm2, β=-0.12.",
        "answer": "the corrosion potential for this reaction is -0.169 v.",
        "type": 1,
        "wrong_answers_1": "the corrosion potential for this reaction is -0.149 v.",
        "wrong_answers_2": "the corrosion potential for this reaction is -0.163 v.",
        "wrong_answers_3": "the corrosion potential for this reaction is -0.155 v."
    },
    {
        "idx": 990,
        "question": "Briefly describe the phenomenon of passivity.",
        "answer": "Passivity is the loss of chemical reactivity, under particular environmental conditions, of normally active metals and alloys.",
        "type": 4,
        "wrong_answers_1": "Distinctive features: relatively low density, high melting temperatures, and high strengths are possible. Limitation: because of chemical reactivity with other materials at elevated temperatures, these alloys are expensive to refine. Applications: aircraft structures, space vehicles, and in chemical and petroleum industries.",
        "wrong_answers_2": "Chemical resistance decreases.",
        "wrong_answers_3": "For the Inconel-nickel couple, corrosion is unlikely inasmuch as both alloys appear within the same set of brackets (in both active and passive states)."
    },
    {
        "idx": 991,
        "question": "Name two common types of alloy that passivate.",
        "answer": "Stainless steels and aluminum alloys often passivate.",
        "type": 4,
        "wrong_answers_1": "The four classifications of steels are: Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, and High Alloy Steels (Stainless and Tool).",
        "wrong_answers_2": "Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, High Alloy Steels (Stainless and Tool)",
        "wrong_answers_3": "For the aluminum-cast iron couple, corrosion is possible, and aluminum will corrode."
    },
    {
        "idx": 992,
        "question": "Why does chromium in stainless steels make them more corrosion resistant in many environments than plain carbon steels?",
        "answer": "The chromium in stainless steels causes a very thin and highly adherent surface coating to form over the surface of the alloy, which protects it from further corrosion. For plain carbon steels, rust, instead of this adherent coating, forms.",
        "type": 4,
        "wrong_answers_1": "The chromium in stainless steels causes a very thin and highly adherent surface coating to form over the surface of the alloy, which protects it from further corrosion. For plain carbon steels, rust, instead of this adherent coating, forms.",
        "wrong_answers_2": "The four classifications of steels are: Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, and High Alloy Steels (Stainless and Tool).",
        "wrong_answers_3": "Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, High Alloy Steels (Stainless and Tool)"
    },
    {
        "idx": 993,
        "question": "Briefly explain why cold-worked metals are more susceptible to corrosion than noncold-worked metals.",
        "answer": "Cold-worked metals are more susceptible to corrosion than noncold-worked metals because of the increased dislocation density for the latter. The region in the vicinity of a dislocation that intersects the surface is at a higher energy state, and, therefore, is more readily attacked by a corrosive solution.",
        "type": 4,
        "wrong_answers_1": "The surface energy for a crystallographic plane will depend on its packing density-that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.55, the planar densities for BCC (100) and (110) are \\frac{3}{16 R^{2}} and \\frac{3}{8 R^{2} \\sqrt{2}}, respectively-that is \\frac{0.19}{R^{2}} and \\frac{0.27}{R^{2}}. Thus, since the planar density for (110) is greater, it will have the lower surface energy.",
        "wrong_answers_2": "The surface energy for a crystallographic plane will depend on its packing density [i.e., the planar density (Section 3.11)] - that is, the higher the packing density, the greater the number of nearest-neighbor atoms, and the more atomic bonds in that plane that are satisfied, and, consequently, the lower the surface energy. From the solution to Problem 3.54, planar densities for FCC (100) and (111) planes are \\frac{1}{4 R^{2}} and \\frac{1}{2 R^{2} \\sqrt{3}}, respectively - that is \\frac{0.25}{R^{2}} and \\frac{0.29}{R^{2}} (where R is the atomic radius). Thus, since the planar density for (111) is greater, it will have the lower surface energy.",
        "wrong_answers_3": "Crystalline ceramics are harder yet more brittle than metals because they (ceramics) have fewer slip systems, and, therefore, dislocation motion is highly restricted."
    },
    {
        "idx": 994,
        "question": "Briefly explain why, for a small anode-to-cathode area ratio, the corrosion rate will be higher than for a large ratio.",
        "answer": "For a small anode-to-cathode area ratio, the corrosion rate will be higher than for A large ratio. The reason for this is that for some given current flow associated with the corrosion reaction, for a small area ratio the current density at the anode will be greater than for a large ratio. The corrosion rate is proportional to the current density (i) according to Equation.",
        "type": 4,
        "wrong_answers_1": "of the possible copolymers, the only one for which there is a restriction on the ratio of repeat unit types is alternating; the ratio must be 1:1. therefore, on the basis of the result in part (a), the possibilities for this copolymer are random, graft, and block.",
        "wrong_answers_2": "A sacrificial anode is electrically coupled to the metal piece to be protected, which anode is also situated in the corrosion environment. The sacrificial anode is a metal or alloy that is chemically more reactive in the particular environment. It (the anode) preferentially oxidizes, and, upon giving up electrons to the other metal, protects it from electrochemical corrosion.",
        "wrong_answers_3": "When a current arises from a flow of electrons, the conduction is termed electronic; for \\\\ ionic conduction, the current results from the net motion of charged ions."
    },
    {
        "idx": 995,
        "question": "What are inhibitors?",
        "answer": "Inhibitors are substances that, when added to a corrosive environment in relatively low concentrations, decrease the environment's corrosiveness.",
        "type": 4,
        "wrong_answers_1": "A sacrificial anode is electrically coupled to the metal piece to be protected, which anode is also situated in the corrosion environment. The sacrificial anode is a metal or alloy that is chemically more reactive in the particular environment. It (the anode) preferentially oxidizes, and, upon giving up electrons to the other metal, protects it from electrochemical corrosion.",
        "wrong_answers_2": "Increasing the concentration of oxygen in the ambient environment will increase the concentration gradient, thereby increasing the diffusion flux of oxygen.",
        "wrong_answers_3": "Decrease in strength perpendicular to the direction of drawing."
    },
    {
        "idx": 996,
        "question": "What possible mechanisms account for their effectiveness?",
        "answer": "Possible mechanisms that account for the effectiveness of inhibitors are: (1) elimination of a chemically active species in the solution; (2) attachment of inhibitor molecules to the corroding surface so as to interfere with either the oxidation or reduction reaction; and (3) the formation of a very thin and protective coating on the corroding surface.",
        "type": 4,
        "wrong_answers_1": "The chromium in stainless steels causes a very thin and highly adherent surface coating to form over the surface of the alloy, which protects it from further corrosion. For plain carbon steels, rust, instead of this adherent coating, forms.",
        "wrong_answers_2": "The chromium in stainless steels causes a very thin and highly adherent surface coating to form over the surface of the alloy, which protects it from further corrosion. For plain carbon steels, rust, instead of this adherent coating, forms.",
        "wrong_answers_3": "Activation polarization is rate controlling when the reaction rate is low and/or the concentration of active species in the liquid solution is high."
    },
    {
        "idx": 997,
        "question": "Briefly describe the sacrificial anode technique used for galvanic protection.",
        "answer": "A sacrificial anode is electrically coupled to the metal piece to be protected, which anode is also situated in the corrosion environment. The sacrificial anode is a metal or alloy that is chemically more reactive in the particular environment. It (the anode) preferentially oxidizes, and, upon giving up electrons to the other metal, protects it from electrochemical corrosion.",
        "type": 4,
        "wrong_answers_1": "Oxidation occurs at the anode; reduction at the cathode.",
        "wrong_answers_2": "For a small anode-to-cathode area ratio, the corrosion rate will be higher than for A large ratio. The reason for this is that for some given current flow associated with the corrosion reaction, for a small area ratio the current density at the anode will be greater than for a large ratio. The corrosion rate is proportional to the current density (i) according to Equation.",
        "wrong_answers_3": "For brass, the bonding is metallic since it is a metal alloy."
    },
    {
        "idx": 998,
        "question": "Briefly describe the impressed current technique used for galvanic protection.",
        "answer": "An impressed current from an external dc power source provides excess electrons to the metallic structure to be protected.",
        "type": 4,
        "wrong_answers_1": "When a current arises from a flow of electrons, the conduction is termed electronic; for \\\\ ionic conduction, the current results from the net motion of charged ions.",
        "wrong_answers_2": "Oxidation is the process by which an atom gives up an electron (or electrons) to become a cation. Reduction is the process by which an atom acquires an extra electron (or electrons) and becomes an anion.",
        "wrong_answers_3": "For tungsten, the bonding is metallic since it is a metallic element from the periodic table."
    },
    {
        "idx": 999,
        "question": "Compute the electrical conductivity of a 5.1-mm (0.2-in.) diameter cylindrical silicon specimen 51mm (2 in.) long in which a current of 0.1 A passes in an axial direction. A voltage of 12.5V is measured across two probes that are separated by 38mm (1.5 in.).",
        "answer": "the electrical conductivity is σ = 14.9 (ω·m)^-1.",
        "type": 1,
        "wrong_answers_1": "the electrical conductivity is σ = 13.1 (ω·m)^-1.",
        "wrong_answers_2": "the electrical conductivity is σ = 14.4 (ω·m)^-1.",
        "wrong_answers_3": "the electrical conductivity is σ = 14.3 (ω·m)^-1."
    },
    {
        "idx": 1000,
        "question": "Compute the resistance over the entire 51mm (2 in.) of the specimen.",
        "answer": "the resistance is r = 168 ω.",
        "type": 1,
        "wrong_answers_1": "the resistance is r = 153 ω.",
        "wrong_answers_2": "the resistance is r = 146 ω.",
        "wrong_answers_3": "the resistance is r = 179 ω."
    },
    {
        "idx": 1001,
        "question": "Calculate the drift velocity of electrons in germanium at room temperature and when the magnitude of the electric field is 1000 V/m.",
        "answer": "the drift velocity of electrons in germanium is 380 m/s.",
        "type": 1,
        "wrong_answers_1": "the drift velocity of electrons in germanium is 431 m/s.",
        "wrong_answers_2": "the drift velocity of electrons in germanium is 435 m/s.",
        "wrong_answers_3": "the drift velocity of electrons in germanium is 397 m/s."
    },
    {
        "idx": 1002,
        "question": "Under these circumstances, how long does it take an electron to traverse a 25-mm (1-in.) length of crystal?",
        "answer": "the time required for an electron to traverse a 25-mm length of crystal is 6.6 × 10^-5 s.",
        "type": 1,
        "wrong_answers_1": "the time required for an electron to traverse a 28-mm length of crystal is 7.2 × 10^-5 s.",
        "wrong_answers_2": "the time required for an electron to traverse a 27-mm length of crystal is 7.5 × 10^-5 s.",
        "wrong_answers_3": "the time required for an electron to traverse a 28-mm length of crystal is 6.3 × 10^-5 s."
    },
    {
        "idx": 1003,
        "question": "At room temperature the electrical conductivity and the electron mobility for copper are 6.0 x 10^7 (Ω·m)^-1 and 0.0030 m^2/V·s, respectively. Compute the number of free electrons per cubic meter for copper at room temperature.",
        "answer": "the number of free electrons per cubic meter for copper at room temperature is 1.25 x 10^29 m^-3.",
        "type": 1,
        "wrong_answers_1": "the number of free electrons per cubic meter for copper at room temperature is 1.40 x 10^29 m^-3.",
        "wrong_answers_2": "the number of free electrons per cubic meter for copper at room temperature is 1.16 x 10^29 m^-3.",
        "wrong_answers_3": "the number of free electrons per cubic meter for copper at room temperature is 1.09 x 10^29 m^-3."
    },
    {
        "idx": 1004,
        "question": "At room temperature the electrical conductivity and the electron mobility for copper are 6.0 x 10^7 (Ω·m)^-1 and 0.0030 m^2/V·s, respectively. What is the number of free electrons per copper atom? Assume a density of 8.9 g/cm^3.",
        "answer": "the number of free electrons per copper atom is 1.48.",
        "type": 1,
        "wrong_answers_1": "the number of free electrons per copper atom is 1.57.",
        "wrong_answers_2": "the number of free electrons per copper atom is 1.41.",
        "wrong_answers_3": "the number of free electrons per copper atom is 1.35."
    },
    {
        "idx": 1005,
        "question": "At room temperature the electrical conductivity of \\mathrm{PbTe} is 500(\\Omega·m)^{-1}, whereas the electron and hole mobilities are 0.16 and 0.075{m}^{2} / V·s, respectively. Compute the intrinsic carrier concentration for \\mathrm{PbTe} at room temperature.",
        "answer": "the intrinsic carrier concentration for \\mathrm{pbte} at room temperature is 1.33 × 10^{22}{m}^{-3}.",
        "type": 1,
        "wrong_answers_1": "the intrinsic carrier concentration for \\mathrm{pbte} at room temperature is 1.41 × 10^{22}{m}^{-3}.",
        "wrong_answers_2": "the intrinsic carrier concentration for \\mathrm{pbte} at room temperature is 1.44 × 10^{22}{m}^{-3}.",
        "wrong_answers_3": "the intrinsic carrier concentration for \\mathrm{pbte} at room temperature is 1.14 × 10^{22}{m}^{-3}."
    },
    {
        "idx": 1006,
        "question": "Define the term 'intrinsic' as it pertains to semiconducting materials and provide an example.",
        "answer": "intrinsic--high purity (undoped) Si, GaAs, CdS, etc.",
        "type": 4,
        "wrong_answers_1": "compound--GaAs, InP, CdS, etc.",
        "wrong_answers_2": "elemental--Ge and Si.",
        "wrong_answers_3": "extrinsic--P-doped Ge, B-doped Si, S-doped GaP, etc."
    },
    {
        "idx": 1007,
        "question": "Define the term 'extrinsic' as it pertains to semiconducting materials and provide an example.",
        "answer": "extrinsic--P-doped Ge, B-doped Si, S-doped GaP, etc.",
        "type": 4,
        "wrong_answers_1": "elemental--Ge and Si.",
        "wrong_answers_2": "intrinsic--high purity (undoped) Si, GaAs, CdS, etc.",
        "wrong_answers_3": "Boron will act as an acceptor in Ge. Since it (B) is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom."
    },
    {
        "idx": 1008,
        "question": "Define the term 'compound' as it pertains to semiconducting materials and provide an example.",
        "answer": "compound--GaAs, InP, CdS, etc.",
        "type": 4,
        "wrong_answers_1": "intrinsic--high purity (undoped) Si, GaAs, CdS, etc.",
        "wrong_answers_2": "Stoichiometric means having exactly the ratio of anions to cations as specified by the chemical formula for the compound.",
        "wrong_answers_3": "The two characteristics of component ions that determine the crystal structure of a ceramic compound are: 1) the magnitude of the electrical charge on each ion, and 2) the relative sizes of the cations and anions."
    },
    {
        "idx": 1009,
        "question": "Define the term 'elemental' as it pertains to semiconducting materials and provide an example.",
        "answer": "elemental--Ge and Si.",
        "type": 4,
        "wrong_answers_1": "Boron will act as an acceptor in Ge. Since it (B) is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom.",
        "wrong_answers_2": "extrinsic--P-doped Ge, B-doped Si, S-doped GaP, etc.",
        "wrong_answers_3": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom."
    },
    {
        "idx": 1010,
        "question": "Explain why no hole is generated by the electron excitation involving a donor impurity atom.",
        "answer": "No hole is generated by an electron excitation involving a donor impurity atom because the excitation comes from a level within the band gap, and thus, no missing electron is created within the normally filled valence band.",
        "type": 4,
        "wrong_answers_1": "No free electron is generated by an electron excitation involving an acceptor impurity atom because the electron is excited from the valence band into the impurity level within the band gap; no free electron is introduced into the conduction band.",
        "wrong_answers_2": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities.",
        "wrong_answers_3": "For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid."
    },
    {
        "idx": 1011,
        "question": "Explain why no free electron is generated by the electron excitation involving an acceptor impurity atom.",
        "answer": "No free electron is generated by an electron excitation involving an acceptor impurity atom because the electron is excited from the valence band into the impurity level within the band gap; no free electron is introduced into the conduction band.",
        "type": 4,
        "wrong_answers_1": "No hole is generated by an electron excitation involving a donor impurity atom because the excitation comes from a level within the band gap, and thus, no missing electron is created within the normally filled valence band.",
        "wrong_answers_2": "For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid.",
        "wrong_answers_3": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities."
    },
    {
        "idx": 1012,
        "question": "Germanium to which 5 × 10^22 m^-3 Sb atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the Sb atoms may be thought of as being ionized (i.e., one charge carrier exists for each Sb atom). Is this material n-type or p-type?",
        "answer": "this germanium material to which has been added 5 × 10^22 m^-3 sb atoms is n-type since sb is a donor in ge.",
        "type": 4,
        "wrong_answers_1": "this germanium material to which has been added 10^24 m^-3 as atoms is n-type since as is a donor in ge.",
        "wrong_answers_2": "Sulfur will act as a donor in InSb. Since S is from group VIA of the periodic table, it will substitute for Sb; an S atom has one more valence electron than an Sb atom.",
        "wrong_answers_3": "Boron will act as an acceptor in Ge. Since it (B) is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom."
    },
    {
        "idx": 1013,
        "question": "Germanium to which 5 × 10^22 m^-3 Sb atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the Sb atoms may be thought of as being ionized (i.e., one charge carrier exists for each Sb atom). Calculate the electrical conductivity of this material, assuming electron and hole mobilities of 0.1 and 0.05 m^2/V-s, respectively.",
        "answer": "the electrical conductivity of this material is 800 (ω·m)^-1.",
        "type": 1,
        "wrong_answers_1": "the electrical conductivity of this material is 918 (ω·m)^-1.",
        "wrong_answers_2": "the electrical conductivity of this material is 765 (ω·m)^-1.",
        "wrong_answers_3": "the electrical conductivity of this material is 888 (ω·m)^-1."
    },
    {
        "idx": 1014,
        "question": "The following electrical characteristics have been determined for both intrinsic and p-type extrinsic indium phosphide ( \\mathrm{InP} ) at room temperature:\n\\begin{tabular}{lccc}\n\\hline & \\boldsymbol{\\sigma}\\left(\\boldsymbol{\\Omega}·\\mathbf{m}\\right)^{-1} & \\mathbf{n}\\left(\\mathbf{m}^{-3}\\right) & \\mathbf{p}\\left(\\mathbf{m}^{-3}\\right) \\\\\n\\hline Intrinsic & 2.5 × 10^{-6} & 3.0 × 10^{13} & 3.0 × 10^{12} \\\\\nExtrinsic (n-type) & 3.6 × 10^{-5} & 4.5 × 10^{14} & 2.0 × 10^{12} \\\\\n\\hline\n\\end{tabular}\nCalculate electron and hole mobilities.",
        "answer": "the electron and hole mobilities for \\mathrm{inp} are \\mu_{e} = 0.50 \\, m^{2}/\\mathrm{v}\\cdots and \\mu_{h} = 0.02 \\, m^{2}/\\mathrm{v}\\cdots.",
        "type": 1,
        "wrong_answers_1": "the electron and hole mobilities for \\mathrm{inp} are \\mu_{e} = 0.47 \\, m^{2}/\\mathrm{v}\\cdots and \\mu_{h} = 0.02 \\, m^{2}/\\mathrm{v}\\cdots.",
        "wrong_answers_2": "the electron and hole mobilities for \\mathrm{inp} are \\mu_{e} = 0.44 \\, m^{2}/\\mathrm{v}\\cdots and \\mu_{h} = 0.02 \\, m^{2}/\\mathrm{v}\\cdots.",
        "wrong_answers_3": "the electron and hole mobilities for \\mathrm{inp} are \\mu_{e} = 0.43 \\, m^{2}/\\mathrm{v}\\cdots and \\mu_{h} = 0.02 \\, m^{2}/\\mathrm{v}\\cdots."
    },
    {
        "idx": 1015,
        "question": "Some metal alloy is known to have electrical conductivity and electron mobility values of 1.5 × 10^{7} (\\Omega·m)^{-1} and 0.0020{m}^{2} / V·s, respectively. Through a specimen of this alloy that is 35mm thick is passed a current of 45 A. What magnetic field would need to be imposed to yield a Hall voltage of -1.0 × 10^{-7}V ?",
        "answer": "the required magnetic field is 0.58 \\mathrm{tesla}.",
        "type": 1,
        "wrong_answers_1": "the required magnetic field is 0.55 \\mathrm{tesla}.",
        "wrong_answers_2": "the required magnetic field is 0.54 \\mathrm{tesla}.",
        "wrong_answers_3": "the required magnetic field is 0.61 \\mathrm{tesla}."
    },
    {
        "idx": 1016,
        "question": "What is one function that a transistor may perform in an electronic circuit?",
        "answer": "In an electronic circuit, a transistor may be used to amplify an electrical signal.",
        "type": 4,
        "wrong_answers_1": "In an electronic circuit, a transistor may act as a switching device in computers.",
        "wrong_answers_2": "(b) Electronic, ionic, and orientation polarizations would be observed in lead titanate. The lead, titanium, and oxygen would undoubtedly be largely ionic in character. Furthermore, orientation polarization is also possible inasmuch as permanent dipole moments may be induced in the same manner as for BaTiO3. Only electronic polarization is to be found in gaseous neon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Only electronic polarization is to be found in solid diamond; this material does not have molecules with permanent dipole moments, nor is it an ionic material. Both electronic and ionic polarizations will be found in solid KCl, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid NH3. The NH3 molecules have permanent dipole moments that are easily oriented in the liquid state.",
        "wrong_answers_3": "Only electronic polarization is found in gaseous argon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Both electronic and ionic polarizations are found in solid LiF, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid H2O. The H2O molecules have permanent dipole moments that are easily oriented in the liquid state. Only electronic polarization is to be found in solid Si; this material does not have molecules with permanent dipole moments, nor is it an ionic material."
    },
    {
        "idx": 1017,
        "question": "What is another function that a transistor may perform in an electronic circuit?",
        "answer": "In an electronic circuit, a transistor may act as a switching device in computers.",
        "type": 4,
        "wrong_answers_1": "In an electronic circuit, a transistor may be used to amplify an electrical signal.",
        "wrong_answers_2": "(b) Electronic, ionic, and orientation polarizations would be observed in lead titanate. The lead, titanium, and oxygen would undoubtedly be largely ionic in character. Furthermore, orientation polarization is also possible inasmuch as permanent dipole moments may be induced in the same manner as for BaTiO3. Only electronic polarization is to be found in gaseous neon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Only electronic polarization is to be found in solid diamond; this material does not have molecules with permanent dipole moments, nor is it an ionic material. Both electronic and ionic polarizations will be found in solid KCl, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid NH3. The NH3 molecules have permanent dipole moments that are easily oriented in the liquid state.",
        "wrong_answers_3": "Arsenic will act as an acceptor in ZnTe. Since As is from group VA of the periodic table, it will substitute for Te; furthermore, an As atom has one less valence electron than a Te atom."
    },
    {
        "idx": 1018,
        "question": "At temperatures between 775^{\\circ} C\\left(1048 K\\right) and 1100^{\\circ} C\\left(1373 K\\right), the activation energy and preexponential for the diffusion coefficient of \\mathrm{Fe}^{2+} in \\mathrm{FeO} are 102,000 J/ mol and 7.3 × 10^{-8}{m}^{2} / s, respectively. Compute the mobility for an \\mathrm{Fe}^{2+} ion at 1000^{\\circ} C(1273 K).",
        "answer": "the mobility for an \\mathrm{fe}^{2+} ion at 1000^{\\circ} C (1273 k) is 8.64 × 10^{-11}{m}^{2} / \\mathrm{v}·s.",
        "type": 1,
        "wrong_answers_1": "the mobility for an \\mathrm{fe}^{2+} ion at 910^{\\circ} C (1393 k) is 9.45 × 10^{-11}{m}^{2} / \\mathrm{v}·s.",
        "wrong_answers_2": "the mobility for an \\mathrm{fe}^{2+} ion at 940^{\\circ} C (1153 k) is 9.90 × 10^{-11}{m}^{2} / \\mathrm{v}·s.",
        "wrong_answers_3": "the mobility for an \\mathrm{fe}^{2+} ion at 930^{\\circ} C (1433 k) is 7.70 × 10^{-11}{m}^{2} / \\mathrm{v}·s."
    },
    {
        "idx": 1019,
        "question": "A parallel-plate capacitor using a dielectric material having an \\varepsilon_{r} of 2.5 has a plate spacing of 1 mm (0.04 in.). If another material having a dielectric constant of 4.0 is used and the capacitance is to be unchanged, what must be the new spacing between the plates?",
        "answer": "the new spacing between the plates must be 1.6 mm.",
        "type": 1,
        "wrong_answers_1": "the new spacing between the plates must be 1.5 mm.",
        "wrong_answers_2": "the new spacing between the plates must be 1.5 mm.",
        "wrong_answers_3": "the new spacing between the plates must be 1.4 mm."
    },
    {
        "idx": 1020,
        "question": "(a) For each of the three types of polarization, briefly describe the mechanism by which dipoles are induced and/or oriented by the action of an applied electric field.",
        "answer": "(a) For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field.",
        "type": 4,
        "wrong_answers_1": "For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field.",
        "wrong_answers_2": "(b) Electronic, ionic, and orientation polarizations would be observed in lead titanate. The lead, titanium, and oxygen would undoubtedly be largely ionic in character. Furthermore, orientation polarization is also possible inasmuch as permanent dipole moments may be induced in the same manner as for BaTiO3. Only electronic polarization is to be found in gaseous neon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Only electronic polarization is to be found in solid diamond; this material does not have molecules with permanent dipole moments, nor is it an ionic material. Both electronic and ionic polarizations will be found in solid KCl, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid NH3. The NH3 molecules have permanent dipole moments that are easily oriented in the liquid state.",
        "wrong_answers_3": "Only electronic polarization is found in gaseous argon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Both electronic and ionic polarizations are found in solid LiF, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid H2O. The H2O molecules have permanent dipole moments that are easily oriented in the liquid state. Only electronic polarization is to be found in solid Si; this material does not have molecules with permanent dipole moments, nor is it an ionic material."
    },
    {
        "idx": 1021,
        "question": "(b) For solid lead titanate (PbTiO3), gaseous neon, diamond, solid KCl, and liquid NH3 what kind(s) of polarization is (are) possible? Why?",
        "answer": "(b) Electronic, ionic, and orientation polarizations would be observed in lead titanate. The lead, titanium, and oxygen would undoubtedly be largely ionic in character. Furthermore, orientation polarization is also possible inasmuch as permanent dipole moments may be induced in the same manner as for BaTiO3. Only electronic polarization is to be found in gaseous neon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Only electronic polarization is to be found in solid diamond; this material does not have molecules with permanent dipole moments, nor is it an ionic material. Both electronic and ionic polarizations will be found in solid KCl, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid NH3. The NH3 molecules have permanent dipole moments that are easily oriented in the liquid state.",
        "type": 4,
        "wrong_answers_1": "Only electronic polarization is found in gaseous argon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Both electronic and ionic polarizations are found in solid LiF, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid H2O. The H2O molecules have permanent dipole moments that are easily oriented in the liquid state. Only electronic polarization is to be found in solid Si; this material does not have molecules with permanent dipole moments, nor is it an ionic material.",
        "wrong_answers_2": "(a) For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field.",
        "wrong_answers_3": "For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field."
    },
    {
        "idx": 1022,
        "question": "For aluminum, the heat capacity at constant volume Cv at 30 K is 0.81 J/mol-K, and the Debye temperature is 375 K. Estimate the specific heat at 50 K.",
        "answer": "the specific heat at 50 K is 139 J/kg-K.",
        "type": 1,
        "wrong_answers_1": "the specific heat at 48 K is 129 J/kg-K.",
        "wrong_answers_2": "the specific heat at 47 K is 119 J/kg-K.",
        "wrong_answers_3": "the specific heat at 43 K is 144 J/kg-K."
    },
    {
        "idx": 1023,
        "question": "For aluminum, the heat capacity at constant volume Cv at 30 K is 0.81 J/mol-K, and the Debye temperature is 375 K. Estimate the specific heat at 425 K.",
        "answer": "the specific heat at 425 K is 923 J/kg-K.",
        "type": 1,
        "wrong_answers_1": "the specific heat at 401 K is 1060 J/kg-K.",
        "wrong_answers_2": "the specific heat at 402 K is 792 J/kg-K.",
        "wrong_answers_3": "the specific heat at 444 K is 1006 J/kg-K."
    },
    {
        "idx": 1024,
        "question": "(a) Briefly explain why Cv rises with increasing temperature at temperatures near 0 K.",
        "answer": "The reason that Cv rises with increasing temperature at temperatures near 0 K is because, in this temperature range, the allowed vibrational energy levels of the lattice waves are far apart relative to the available thermal energy, and only a portion of the lattice waves may be excited. As temperature increases, more of the lattice waves may be excited by the available thermal energy, and, hence, the ability of the solid to absorb energy (i.e., the magnitude of the heat capacity) increases.",
        "type": 4,
        "wrong_answers_1": "At temperatures far removed from 0 K, Cv becomes independent of temperature because all of the lattice waves have been excited and the energy required to produce an incremental temperature change is nearly constant.",
        "wrong_answers_2": "For some ceramic materials, the thermal conductivity first decreases with rising temperature because the scattering of lattice vibrations increases with temperature. At higher temperatures, the thermal conductivity will increase for some ceramics that are porous because radiant heat transfer across pores may become important, which process increases with rising temperature.",
        "wrong_answers_3": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities."
    },
    {
        "idx": 1025,
        "question": "(b) Briefly explain why Cv becomes virtually independent of temperature at temperatures far removed from 0 K.",
        "answer": "At temperatures far removed from 0 K, Cv becomes independent of temperature because all of the lattice waves have been excited and the energy required to produce an incremental temperature change is nearly constant.",
        "type": 4,
        "wrong_answers_1": "The reason that Cv rises with increasing temperature at temperatures near 0 K is because, in this temperature range, the allowed vibrational energy levels of the lattice waves are far apart relative to the available thermal energy, and only a portion of the lattice waves may be excited. As temperature increases, more of the lattice waves may be excited by the available thermal energy, and, hence, the ability of the solid to absorb energy (i.e., the magnitude of the heat capacity) increases.",
        "wrong_answers_2": "The glass transition temperature is, for a noncrystalline ceramic, that temperature at which there is a change of slope for the specific volume versus temperature curve.",
        "wrong_answers_3": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities."
    },
    {
        "idx": 1026,
        "question": "A 0.1{m}(3.9 in.) rod of a metal elongates 0.2 mm(0.0079 in.) on heating from 20 to 100^{\\circ} C(68 \\mathrm{to} 212^{\\circ} F ). Determine the value of the linear coefficient of thermal expansion for this material.",
        "answer": "the linear coefficient of thermal expansion for this material is 25.0 × 10^{-6} \\, (\\degree c)^{-1}.",
        "type": 1,
        "wrong_answers_1": "the linear coefficient of thermal expansion for this material is 28.4 × 10^{-6} \\, (\\degree c)^{-1}.",
        "wrong_answers_2": "the linear coefficient of thermal expansion for this material is 21.8 × 10^{-6} \\, (\\degree c)^{-1}.",
        "wrong_answers_3": "the linear coefficient of thermal expansion for this material is 26.8 × 10^{-6} \\, (\\degree c)^{-1}."
    },
    {
        "idx": 1027,
        "question": "Briefly explain why the thermal conductivities are higher for crystalline than noncrystalline ceramics.",
        "answer": "Thermal conductivities are higher for crystalline than for noncrystalline ceramics because, for noncrystalline, phonon scattering, and thus the resistance to heat transport, is much more effective due to the highly disordered and irregular atomic structure.",
        "type": 4,
        "wrong_answers_1": "Crystalline ceramics are harder yet more brittle than metals because they (ceramics) have fewer slip systems, and, therefore, dislocation motion is highly restricted.",
        "wrong_answers_2": "Quartz will have a larger conductivity than fused silica because fused silica is noncrystalline (whereas quartz is crystalline) and lattice vibrations are more effectively scattered in noncrystalline materials.",
        "wrong_answers_3": "Increasing the degree of crystallinity of a semicrystalline polymer enhances its thermal conductivity; the vibrations, rotations, etc. of the molecular chains are more effective modes of thermal transport when a crystalline structure prevails."
    },
    {
        "idx": 1028,
        "question": "Briefly explain why metals are typically better thermal conductors than ceramic materials.",
        "answer": "Metals are typically better thermal conductors than are ceramic materials because, for metals, most of the heat is transported by free electrons (of which there are relatively large numbers). In ceramic materials, the primary mode of thermal conduction is via phonons, and phonons are more easily scattered than are free electrons.",
        "type": 4,
        "wrong_answers_1": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities.",
        "wrong_answers_2": "Hexagonal close packed metals are typically more brittle than FCC and BCC metals because there are fewer slip systems in HCP.",
        "wrong_answers_3": "Wires are large-diameter metals."
    },
    {
        "idx": 1029,
        "question": "Briefly explain why porosity decreases the thermal conductivity of ceramic and polymeric materials, rendering them more thermally insulative.",
        "answer": "Porosity decreases the thermal conductivity of ceramic and polymeric materials because the thermal conductivity of a gas phase that occupies pore space is extremely small relative to that of the solid material. Furthermore, contributions from gaseous convection are generally insignificant.",
        "type": 4,
        "wrong_answers_1": "For some ceramic materials, the thermal conductivity first decreases with rising temperature because the scattering of lattice vibrations increases with temperature. At higher temperatures, the thermal conductivity will increase for some ceramics that are porous because radiant heat transfer across pores may become important, which process increases with rising temperature.",
        "wrong_answers_2": "For refractory ceramic materials, three characteristics that improve with increasing porosity are (1) decreased thermal expansion and contraction upon thermal cycling, (2) improved thermal insulation, and (3) improved resistance to thermal shock.",
        "wrong_answers_3": "The two thermal properties of a liquid medium that influence its quenching effectiveness are thermal conductivity and heat capacity."
    },
    {
        "idx": 1030,
        "question": "Briefly explain how the degree of crystallinity affects the thermal conductivity of polymeric materials and why.",
        "answer": "Increasing the degree of crystallinity of a semicrystalline polymer enhances its thermal conductivity; the vibrations, rotations, etc. of the molecular chains are more effective modes of thermal transport when a crystalline structure prevails.",
        "type": 4,
        "wrong_answers_1": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "wrong_answers_2": "The linear polyethylene will have the larger conductivity than the branched polyethylene because the former will have the higher degree of crystallinity. Linear polymers have higher degrees of crystallinity than branched polymers. Since heat transfer is accomplished by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "wrong_answers_3": "The two thermal properties of a liquid medium that influence its quenching effectiveness are thermal conductivity and heat capacity."
    },
    {
        "idx": 1031,
        "question": "For some ceramic materials, why does the thermal conductivity first decrease and then increase with rising temperature?",
        "answer": "For some ceramic materials, the thermal conductivity first decreases with rising temperature because the scattering of lattice vibrations increases with temperature. At higher temperatures, the thermal conductivity will increase for some ceramics that are porous because radiant heat transfer across pores may become important, which process increases with rising temperature.",
        "type": 4,
        "wrong_answers_1": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "wrong_answers_2": "The reason that Cv rises with increasing temperature at temperatures near 0 K is because, in this temperature range, the allowed vibrational energy levels of the lattice waves are far apart relative to the available thermal energy, and only a portion of the lattice waves may be excited. As temperature increases, more of the lattice waves may be excited by the available thermal energy, and, hence, the ability of the solid to absorb energy (i.e., the magnitude of the heat capacity) increases.",
        "wrong_answers_3": "The linear polyethylene will have the larger conductivity than the branched polyethylene because the former will have the higher degree of crystallinity. Linear polymers have higher degrees of crystallinity than branched polymers. Since heat transfer is accomplished by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity."
    },
    {
        "idx": 1032,
        "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Pure copper; aluminum bronze (95 wt% Cu-5 wt% Al).",
        "answer": "Pure copper will have a larger conductivity than aluminum bronze because the impurity atoms in the latter will lead to a greater degree of free electron scattering.",
        "type": 4,
        "wrong_answers_1": "Pure silver will have a larger conductivity than sterling silver because the impurity atoms in the latter will lead to a greater degree of free electron scattering.",
        "wrong_answers_2": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "wrong_answers_3": "The linear polyethylene will have the larger conductivity than the branched polyethylene because the former will have the higher degree of crystallinity. Linear polymers have higher degrees of crystallinity than branched polymers. Since heat transfer is accomplished by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity."
    },
    {
        "idx": 1033,
        "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Fused silica; quartz.",
        "answer": "Quartz will have a larger conductivity than fused silica because fused silica is noncrystalline (whereas quartz is crystalline) and lattice vibrations are more effectively scattered in noncrystalline materials.",
        "type": 4,
        "wrong_answers_1": "Polycrystalline silica will have a larger conductivity than fused silica because fused silica is noncrystalline and lattice vibrations are more effectively scattered in noncrystalline materials.",
        "wrong_answers_2": "clay, quartz, and a flux.",
        "wrong_answers_3": "Thermal conductivities are higher for crystalline than for noncrystalline ceramics because, for noncrystalline, phonon scattering, and thus the resistance to heat transport, is much more effective due to the highly disordered and irregular atomic structure."
    },
    {
        "idx": 1034,
        "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Linear polyethylene; branched polyethylene.",
        "answer": "The linear polyethylene will have the larger conductivity than the branched polyethylene because the former will have the higher degree of crystallinity. Linear polymers have higher degrees of crystallinity than branched polymers. Since heat transfer is accomplished by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "type": 4,
        "wrong_answers_1": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "wrong_answers_2": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will most likely have a higher percent crystallinity, and, therefore, a higher melting temperature than the branched polyethylene. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration.",
        "wrong_answers_3": "The isotactic polypropylene will have a larger thermal conductivity than the atactic polypropylene because isotactic polymers have higher degrees of crystallinity. The influence of crystallinity on conductivity is explained in part (c)."
    },
    {
        "idx": 1035,
        "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Random poly(styrene-butadiene) copolymer; alternating poly(styrene-butadiene) copolymer.",
        "answer": "The alternating poly(styrene-butadiene) copolymer will have a higher crystallinity than the random copolymer; alternating copolymers crystallize more easily than random ones. The influence of crystallinity on conductivity is explained in part (c).",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to decide for these two copolymers. The alternating poly(styrene-ethylene) copolymer is more likely to crystallize. Alternating copolymers crystallize more easily than do random copolymers.",
        "wrong_answers_2": "No, it is not possible. The random acrylonitrile-butadiene copolymer will tend to a lower degree of crystallinity than the alternating acrylonitrile-butadiene copolymer inasmuch as random copolymers don't normally crystallize. On this basis only, the alternating material would have a higher modulus inasmuch as tensile modulus increases with degree of crystallinity. On the other hand, the random copolymer has a higher degree of crosslinking (10% versus 5% for the alternating copolymer), and, on this basis only would have the higher tensile modulus-an increase in crosslinking leads to an increase in E. Thus, this determination is not possible; with regard to degree of crystallinity the alternating material has the higher E, whereas the random copolymer would have a higher E value on the basis of degree of crosslinking.",
        "wrong_answers_3": "The isotactic polypropylene will have a larger thermal conductivity than the atactic polypropylene because isotactic polymers have higher degrees of crystallinity. The influence of crystallinity on conductivity is explained in part (c)."
    },
    {
        "idx": 1036,
        "question": "Explain the two sources of magnetic moments for electrons.",
        "answer": "The two sources of magnetic moments for electrons are the electron's orbital motion around the nucleus, and also, its spin.",
        "type": 4,
        "wrong_answers_1": "All atoms do not have a net magnetic moment. If an atom has completely filled electron shells or subshells, there will be a cancellation of both orbital and spin magnetic moments.",
        "wrong_answers_2": "Each electron will have a net magnetic moment from spin, and possibly, orbital contributions, which do not cancel for an isolated atom.",
        "wrong_answers_3": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions--domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.\nFor paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains."
    },
    {
        "idx": 1037,
        "question": "Do all electrons have a net magnetic moment? Why or why not?",
        "answer": "Each electron will have a net magnetic moment from spin, and possibly, orbital contributions, which do not cancel for an isolated atom.",
        "type": 4,
        "wrong_answers_1": "All atoms do not have a net magnetic moment. If an atom has completely filled electron shells or subshells, there will be a cancellation of both orbital and spin magnetic moments.",
        "wrong_answers_2": "The ms quantum number designates the spin moment on each electron.",
        "wrong_answers_3": "The two sources of magnetic moments for electrons are the electron's orbital motion around the nucleus, and also, its spin."
    },
    {
        "idx": 1038,
        "question": "Do all atoms have a net magnetic moment? Why or why not?",
        "answer": "All atoms do not have a net magnetic moment. If an atom has completely filled electron shells or subshells, there will be a cancellation of both orbital and spin magnetic moments.",
        "type": 4,
        "wrong_answers_1": "Each electron will have a net magnetic moment from spin, and possibly, orbital contributions, which do not cancel for an isolated atom.",
        "wrong_answers_2": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions--domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.\nFor paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains.",
        "wrong_answers_3": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions - domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.For paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains."
    },
    {
        "idx": 1039,
        "question": "Confirm that there are 2.2 Bohr magnetons associated with each iron atom, given that the saturation magnetization is 1.70 × 10^{6} \\mathrm{~A} / m, that iron has a BCC crystal structure, and that the unit cell edge length is 0.2866 nm.",
        "answer": "2.16 bohr magnetons/atom",
        "type": 1,
        "wrong_answers_1": "1.98 bohr magnetons/atom",
        "wrong_answers_2": "1.99 bohr magnetons/atom",
        "wrong_answers_3": "1.97 bohr magnetons/atom"
    },
    {
        "idx": 1040,
        "question": "There is associated with each atom in paramagnetic and ferromagnetic materials a net magnetic moment. Explain why ferromagnetic materials can be permanently magnetized whereas paramagnetic ones cannot.",
        "answer": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions--domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.\nFor paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains.",
        "type": 4,
        "wrong_answers_1": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions - domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.For paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains.",
        "wrong_answers_2": "All atoms do not have a net magnetic moment. If an atom has completely filled electron shells or subshells, there will be a cancellation of both orbital and spin magnetic moments.",
        "wrong_answers_3": "For type I superconductors, with increasing magnetic field the material is completely diamagnetic and superconductive below H_{C}, while at H_{C} conduction becomes normal and complete magnetic flux penetration takes place. On the other hand, for type II superconductors upon increasing the magnitude of the magnetic field, the transition from the superconducting to normal conducting states is gradual between lower-critical and upper-critical fields; so also is magnetic flux penetration gradual. Furthermore, type II superconductors generally have higher critical temperatures and critical magnetic fields."
    },
    {
        "idx": 1041,
        "question": "The formula for yttrium iron garnet \\left(\\mathrm{Y}_{3} \\mathrm{Fe}_{5} \\mathrm{O}_{12}\\right) may be written in the form \\mathrm{Y}_{3}^{+} \\mathrm{Fe}_{4}^{a} \\mathrm{Fe}_{d}^{a} \\mathrm{O}_{12}, where the superscripts a, c, and d represent different sites on which the Y^{3+} and \\mathrm{Fe}^{3+} ions are located. The spin magnetic moments for the Y^{3+} and \\mathrm{Fe}^{3+}\n moments for the Y^{3+} and \\mathrm{Fe}^{3-} ions positioned in the a and c sites are oriented parallel to one another and antiparallel to the \\mathrm{Fe}^{3+} ions in d sites. Compute the number of Bohr magnetons associated with each Y^{3+} ion, given the following information: (1) each unit cell consists of eight formula \\left(Y_{3} \\mathrm{Fe}_{5} \\mathrm{O}_{12}\\right), units; (2) the unit cell is cubic with an edge length of 1.2376nm; (3) the saturation magnetization for this material is 1.0 × 10^{4} \\mathrm{~A} / m; and (4) assume that there are 5 Bohr magnetons associated with each \\mathrm{Fe}^{3+} ion.",
        "answer": "the number of bohr magnetons associated with each y^{3+} ion is 1.75 \\, \\text{bm}.",
        "type": 1,
        "wrong_answers_1": "the number of bohr magnetons associated with each y^{3+} ion is 2.00 \\, \\text{bm}.",
        "wrong_answers_2": "the number of bohr magnetons associated with each y^{3+} ion is 1.91 \\, \\text{bm}.",
        "wrong_answers_3": "the number of bohr magnetons associated with each y^{3+} ion is 1.97 \\, \\text{bm}."
    },
    {
        "idx": 1042,
        "question": "Why does the magnitude of the saturation magnetization decrease with increasing temperature for ferromagnetic materials?",
        "answer": "The saturation magnetization decreases with increasing temperature because the atomic thermal vibrational motions counteract the coupling forces between the adjacent atomic dipole moments, causing some magnetic dipole misalignment.",
        "type": 4,
        "wrong_answers_1": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions - domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.For paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains.",
        "wrong_answers_2": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions--domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.\nFor paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains.",
        "wrong_answers_3": "Ferromagnetic behavior ceases above the Curie temperature because the atomic thermal vibrations are sufficiently violent so as to completely destroy the mutual spin coupling forces."
    },
    {
        "idx": 1043,
        "question": "Why does ferromagnetic behavior cease above the Curie temperature for ferromagnetic materials?",
        "answer": "Ferromagnetic behavior ceases above the Curie temperature because the atomic thermal vibrations are sufficiently violent so as to completely destroy the mutual spin coupling forces.",
        "type": 4,
        "wrong_answers_1": "The saturation magnetization decreases with increasing temperature because the atomic thermal vibrational motions counteract the coupling forces between the adjacent atomic dipole moments, causing some magnetic dipole misalignment.",
        "wrong_answers_2": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "wrong_answers_3": "For some ceramic materials, the thermal conductivity first decreases with rising temperature because the scattering of lattice vibrations increases with temperature. At higher temperatures, the thermal conductivity will increase for some ceramics that are porous because radiant heat transfer across pores may become important, which process increases with rising temperature."
    },
    {
        "idx": 1044,
        "question": "An iron bar magnet having a coercivity of 4000 \\mathrm{~A} / m is to be demagnetized. If the bar is inserted within a cylindrical wire coil 0.15{m} long and having 100 turns, what electric current is required to generate the necessary magnetic field?",
        "answer": "the required electric current is 6 \\mathrm{a}.",
        "type": 1,
        "wrong_answers_1": "the required electric current is 6 \\mathrm{a}.",
        "wrong_answers_2": "the required electric current is 6 \\mathrm{a}.",
        "wrong_answers_3": "the required electric current is 7 \\mathrm{a}."
    },
    {
        "idx": 1045,
        "question": "Visible light having a wavelength of 6 x 10^-7 m appears orange. Compute the frequency of a photon of this light.",
        "answer": "the frequency of the photon is 5 x 10^14 s^-1.",
        "type": 1,
        "wrong_answers_1": "the frequency of the photon is 4 x 10^14 s^-1.",
        "wrong_answers_2": "the frequency of the photon is 5 x 10^14 s^-1.",
        "wrong_answers_3": "the frequency of the photon is 5 x 10^14 s^-1."
    },
    {
        "idx": 1046,
        "question": "Visible light having a wavelength of 6 x 10^-7 m appears orange. Compute the energy of a photon of this light.",
        "answer": "the energy of the photon is 3.31 x 10^-19 j (2.07 ev).",
        "type": 1,
        "wrong_answers_1": "the energy of the photon is 3.05 x 10^-19 j (2.35 ev).",
        "wrong_answers_2": "the energy of the photon is 3.70 x 10^-19 j (1.85 ev).",
        "wrong_answers_3": "the energy of the photon is 3.42 x 10^-19 j (1.92 ev)."
    },
    {
        "idx": 1047,
        "question": "What are the characteristics of opaque materials in terms of their appearance and light transmittance?",
        "answer": "Opaque materials are impervious to light transmission; it is not possible to see through them.",
        "type": 4,
        "wrong_answers_1": "Opaque materials are impervious to light transmission; it is not possible to see through them.",
        "wrong_answers_2": "An opaque material is impervious to light transmission.",
        "wrong_answers_3": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them."
    },
    {
        "idx": 1048,
        "question": "What are the characteristics of translucent materials in terms of their appearance and light transmittance?",
        "answer": "Light is transmitted diffusely through translucent materials (there is some internal light scattering). Objects are not clearly distinguishable when viewed through a translucent material.",
        "type": 4,
        "wrong_answers_1": "Light is transmitted diffusely through translucent materials (there is some internal light scattering). Objects are not clearly distinguishable when viewed through a translucent material.",
        "wrong_answers_2": "A translucent material transmits light diffusely.",
        "wrong_answers_3": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them."
    },
    {
        "idx": 1049,
        "question": "What are the characteristics of transparent materials in terms of their appearance and light transmittance?",
        "answer": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them.",
        "type": 4,
        "wrong_answers_1": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them.",
        "wrong_answers_2": "Opaque materials are impervious to light transmission; it is not possible to see through them.",
        "wrong_answers_3": "Opaque materials are impervious to light transmission; it is not possible to see through them."
    },
    {
        "idx": 1050,
        "question": "In ionic materials, how does the size of the component ions affect the extent of electronic polarization?",
        "answer": "In ionic materials, the larger the size of the component ions the greater the degree of electronic polarization.",
        "type": 4,
        "wrong_answers_1": "The two characteristics of component ions that determine the crystal structure of a ceramic compound are: 1) the magnitude of the electrical charge on each ion, and 2) the relative sizes of the cations and anions.",
        "wrong_answers_2": "For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field.",
        "wrong_answers_3": "(a) For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field."
    },
    {
        "idx": 1051,
        "question": "Can a material have an index of refraction less than unity? Why or why not?",
        "answer": "In order for a material to have an index of refraction less than unity, the velocity of light in the material (v) would necessarily have to be greater than the velocity of light in a vacuum. This is not possible.",
        "type": 4,
        "wrong_answers_1": "In order for a material to have an index of refraction less than unity, the velocity of light in the material (v) would necessarily have to be greater than the velocity of light in a vacuum . This is not possible.",
        "wrong_answers_2": "The drift velocity of a free electron is the average electron velocity in the direction of the force imposed by an electric field.",
        "wrong_answers_3": "No, it is not possible. Linear polymers have higher degrees of crystallization (and higher tensile moduli) than branched polymers - on this basis, the PVC material should have the higher value of E. On the other hand, PVC has a more complex repeat unit structure than does polyethylene, which means that, on this basis, the PE would have a higher degree of crystallinity and also a greater tensile modulus. Also, tensile modulus is relatively independent of number-average molecular weight. Therefore, this determination is not possible since it is not possible to determine which of the two materials has the greater degree of crystallinity."
    },
    {
        "idx": 1052,
        "question": "Briefly describe the phenomenon of dispersion in a transparent medium.",
        "answer": "Dispersion in a transparent medium is the phenomenon wherein the index of refraction varies slightly with the wavelength of the electromagnetic radiation.",
        "type": 4,
        "wrong_answers_1": "The characteristic color of a transparent nonmetal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is transmitted through the material.",
        "wrong_answers_2": "In order for a material to have an index of refraction less than unity, the velocity of light in the material (v) would necessarily have to be greater than the velocity of light in a vacuum . This is not possible.",
        "wrong_answers_3": "In order for a material to have an index of refraction less than unity, the velocity of light in the material (v) would necessarily have to be greater than the velocity of light in a vacuum. This is not possible."
    },
    {
        "idx": 1053,
        "question": "Briefly explain how reflection losses of transparent materials are minimized by thin surface coatings.",
        "answer": "The thickness and dielectric constant of a thin surface coating are selected such that there is destructive interference between the light beam that is reflected from the lens-coating interface and the light beam that is reflected from the coating-air interface; thus, the net intensity of the total reflected beam is very low.",
        "type": 4,
        "wrong_answers_1": "The characteristic color of a metal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is reflected.",
        "wrong_answers_2": "The chromium in stainless steels causes a very thin and highly adherent surface coating to form over the surface of the alloy, which protects it from further corrosion. For plain carbon steels, rust, instead of this adherent coating, forms.",
        "wrong_answers_3": "The chromium in stainless steels causes a very thin and highly adherent surface coating to form over the surface of the alloy, which protects it from further corrosion. For plain carbon steels, rust, instead of this adherent coating, forms."
    },
    {
        "idx": 1054,
        "question": "The index of refraction of corundum \\left(\\mathrm{Al}_{2} \\mathrm{O}_{3}\\right) is anisotropic. Suppose that visible light is passing from one grain to another of different crystallographic orientation and at normal incidence to the grain boundary. Calculate the reflectivity at the boundary if the indices of refraction for the two grains are 1.757 and 1.779 in the direction of light propagation.",
        "answer": "the reflectivity at the boundary is 3.87 × 10^{-5}.",
        "type": 1,
        "wrong_answers_1": "the reflectivity at the boundary is 4.43 × 10^{-5}.",
        "wrong_answers_2": "the reflectivity at the boundary is 4.31 × 10^{-5}.",
        "wrong_answers_3": "the reflectivity at the boundary is 4.16 × 10^{-5}."
    },
    {
        "idx": 1055,
        "question": "Briefly explain what determines the characteristic color of a metal.",
        "answer": "The characteristic color of a metal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is reflected.",
        "type": 4,
        "wrong_answers_1": "The characteristic color of a transparent nonmetal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is transmitted through the material.",
        "wrong_answers_2": "The thickness and dielectric constant of a thin surface coating are selected such that there is destructive interference between the light beam that is reflected from the lens-coating interface and the light beam that is reflected from the coating-air interface; thus, the net intensity of the total reflected beam is very low.",
        "wrong_answers_3": "For a transparent material that appears colorless, any absorption within its interior is the same for all visible wavelengths. On the other hand, if there is any selective absorption of visible light (usually by electron excitations), the material will appear colored, its color being dependent on the frequency distribution of the transmitted light beam."
    },
    {
        "idx": 1056,
        "question": "Briefly explain what determines the characteristic color of a transparent nonmetal.",
        "answer": "The characteristic color of a transparent nonmetal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is transmitted through the material.",
        "type": 4,
        "wrong_answers_1": "The characteristic color of a metal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is reflected.",
        "wrong_answers_2": "For a transparent material that appears colorless, any absorption within its interior is the same for all visible wavelengths. On the other hand, if there is any selective absorption of visible light (usually by electron excitations), the material will appear colored, its color being dependent on the frequency distribution of the transmitted light beam.",
        "wrong_answers_3": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them."
    },
    {
        "idx": 1057,
        "question": "Briefly explain why some transparent materials appear colored while others are colorless.",
        "answer": "For a transparent material that appears colorless, any absorption within its interior is the same for all visible wavelengths. On the other hand, if there is any selective absorption of visible light (usually by electron excitations), the material will appear colored, its color being dependent on the frequency distribution of the transmitted light beam.",
        "type": 4,
        "wrong_answers_1": "The characteristic color of a transparent nonmetal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is transmitted through the material.",
        "wrong_answers_2": "A transparent material transmits light with relatively little absorption.",
        "wrong_answers_3": "The characteristic color of a metal is determined by the distribution of wavelengths of the nonabsorbed light radiation that is reflected."
    },
    {
        "idx": 1058,
        "question": "Briefly explain why amorphous polymers are transparent, while predominantly crystalline polymers appear opaque or, at best, translucent.",
        "answer": "Amorphous polymers are normally transparent because there is no scattering of a light beam within the material. However, for semicrystalline polymers, visible light will be scattered at boundaries between amorphous and crystalline regions since they have different indices of refraction. This leads to translucency or, for extensive scattering, opacity, except for semicrystalline polymers having very small crystallites.",
        "type": 4,
        "wrong_answers_1": "For a transparent material that appears colorless, any absorption within its interior is the same for all visible wavelengths. On the other hand, if there is any selective absorption of visible light (usually by electron excitations), the material will appear colored, its color being dependent on the frequency distribution of the transmitted light beam.",
        "wrong_answers_2": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked.",
        "wrong_answers_3": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked."
    },
    {
        "idx": 1059,
        "question": "Zinc has five naturally occurring isotopes: 48.63% of { }^{64} Zn with an atomic weight of 63.929 amu; 27.90% of { }^{66} Zn with an atomic weight of 65.926 amu^{\\text {; }} 4.10% of { }^{67} Zn with an atomic weight of 66.927 amu ; 18.75% of { }^{68} Zn with an atomic weight of 67.925 amu ; and 0.62% of { }^{70} Zn with an atomic weight of 69.925 amu. Calculate the average atomic weight of Zn.",
        "answer": "the average atomic weight of \\mathrm{zn} is 65.400 amu.",
        "type": 1,
        "wrong_answers_1": "the average atomic weight of \\mathrm{zn} is 67.716 amu.",
        "wrong_answers_2": "the average atomic weight of \\mathrm{zn} is 72.776 amu.",
        "wrong_answers_3": "the average atomic weight of \\mathrm{zn} is 62.203 amu."
    },
    {
        "idx": 1060,
        "question": "Indium has two naturally occurring isotopes: { }^{113} In with an atomic weight of 112.904 amu, and { }^{115} \\mathrm{In} with an atomic weight of 114.904 amu. If the average atomic weight for In is 114.818 amu, calculate the fraction-of-occurrences of these two isotopes.",
        "answer": "the fraction-of-occurrence of { }^{113}in is 0.043.   the fraction-of-occurrence of { }^{115}in is 0.957.",
        "type": 1,
        "wrong_answers_1": "the fraction-of-occurrence of { }^{103}in is 0.040.   the fraction-of-occurrence of { }^{109}in is 1.087.",
        "wrong_answers_2": "the fraction-of-occurrence of { }^{129}in is 0.049.   the fraction-of-occurrence of { }^{129}in is 1.066.",
        "wrong_answers_3": "the fraction-of-occurrence of { }^{107}in is 0.047.   the fraction-of-occurrence of { }^{125}in is 0.825."
    },
    {
        "idx": 1061,
        "question": "Potassium iodide (KI) exhibits predominantly ionic bonding. The K+ ion has an electron structure that is identical to which inert gas?",
        "answer": "The K+ ion is just a potassium atom that has lost one electron; therefore, it has an electron configuration the same as argon.",
        "type": 4,
        "wrong_answers_1": "The Cl- ion has an electron configuration the same as argon.",
        "wrong_answers_2": "The I- ion is an iodine atom that has acquired one extra electron; therefore, it has an electron configuration the same as xenon.",
        "wrong_answers_3": "The Na+ ion has an electron configuration the same as neon."
    },
    {
        "idx": 1062,
        "question": "Potassium iodide (KI) exhibits predominantly ionic bonding. The I- ion has an electron structure that is identical to which inert gas?",
        "answer": "The I- ion is an iodine atom that has acquired one extra electron; therefore, it has an electron configuration the same as xenon.",
        "type": 4,
        "wrong_answers_1": "The K+ ion is just a potassium atom that has lost one electron; therefore, it has an electron configuration the same as argon.",
        "wrong_answers_2": "The Cl- ion has an electron configuration the same as argon.",
        "wrong_answers_3": "The Na+ ion has an electron configuration the same as neon."
    },
    {
        "idx": 1063,
        "question": "With regard to electron configuration, what do all the elements in Group IIA of the periodic table have in common?",
        "answer": "Each of the elements in Group IIA has two s electrons.",
        "type": 4,
        "wrong_answers_1": "Each of the elements in Group VIIA has five p electrons.",
        "wrong_answers_2": "Three variables that determine the microstructure of an alloy are (1) the alloying elements present, (2) the concentrations of these alloying elements, and (3) the heat treatment of the alloy.",
        "wrong_answers_3": "For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid."
    },
    {
        "idx": 1064,
        "question": "To what group in the periodic table would an element with atomic number 112 belong?",
        "answer": "From the periodic table the element having atomic number 112 would belong to group IIB. According to the periodic table, Ds, having an atomic number of 110 lies below Pt in the periodic table and in the right-most column of group VIII. Moving two columns to the right puts element 112 under Hg and in group IIB. This element has been artificially created and given the name Copernicium with the symbol Cn. It was named after Nicolaus Copernicus, the Polish scientist who proposed that the earth moves around the sun (and not vice versa).",
        "type": 4,
        "wrong_answers_1": "From the periodic table the element having atomic number 114 would belong to group IVA. Ds, having an atomic number of 110 lies below Pt in the periodic table and in the rightmost column of group VIII. Moving four columns to the right puts element 114 under \\mathrm{Pb} and in group IVA.",
        "wrong_answers_2": "For tungsten, the bonding is metallic since it is a metallic element from the periodic table.",
        "wrong_answers_3": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom."
    },
    {
        "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{5}) electron configuration is that of a halogen because it is one electron deficient from having a filled (p) subshell.",
        "idx": 1065,
        "type": 4,
        "wrong_answers_1": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}) electron configuration is that of an inert gas because of filled (3 s) and (3 p) subshells."
    },
    {
        "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "idx": 1066,
        "type": 4,
        "wrong_answers_1": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{5} 5 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron."
    },
    {
        "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6}) electron configuration is that of an inert gas because of filled (4 s) and (4 p) subshells.",
        "idx": 1067,
        "type": 4,
        "wrong_answers_1": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6}) electron configuration is that of an inert gas because of filled (3 s) and (3 p) subshells.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{5} 5 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron."
    },
    {
        "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron.",
        "idx": 1068,
        "type": 4,
        "wrong_answers_1": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell."
    },
    {
        "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{5} 5 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{5} 5 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "idx": 1069,
        "type": 4,
        "wrong_answers_1": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron."
    },
    {
        "question": "Without consulting Figure or Table, determine whether the electron configuration (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) is an inert gas, a halogen, an alkali metal, an alkaline earth metal, or a transition metal. Justify your choice.",
        "answer": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2}) electron configuration is that of an alkaline earth metal because of two (s) electrons.",
        "idx": 1070,
        "type": 4,
        "wrong_answers_1": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron.",
        "wrong_answers_2": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 4 s^{1}) electron configuration is that of an alkali metal because of a single (s) electron.",
        "wrong_answers_3": "The (1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7} 4 s^{2}) electron configuration is that of a transition metal because of an incomplete (d) subshell."
    },
    {
        "question": "Differentiate (E_{N}) with respect to (r), and then set the resulting expression equal to zero, because the curve of (E_{N}) versus (r) is a minimum at (E_{0}).",
        "answer": "The derivative of (E_{N}) with respect to (r) is set to zero to find the minimum.",
        "idx": 1071,
        "type": 1
    },
    {
        "question": "Solve for (r) in terms of (A, B), and (n), which yields (r_{0}), the equilibrium interionic spacing.",
        "answer": "The equilibrium interionic spacing (r_{0}) is solved in terms of (A, B), and (n).",
        "idx": 1072,
        "type": 1,
        "wrong_answers_1": "The equilibrium interionic spacing (r_{0}) is solved in terms of (A, B), and (n).",
        "wrong_answers_2": "The equilibrium interionic spacing (r_{0}) is solved in terms of (A, B), and (n).",
        "wrong_answers_3": "The equilibrium interionic spacing (r_{0}) is solved in terms of (A, B), and (n)."
    },
    {
        "question": "Determine the expression for (E_{0}) by substitution of (r_{0}) into Equation.",
        "answer": "The bonding energy (e_{0}) is given by: (e_{0}=-\frac{a}{\\left(\frac{a}{nb}\right)^{1 /(1-n)}}+\frac{b}{\\left(\frac{a}{nb}\right)^{n /(1-n)}})",
        "idx": 1073,
        "type": 1,
        "wrong_answers_1": "The bonding energy (e_{0}) is given by: (e_{0}=-\frac{a}{\\left(\frac{a}{nb}\right)^{1 /(1-n)}}+\frac{b}{\\left(\frac{a}{nb}\right)^{n /(1-n)}})",
        "wrong_answers_2": "The bonding energy (e_{0}) is given by: (e_{0}=-\frac{a}{\\left(\frac{a}{nb}\right)^{1 /(1-n)}}+\frac{b}{\\left(\frac{a}{nb}\right)^{n /(1-n)}})",
        "wrong_answers_3": "The bonding energy (e_{0}) is given by: (e_{0}=-\frac{a}{\\left(\frac{a}{nb}\right)^{1 /(1-n)}}+\frac{b}{\\left(\frac{a}{nb}\right)^{n /(1-n)}})"
    },
    {
        "idx": 1074,
        "question": "Explain why hydrogen fluoride (HF) has a higher boiling temperature than hydrogen \\\\ chloride (\\mathrm{HCl})\\left(19.4 \\mathrm{vs} .-85^{\\circ} C\\right), even though \\mathrm{HF} has a lower molecular weight.",
        "answer": "The intermolecular bonding for \\mathrm{HF} is hydrogen, whereas for \\mathrm{HCl}, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher melting temperature.",
        "type": 4,
        "wrong_answers_1": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)",
        "wrong_answers_2": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)",
        "wrong_answers_3": "For nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.)"
    },
    {
        "idx": 1075,
        "question": "What type(s) of bonding would be expected for solid xenon?",
        "answer": "For solid xenon, the bonding is van der Waals since xenon is an inert gas.",
        "type": 4,
        "wrong_answers_1": "For solid xenon, the bonding is van der Waals since xenon is an inert gas.",
        "wrong_answers_2": "The intermolecular bonding for \\mathrm{HF} is hydrogen, whereas for \\mathrm{HCl}, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher melting temperature.",
        "wrong_answers_3": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)"
    },
    {
        "idx": 1076,
        "question": "What type(s) of bonding would be expected for calcium fluoride (CaF2)?",
        "answer": "For CaF2, the bonding is predominantly ionic (but with some slight covalent character) on the basis of the relative positions of Ca and F in the periodic table.",
        "type": 4,
        "wrong_answers_1": "For AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of Al and P in the periodic table.",
        "wrong_answers_2": "For BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of Ba and S in the periodic table.",
        "wrong_answers_3": "For CdTe, the bonding is predominantly covalent (with some slight ionic character) on the basis of the relative positions of Cd and Te in the periodic table."
    },
    {
        "idx": 1077,
        "question": "What type(s) of bonding would be expected for bronze?",
        "answer": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin).",
        "type": 4,
        "wrong_answers_1": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin).",
        "wrong_answers_2": "For brass, the bonding is metallic since it is a metal alloy.",
        "wrong_answers_3": "For tungsten, the bonding is metallic since it is a metallic element from the periodic table."
    },
    {
        "idx": 1078,
        "question": "What type(s) of bonding would be expected for cadmium telluride (CdTe)?",
        "answer": "For CdTe, the bonding is predominantly covalent (with some slight ionic character) on the basis of the relative positions of Cd and Te in the periodic table.",
        "type": 4,
        "wrong_answers_1": "For CaF2, the bonding is predominantly ionic (but with some slight covalent character) on the basis of the relative positions of Ca and F in the periodic table.",
        "wrong_answers_2": "For AlP the bonding is predominantly covalent (but with some ionic character) on the basis of the relative positions of Al and P in the periodic table.",
        "wrong_answers_3": "For BaS, the bonding is predominantly ionic (but with some covalent character) on the basis of the relative positions of Ba and S in the periodic table."
    },
    {
        "idx": 1079,
        "question": "What type(s) of bonding would be expected for rubber?",
        "answer": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)",
        "type": 4,
        "wrong_answers_1": "For rubber, the bonding is covalent with some van der Waals. (Rubber is composed primarily of carbon and hydrogen atoms.)",
        "wrong_answers_2": "For nylon, the bonding is covalent with perhaps some van der Waals. (Nylon is composed primarily of carbon and hydrogen.)",
        "wrong_answers_3": "The intermolecular bonding for \\mathrm{HF} is hydrogen, whereas for \\mathrm{HCl}, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher melting temperature."
    },
    {
        "idx": 1080,
        "question": "What type(s) of bonding would be expected for tungsten?",
        "answer": "For tungsten, the bonding is metallic since it is a metallic element from the periodic table.",
        "type": 4,
        "wrong_answers_1": "For brass, the bonding is metallic since it is a metal alloy.",
        "wrong_answers_2": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin).",
        "wrong_answers_3": "For bronze, the bonding is metallic since it is a metal alloy (composed of copper and tin)."
    },
    {
        "idx": 1081,
        "question": "Which of the following electron configurations is for an inert gas? \\\\ (A) 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} \\\\ (B) 1 s^{2} 2 s^{2} 2 p^{6} s 3 s^{2} \\\\ (C) 1 s^{2} 2 s^{2} 2 p^{6} .3 s^{2} 3 p^{6} 4 s^{1} \\\\ (D) 1 s^{2} 2 s^{2} 2 p^{6}·3 s^{2} 3 p^{6} 3 d^{2} 4 s^{2}",
        "answer": "The correct answer is A. The 1 s^{2} 2 s^{2} 2 p^{6} 13 s^{2} 3 p^{6} electron configuration is that of an inert gas because of filled 3 s and 3 p subshells.",
        "type": 2
    },
    {
        "idx": 1082,
        "question": "What type(s) of bonding would be expected for bronze (a copper-tin alloy)?\n(A) Ionic bonding\n(B) Metallic bonding\n(C) Covalent bonding with some van der Waals bonding\n(D) van der Waals bonding",
        "answer": "The correct answer is B. For bronze, the bonding is metallic because it is a metal alloy.",
        "type": 2
    },
    {
        "idx": 1083,
        "question": "What type(s) of bonding would be expected for rubber?\n(A) Ionic bonding\n(B) Metallic bonding\n(C) Covalent bonding with some van der Waals bonding\n(D) van der Waals bonding",
        "answer": "FE}\nThe correct answer is C. For rubber, the bonding is covalent with some van der Waals bonding. (Rubber is composed primarily of carbon and hydrogen atoms.)",
        "type": 2
    },
    {
        "idx": 1084,
        "question": "If the atomic radius of lead is 0.175nm, calculate the volume of its unit cell in cubic meters.",
        "answer": "the volume of the unit cell is 1.213 × 10^{-28} m^{3}.",
        "type": 1,
        "wrong_answers_1": "the volume of the unit cell is 1.256 × 10^{-28} m^{3}.",
        "wrong_answers_2": "the volume of the unit cell is 1.153 × 10^{-28} m^{3}.",
        "wrong_answers_3": "the volume of the unit cell is 1.044 × 10^{-28} m^{3}."
    },
    {
        "idx": 1085,
        "question": "Calculate the radius of a palladium (P d) atom, given that P d has an FCC crystal structure, a density of 12.0g / {cm}^{3}, and an atomic weight of 106.4g / mol.",
        "answer": "the radius of a palladium (pd) atom is 0.138nm.",
        "type": 1,
        "wrong_answers_1": "the radius of a palladium (pd) atom is 0.124nm.",
        "wrong_answers_2": "the radius of a palladium (pd) atom is 0.128nm.",
        "wrong_answers_3": "the radius of a palladium (pd) atom is 0.119nm."
    },
    {
        "idx": 1086,
        "question": "Calculate the radius of a tantalum (Ta) atom, given that Ta has a BCC crystal structure, a density of 16.6g / {cm}^{3}, and an atomic weight of 180.9g / mol.",
        "answer": "the radius of a tantalum (ta) atom is 0.143 \\, \\text{nm}.",
        "type": 1,
        "wrong_answers_1": "the radius of a tantalum (ta) atom is 0.123 \\, \\text{nm}.",
        "wrong_answers_2": "the radius of a tantalum (ta) atom is 0.148 \\, \\text{nm}.",
        "wrong_answers_3": "the radius of a tantalum (ta) atom is 0.159 \\, \\text{nm}."
    },
    {
        "idx": 1087,
        "question": "Titanium (Ti) has an HCP crystal structure and a density of 4.51 g/cm^3. What is the volume of its unit cell in cubic meters?",
        "answer": "the volume of the ti unit cell is 1.058 x 10^-28 m^3.",
        "type": 1,
        "wrong_answers_1": "the volume of the ti unit cell is 1.145 x 10^-28 m^3.",
        "wrong_answers_2": "the volume of the ti unit cell is 1.192 x 10^-28 m^3.",
        "wrong_answers_3": "the volume of the ti unit cell is 0.984 x 10^-28 m^3."
    },
    {
        "idx": 1088,
        "question": "Titanium (Ti) has an HCP crystal structure and a density of 4.51 g/cm^3. If the c/a ratio is 1.58, compute the values of c and a.",
        "answer": "the values of a and c are 0.296 nm and 0.468 nm, respectively.",
        "type": 1,
        "wrong_answers_1": "the values of a and c are 0.318 nm and 0.493 nm, respectively.",
        "wrong_answers_2": "the values of a and c are 0.323 nm and 0.454 nm, respectively.",
        "wrong_answers_3": "the values of a and c are 0.309 nm and 0.429 nm, respectively."
    },
    {
        "idx": 1089,
        "question": "Magnesium (Mg) has an HCP crystal structure and a density of 1.74 g/cm^3. What is the volume of its unit cell in cubic centimeters?",
        "answer": "the volume of the unit cell is 1.39 × 10^-22 cm^3.",
        "type": 1,
        "wrong_answers_1": "the volume of the unit cell is 1.18 × 10^-22 cm^3.",
        "wrong_answers_2": "the volume of the unit cell is 1.49 × 10^-22 cm^3.",
        "wrong_answers_3": "the volume of the unit cell is 1.24 × 10^-22 cm^3."
    },
    {
        "idx": 1090,
        "question": "Magnesium (Mg) has an HCP crystal structure and a density of 1.74 g/cm^3. If the c/a ratio is 1.624, compute the values of c and a.",
        "answer": "the values of a and c are 0.321 nm and 0.521 nm, respectively.",
        "type": 1,
        "wrong_answers_1": "the values of a and c are 0.339 nm and 0.480 nm, respectively.",
        "wrong_answers_2": "the values of a and c are 0.309 nm and 0.537 nm, respectively.",
        "wrong_answers_3": "the values of a and c are 0.285 nm and 0.570 nm, respectively."
    },
    {
        "idx": 1091,
        "question": "The unit cell for uranium (U) has orthorhombic symmetry, with a, b, and c lattice parameters of 0.286,0.587, and 0.495nm, respectively. If its density, atomic weight, and atomic radius are 19.05g / {cm}^{3}, 238.03g / mol, and 0.1385nm, respectively, compute the atomic packing factor.\n\\title{",
        "answer": "the atomic packing factor (apf) for uranium is 0.536.",
        "type": 1,
        "wrong_answers_1": "the atomic packing factor (apf) for uranium is 0.495.",
        "wrong_answers_2": "the atomic packing factor (apf) for uranium is 0.472.",
        "wrong_answers_3": "the atomic packing factor (apf) for uranium is 0.483."
    },
    {
        "idx": 1092,
        "question": "Indium (In) has a tetragonal unit cell for which the a and c lattice parameters are 0.459 and 0.495 nm, respectively. If the atomic packing factor and atomic radius are 0.693 and 0.1625 nm, respectively, determine the number of atoms in each unit cell.",
        "answer": "4.0 atoms/unit cell",
        "type": 1,
        "wrong_answers_1": "3.7 atoms/unit cell",
        "wrong_answers_2": "4.2 atoms/unit cell",
        "wrong_answers_3": "4.2 atoms/unit cell"
    },
    {
        "idx": 1093,
        "question": "Indium (In) has a tetragonal unit cell for which the a and c lattice parameters are 0.459 and 0.495 nm, respectively. The atomic weight of indium is 114.82 g/mol; compute its theoretical density.",
        "answer": "7.31 g/cm^3",
        "type": 1,
        "wrong_answers_1": "8.40 g/cm^3",
        "wrong_answers_2": "8.28 g/cm^3",
        "wrong_answers_3": "7.02 g/cm^3"
    },
    {
        "question": "Beryllium (Be) has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.568. If the radius of the Be atom is 0.1143 nm, determine the unit cell volume.",
        "answer": "4.87 × 10^{-23} cm^3",
        "idx": 1094,
        "type": 1,
        "wrong_answers_1": "4.29 × 10^{-23} cm^3",
        "wrong_answers_2": "5.17 × 10^{-23} cm^3",
        "wrong_answers_3": "5.35 × 10^{-23} cm^3"
    },
    {
        "question": "Beryllium (Be) has an HCP unit cell for which the ratio of the lattice parameters c/a is 1.568. If the radius of the Be atom is 0.1143nm, calculate the theoretical density of Be and compare it with the literature value.",
        "answer": "the theoretical density of be is 1.84 g/cm^3, and the literature value is 1.85 g/cm^3",
        "idx": 1095,
        "type": 1,
        "wrong_answers_1": "the theoretical density of be is 1.75 g/cm^3, and the literature value is 1.94 g/cm^3",
        "wrong_answers_2": "the theoretical density of be is 1.74 g/cm^3, and the literature value is 2.05 g/cm^3",
        "wrong_answers_3": "the theoretical density of be is 2.11 g/cm^3, and the literature value is 1.96 g/cm^3"
    },
    {
        "idx": 1096,
        "question": "Magnesium ( Mg ) has an HCP crystal structure, a c/a ratio of 1.624 , and a density of 1.74 g / {cm}^{3}. Compute the atomic radius for Mg.",
        "answer": "the atomic radius for \\mathrm{mg} is 0.160\\;nm.",
        "type": 1,
        "wrong_answers_1": "the atomic radius for \\mathrm{mg} is 0.139\\;nm.",
        "wrong_answers_2": "the atomic radius for \\mathrm{mg} is 0.148\\;nm.",
        "wrong_answers_3": "the atomic radius for \\mathrm{mg} is 0.140\\;nm."
    },
    {
        "idx": 1097,
        "question": "Cobalt (Co) has an HCP crystal structure, an atomic radius of 0.1253nm, and a c/a ratio of 1.623. Compute the volume of the unit cell for Co.",
        "answer": "the volume of the unit cell for co is 6.64 × 10^{-23} {cm}^3 or 6.64 × 10^{-2} nm^3.",
        "type": 1,
        "wrong_answers_1": "the volume of the unit cell for co is 7.53 × 10^{-23} {cm}^3 or 6.33 × 10^{-2} nm^3.",
        "wrong_answers_2": "the volume of the unit cell for co is 5.73 × 10^{-23} {cm}^3 or 5.75 × 10^{-2} nm^3.",
        "wrong_answers_3": "the volume of the unit cell for co is 6.96 × 10^{-23} {cm}^3 or 7.27 × 10^{-2} nm^3."
    },
    {
        "idx": 1098,
        "question": "One crystalline form of silica \\left(\\mathrm{SiO}_{2}\\right) has a cubic unit cell, and from x-ray diffraction data it is known that the cell edge length is 0.700nm. If the measured density is 2.32g / {cm}^{3}, how many \\mathrm{Si}^{4+} and \\mathrm{O}^{2}-ions are there per unit cell?",
        "answer": "there are 8 \\mathrm{si}^{4+} ions and 16 \\mathrm{o}^{2-} ions per \\mathrm{sio}_{2} unit cell.",
        "type": 1,
        "wrong_answers_1": "there are 8 \\mathrm{si}^{4+} ions and 18 \\mathrm{o}^{2-} ions per \\mathrm{sio}_{2} unit cell.",
        "wrong_answers_2": "there are 7 \\mathrm{si}^{4+} ions and 17 \\mathrm{o}^{2-} ions per \\mathrm{sio}_{2} unit cell.",
        "wrong_answers_3": "there are 7 \\mathrm{si}^{4+} ions and 17 \\mathrm{o}^{2-} ions per \\mathrm{sio}_{2} unit cell."
    },
    {
        "idx": 1099,
        "question": "A hypothetical AX type of ceramic material is known to have a density of 2.10g / {cm}^{3} and a unit cell of cubic symmetry with a cell edge length of 0.57nm. The atomic weights of the A and X elements are 28.5 and 30.0g / mol, respectively. On the basis of this information, which of the following crystal structures is (are) possible for this material: sodium chloride, cesium chloride, or zinc blende? Justify your choice(s).",
        "answer": "of the three possible crystal structures, only sodium chloride and zinc blende have four formula units per unit cell, and therefore, are possibilities.",
        "type": 2
    },
    {
        "idx": 1100,
        "question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. Which type of interstitial site will the Fe2+ ions occupy? Why?",
        "answer": "From Table, the cation-anion radius ratio is 0.077 nm / 0.140 nm = 0.550. Since this ratio is between 0.414 and 0.732, the Fe2+ ions will occupy octahedral sites.",
        "type": 4,
        "wrong_answers_1": "The titanium-oxygen ionic radius ratio is 0.0601 nm / 0.140 nm = 0.436. Since this ratio is between 0.414 and 0.723, the Ti4+ ions will also occupy octahedral sites.",
        "wrong_answers_2": "Since both Fe2+ and Ti4+ ions occupy octahedral sites, no tetrahedral sites will be occupied.",
        "wrong_answers_3": "For MgO, the ionic radii of the Mn2+ and Mg2+ are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.072 nm - 0.067 nm) / 0.072 nm × 100 = 6.9%, which value is within the acceptable range for a high degree of solubility."
    },
    {
        "idx": 1101,
        "question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. Which type of interstitial site will the Ti4+ ions occupy? Why?",
        "answer": "The titanium-oxygen ionic radius ratio is 0.0601 nm / 0.140 nm = 0.436. Since this ratio is between 0.414 and 0.723, the Ti4+ ions will also occupy octahedral sites.",
        "type": 4,
        "wrong_answers_1": "From Table, the cation-anion radius ratio is 0.077 nm / 0.140 nm = 0.550. Since this ratio is between 0.414 and 0.732, the Fe2+ ions will occupy octahedral sites.",
        "wrong_answers_2": "Since both Fe2+ and Ti4+ ions occupy octahedral sites, no tetrahedral sites will be occupied.",
        "wrong_answers_3": "For MgO, the ionic radii of the Mn2+ and Mg2+ are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.072 nm - 0.067 nm) / 0.072 nm × 100 = 6.9%, which value is within the acceptable range for a high degree of solubility."
    },
    {
        "idx": 1102,
        "question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. What fraction of the total tetrahedral sites will be occupied?",
        "answer": "Since both Fe2+ and Ti4+ ions occupy octahedral sites, no tetrahedral sites will be occupied.",
        "type": 4,
        "wrong_answers_1": "The titanium-oxygen ionic radius ratio is 0.0601 nm / 0.140 nm = 0.436. Since this ratio is between 0.414 and 0.723, the Ti4+ ions will also occupy octahedral sites.",
        "wrong_answers_2": "From Table, the cation-anion radius ratio is 0.077 nm / 0.140 nm = 0.550. Since this ratio is between 0.414 and 0.732, the Fe2+ ions will occupy octahedral sites.",
        "wrong_answers_3": "Only one-half of the tetrahedral positions will be occupied because there are two tetrahedral sites per anion, and yet only one cation per anion."
    },
    {
        "idx": 1103,
        "question": "Iron titanate, FeTiO3, forms in the ilmenite crystal structure that consists of an HCP arrangement of O2- ions. What fraction of the total octahedral sites will be occupied?",
        "answer": "For every FeTiO3 formula unit, there are three O2- ions, and, therefore, three octahedral sites; since there is one ion each of Fe2+ and Ti4+ per formula unit, two-thirds of these octahedral sites will be occupied.",
        "type": 1,
        "wrong_answers_1": "For every FeTiO3 formula unit, there are three O2- ions, and, therefore, three octahedral sites; since there is one ion each of Fe2+ and Ti4+ per formula unit, two-thirds of these octahedral sites will be occupied.",
        "wrong_answers_2": "For every FeTiO3 formula unit, there are three O2- ions, and, therefore, three octahedral sites; since there is one ion each of Fe2+ and Ti4+ per formula unit, two-thirds of these octahedral sites will be occupied.",
        "wrong_answers_3": "For every FeTiO3 formula unit, there are three O2- ions, and, therefore, three octahedral sites; since there is one ion each of Fe2+ and Ti3+ per formula unit, two-thirds of these octahedral sites will be occupied."
    },
    {
        "question": "To what equation does the expression for interplanar spacing (d_{\text {hkl }}) reduce for crystals having cubic symmetry, given that (a,(b), and c are the lattice parameters?",
        "answer": "For the crystals having cubic symmetry, a=b=c. Making this substitution into the given equation leads to (\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{a^{2}} = \frac{h^{2}+k^{2}+l^{2}}{a^{2}}).",
        "idx": 1104,
        "type": 1,
        "wrong_answers_1": "For the crystals having cubic symmetry, a=b=c. Making this substitution into the given equation leads to (\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{a^{2}} = \frac{h^{2}+k^{2}+l^{2}}{a^{2}}).",
        "wrong_answers_2": "For the crystals having cubic symmetry, a=b=c. Making this substitution into the given equation leads to (\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{a^{2}} = \frac{h^{2}+k^{2}+l^{2}}{a^{2}}).",
        "wrong_answers_3": "For the crystals having cubic symmetry, a=b=c. Making this substitution into the given equation leads to (\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{a^{2}} = \frac{h^{2}+k^{2}+l^{2}}{a^{2}})."
    },
    {
        "question": "To what equation does the expression for interplanar spacing (d_{\text {hkl }}) reduce for crystals having tetragonal symmetry, given that a, b and c are the lattice parameters?",
        "answer": "For crystals having tetragonal symmetry, a=b \neq c. Replacing b with a in the given equation leads to (\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}} = \frac{h^{2}+k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}}).",
        "idx": 1105,
        "type": 1,
        "wrong_answers_1": "For crystals having tetragonal symmetry, a=b \neq c. Replacing b with a in the given equation leads to (\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}} = \frac{h^{2}+k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}}).",
        "wrong_answers_2": "For crystals having tetragonal symmetry, a=b \neq c. Replacing b with a in the given equation leads to (\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}} = \frac{h^{2}+k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}}).",
        "wrong_answers_3": "For crystals having tetragonal symmetry, a=b \neq c. Replacing b with a in the given equation leads to (\frac{1}{d_{h k l}^{2}}=\frac{h^{2}}{a^{2}}+\frac{k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}} = \frac{h^{2}+k^{2}}{a^{2}}+\frac{l^{2}}{c^{2}})."
    },
    {
        "idx": 1106,
        "question": "The number-average molecular weight of a polystyrene is 500,000g / mol. Compute the degree of polymerization.\n\\title{",
        "answer": "the degree of polymerization is 4800.",
        "type": 1,
        "wrong_answers_1": "the degree of polymerization is 4440.",
        "wrong_answers_2": "the degree of polymerization is 4610.",
        "wrong_answers_3": "the degree of polymerization is 4360."
    },
    {
        "idx": 1107,
        "question": "High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen. Determine the concentration of Cl (in wt%) that must be added if this substitution occurs for 8% of all the original hydrogen atoms.",
        "answer": "the concentration of Cl that must be added is 29.0 wt%.",
        "type": 1,
        "wrong_answers_1": "the concentration of Cl that must be added is 29.9 wt%.",
        "wrong_answers_2": "the concentration of Cl that must be added is 25.3 wt%.",
        "wrong_answers_3": "the concentration of Cl that must be added is 30.5 wt%."
    },
    {
        "idx": 1108,
        "question": "High-density polyethylene may be chlorinated by inducing the random substitution of chlorine atoms for hydrogen. In what ways does this chlorinated polyethylene differ from poly(vinyl chloride)?",
        "answer": "chlorinated polyethylene differs from poly(vinyl chloride) in that, for pvc, (1) 25% of the side-bonding sites are substituted with Cl, and (2) the substitution is probably much less random.",
        "type": 4,
        "wrong_answers_1": "chlorinated polyethylene differs from poly(vinyl chloride) in that, for pvc, (1) 25% of the side-bonding sites are substituted with Cl, and (2) the substitution is probably much less random.",
        "wrong_answers_2": "Yes, for these two polymers it is possible to decide. The linear and syndiotactic poly(vinyl chloride) is more likely to crystallize; the phenyl side-group for polystyrene is bulkier than the Cl side-group for poly(vinyl chloride). Syndiotactic and isotactic isomers are equally likely to crystallize.",
        "wrong_answers_3": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity."
    },
    {
        "idx": 1109,
        "question": "Compare thermoplastic and thermosetting polymers on the basis of mechanical characteristics upon heating.",
        "answer": "Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers harden upon heating, while further heating will not lead to softening.",
        "type": 4,
        "wrong_answers_1": "Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers harden upon heating, while further heating will not lead to softening.",
        "wrong_answers_2": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked.",
        "wrong_answers_3": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked."
    },
    {
        "idx": 1110,
        "question": "Compare thermoplastic and thermosetting polymers according to possible molecular structures.",
        "answer": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked.",
        "type": 4,
        "wrong_answers_1": "Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked.",
        "wrong_answers_2": "No, it is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize.",
        "wrong_answers_3": "Yes, it is possible. The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's crystallinity leads to an increase in its tensile modulus. In addition, tensile modulus is independent of molecular weight - the atactic/branched material has the higher molecular weight."
    },
    {
        "idx": 1111,
        "question": "The number-average molecular weight of a poly(acrylonitrile-butadiene) alternating copolymer is 1,000,000g / mol; determine the average number of acrylonitrile and butadiene repeat units per molecule.",
        "answer": "the average number of acrylonitrile and butadiene repeat units per molecule is 9333.",
        "type": 1,
        "wrong_answers_1": "the average number of acrylonitrile and butadiene repeat units per molecule is 9673.",
        "wrong_answers_2": "the average number of acrylonitrile and butadiene repeat units per molecule is 8773.",
        "wrong_answers_3": "the average number of acrylonitrile and butadiene repeat units per molecule is 10253."
    },
    {
        "idx": 1112,
        "question": "Calculate the number-average molecular weight of a random poly(isobutylene-isoprene) copolymer in which the fraction of isobutylene repeat units is 0.25 ; assume that this concentration corresponds to a degree of polymerization of 1500 .",
        "answer": "the number-average molecular weight of the poly(isobutylene-isoprene) copolymer is \\bar{m}_{n} = 97,700g/mol.",
        "type": 1,
        "wrong_answers_1": "the number-average molecular weight of the poly(isobutylene-isoprene) copolymer is \\bar{m}_{n} = 112260g/mol.",
        "wrong_answers_2": "the number-average molecular weight of the poly(isobutylene-isoprene) copolymer is \\bar{m}_{n} = 103582g/mol.",
        "wrong_answers_3": "the number-average molecular weight of the poly(isobutylene-isoprene) copolymer is \\bar{m}_{n} = 89503g/mol."
    },
    {
        "idx": 1113,
        "question": "An alternating copolymer is known to have a number-average molecular weight of 100,000g / mol and a degree of polymerization of 2210 . If one of the repeat units is ethylene, which of styrene, propylene, tetrafluoroethylene, and vinyl chloride is the other repeat unit? Why?",
        "answer": "the other repeat unit is vinyl chloride, with a molecular weight of 62.49 \\mathrm{g/mol}. this matches the calculated value of 62.45 \\mathrm{g/mol} for the unknown repeat unit.",
        "type": 2
    },
    {
        "idx": 1114,
        "question": "(a) Determine the ratio of butadiene to acrylonitrile repeat units in a copolymer having a number-average molecular weight of 250,000 g/mol and a degree of polymerization of 4640.",
        "answer": "the ratio of butadiene to acrylonitrile repeat units in the copolymer is fb/fa = 4.0.",
        "type": 1,
        "wrong_answers_1": "the ratio of butadiene to acrylonitrile repeat units in the copolymer is fb/fa = 4.2.",
        "wrong_answers_2": "the ratio of butadiene to acrylonitrile repeat units in the copolymer is fb/fa = 3.8.",
        "wrong_answers_3": "the ratio of butadiene to acrylonitrile repeat units in the copolymer is fb/fa = 3.5."
    },
    {
        "idx": 1115,
        "question": "(b) Which type(s) of copolymer(s) will this copolymer be, considering the following possibilities: random, alternating, graft, and block? Why?",
        "answer": "of the possible copolymers, the only one for which there is a restriction on the ratio of repeat unit types is alternating; the ratio must be 1:1. therefore, on the basis of the result in part (a), the possibilities for this copolymer are random, graft, and block.",
        "type": 4,
        "wrong_answers_1": "No, it is not possible. The random acrylonitrile-butadiene copolymer will tend to a lower degree of crystallinity than the alternating acrylonitrile-butadiene copolymer inasmuch as random copolymers don't normally crystallize. On this basis only, the alternating material would have a higher modulus inasmuch as tensile modulus increases with degree of crystallinity. On the other hand, the random copolymer has a higher degree of crosslinking (10% versus 5% for the alternating copolymer), and, on this basis only would have the higher tensile modulus-an increase in crosslinking leads to an increase in E. Thus, this determination is not possible; with regard to degree of crystallinity the alternating material has the higher E, whereas the random copolymer would have a higher E value on the basis of degree of crosslinking.",
        "wrong_answers_2": "The alternating poly(styrene-butadiene) copolymer will have a higher crystallinity than the random copolymer; alternating copolymers crystallize more easily than random ones. The influence of crystallinity on conductivity is explained in part (c).",
        "wrong_answers_3": "Yes, it is possible to decide for these two copolymers. The block poly(acrylonitrileisoprene) copolymer is more likely to crystallize than the graft poly(chloroprene-isobutylene) copolymer. Block copolymers crystallize more easily than graft ones."
    },
    {
        "idx": 1116,
        "question": "Crosslinked copolymers consisting of 35 wt% ethylene and 65 wt% propylene may have elastic properties similar to those for natural rubber. For a copolymer of this composition, determine the fraction of both repeat unit types.\n\\title{",
        "answer": "the fraction of the ethylene repeat unit is 0.45. the fraction of the propylene repeat unit is 0.55.",
        "type": 1,
        "wrong_answers_1": "the fraction of the ethylene repeat unit is 0.41. the fraction of the propylene repeat unit is 0.61.",
        "wrong_answers_2": "the fraction of the ethylene repeat unit is 0.47. the fraction of the propylene repeat unit is 0.59.",
        "wrong_answers_3": "the fraction of the ethylene repeat unit is 0.52. the fraction of the propylene repeat unit is 0.62."
    },
    {
        "idx": 1117,
        "question": "For the following pairs of polymers, do the following: (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (a) Linear and atactic poly(vinyl chloride); linear and isotactic polypropylene",
        "answer": "No, it is not possible to decide for these two polymers. On the basis of tacticity, the isotactic PP is more likely to crystallize than the atactic PVC. On the other hand, with regard to side-group bulkiness, the PVC is more likely to crystallize.",
        "type": 4,
        "wrong_answers_1": "Yes, for these two polymers it is possible to decide. The linear and syndiotactic poly(vinyl chloride) is more likely to crystallize; the phenyl side-group for polystyrene is bulkier than the Cl side-group for poly(vinyl chloride). Syndiotactic and isotactic isomers are equally likely to crystallize.",
        "wrong_answers_2": "Yes, it is possible to decide for these two polymers. The linear and isotactic polystyrene is more likely to crystallize than network phenol-formaldehyde; network polymers rarely crystallize, whereas isotactic ones crystallize relatively easily.",
        "wrong_answers_3": "No, it is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize."
    },
    {
        "idx": 1118,
        "question": "For the following pairs of polymers, do the following: (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (b) Linear and syndiotactic polypropylene; crosslinked cis-polyisoprene",
        "answer": "Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is more likely to crystallize than crosslinked cis-isoprene because linear polymers are more likely to crystallize than crosslinked ones.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible. The linear and isotactic polypropylene will have a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. ",
        "wrong_answers_2": "Yes, it is possible to decide for these two polymers. The linear polyethylene is more likely to crystallize. The repeat unit structure for polypropylene is chemically more complicated than is the repeat unit structure for polyethylene. Furthermore, branched structures are less likely to crystallize than are linear structures.",
        "wrong_answers_3": "No, it is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize."
    },
    {
        "idx": 1119,
        "question": "For the following pairs of polymers, do the following: (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (c) Network phenol-formaldehyde; linear and isotactic polystyrene",
        "answer": "Yes, it is possible to decide for these two polymers. The linear and isotactic polystyrene is more likely to crystallize than network phenol-formaldehyde; network polymers rarely crystallize, whereas isotactic ones crystallize relatively easily.",
        "type": 4,
        "wrong_answers_1": "No, it is not possible to decide for these two polymers. Both heavily crosslinked and network polymers are not likely to crystallize.",
        "wrong_answers_2": "Yes, it is possible. The linear and isotactic polypropylene will have a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. ",
        "wrong_answers_3": "Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is more likely to crystallize than crosslinked cis-isoprene because linear polymers are more likely to crystallize than crosslinked ones."
    },
    {
        "idx": 1120,
        "question": "For the following pairs of polymers, do the following: (1) State whether it is possible to determine whether one polymer is more likely to crystallize than the other; (2) if it is possible, note which is the more likely and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. (d) Block poly(acrylonitrile-isoprene) copolymer; graft poly(chloroprene-isobutylene) copolymer",
        "answer": "Yes, it is possible to decide for these two copolymers. The block poly(acrylonitrileisoprene) copolymer is more likely to crystallize than the graft poly(chloroprene-isobutylene) copolymer. Block copolymers crystallize more easily than graft ones.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to decide for these two copolymers. The alternating poly(styrene-ethylene) copolymer is more likely to crystallize. Alternating copolymers crystallize more easily than do random copolymers.",
        "wrong_answers_2": "Yes, it is possible to decide for these two copolymers. The linear and syndiotactic polypropylene is more likely to crystallize than crosslinked cis-isoprene because linear polymers are more likely to crystallize than crosslinked ones.",
        "wrong_answers_3": "The alternating poly(styrene-butadiene) copolymer will have a higher crystallinity than the random copolymer; alternating copolymers crystallize more easily than random ones. The influence of crystallinity on conductivity is explained in part (c)."
    },
    {
        "idx": 1121,
        "question": "For some hypothetical metal, the equilibrium number of vacancies at 900^{\\circ} C is 2.3 × 10^{25} m^{-3}. If the density and atomic weight of this metal are 7.40g / {cm}^{3} and 85.5g / mol, respectively, calculate the fraction of vacancies for this metal at 900^{\\circ} C.",
        "answer": "the fraction of vacancies for this metal at 900^{\\circ} C is 4.41 × 10^{-4}.",
        "type": 1,
        "wrong_answers_1": "the fraction of vacancies for this metal at 785^{\\circ} C is 3.98 × 10^{-4}.",
        "wrong_answers_2": "the fraction of vacancies for this metal at 857^{\\circ} C is 4.78 × 10^{-4}.",
        "wrong_answers_3": "the fraction of vacancies for this metal at 1024^{\\circ} C is 4.55 × 10^{-4}."
    },
    {
        "idx": 1122,
        "question": "(a) Calculate the fraction of atom sites that are vacant for copper (Cu) at its melting temperature of 1084°C (1357 K). Assume an energy for vacancy formation of 0.90 eV/atom.",
        "answer": "the fraction of atom sites that are vacant for copper at 1357 k is 4.56 × 10^-4.",
        "type": 1,
        "wrong_answers_1": "the fraction of atom sites that are vacant for copper at 1546 k is 4.29 × 10^-4.",
        "wrong_answers_2": "the fraction of atom sites that are vacant for copper at 1478 k is 3.91 × 10^-4.",
        "wrong_answers_3": "the fraction of atom sites that are vacant for copper at 1408 k is 4.35 × 10^-4."
    },
    {
        "idx": 1123,
        "question": "(b) Repeat this calculation at room temperature (298 K). Assume an energy for vacancy formation of 0.90 eV/atom.",
        "answer": "the fraction of atom sites that are vacant for copper at 298 k is 6.08 × 10^-16.",
        "type": 1,
        "wrong_answers_1": "the fraction of atom sites that are vacant for copper at 328 k is 5.64 × 10^-16.",
        "wrong_answers_2": "the fraction of atom sites that are vacant for copper at 327 k is 5.17 × 10^-16.",
        "wrong_answers_3": "the fraction of atom sites that are vacant for copper at 280 k is 5.23 × 10^-16."
    },
    {
        "idx": 1124,
        "question": "(c) What is ratio of N_V/N(1357 K) and N_V/N(298 K) ?",
        "answer": "the ratio of n_v/n(1357 k) to n_v/n(298 k) is 7.5 × 10^11.",
        "type": 1,
        "wrong_answers_1": "the ratio of n_v/n(1286 k) to n_v/n(279 k) is 8.0 × 10^11.",
        "wrong_answers_2": "the ratio of n_v/n(1217 k) to n_v/n(319 k) is 8.3 × 10^11.",
        "wrong_answers_3": "the ratio of n_v/n(1508 k) to n_v/n(268 k) is 7.1 × 10^11."
    },
    {
        "idx": 1125,
        "question": "Calculate the number of vacancies per cubic meter in gold (\\mathrm{Au}) at 900^{\\circ} C. The energy for vacancy formation is 0.98 \\mathrm{eV} / atom. Furthermore, the density and atomic weight for Au are 18.63 g / {cm}^{3} (at 900^{\\circ} C ) and 196.9g / mol, respectively.",
        "answer": "the number of vacancies per cubic meter in gold at 900^{\\circ} C is 3.52 × 10^{24}{m}^{-3}.",
        "type": 1,
        "wrong_answers_1": "the number of vacancies per cubic meter in gold at 851^{\\circ} C is 3.28 × 10^{24}{m}^{-3}.",
        "wrong_answers_2": "the number of vacancies per cubic meter in gold at 802^{\\circ} C is 4.04 × 10^{24}{m}^{-3}.",
        "wrong_answers_3": "the number of vacancies per cubic meter in gold at 930^{\\circ} C is 3.82 × 10^{24}{m}^{-3}."
    },
    {
        "idx": 1126,
        "question": "Calculate the fraction of lattice sites that are Schottky defects for cesium chloride at its melting temperature \\left(645^{\\circ} C\\right). Assume an energy for defect formation of 1.86 \\mathrm{eV}.",
        "answer": "the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature is 7.87 × 10^{-6}.",
        "type": 1,
        "wrong_answers_1": "the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature is 8.95 × 10^{-6}.",
        "wrong_answers_2": "the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature is 9.05 × 10^{-6}.",
        "wrong_answers_3": "the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature is 6.73 × 10^{-6}."
    },
    {
        "idx": 1127,
        "question": "In your own words, briefly define the term stoichiometric.",
        "answer": "Stoichiometric means having exactly the ratio of anions to cations as specified by the chemical formula for the compound.",
        "type": 4,
        "wrong_answers_1": "The two characteristics of component ions that determine the crystal structure of a ceramic compound are: 1) the magnitude of the electrical charge on each ion, and 2) the relative sizes of the cations and anions.",
        "wrong_answers_2": "The cations will fill tetrahedral positions since the coordination number for cations is four.",
        "wrong_answers_3": "For a small anode-to-cathode area ratio, the corrosion rate will be higher than for A large ratio. The reason for this is that for some given current flow associated with the corrosion reaction, for a small area ratio the current density at the anode will be greater than for a large ratio. The corrosion rate is proportional to the current density (i) according to Equation."
    },
    {
        "idx": 1128,
        "question": "Under conditions where cupric oxide (CuO) is exposed to reducing atmospheres at elevated temperatures, some of the Cu2+ ions will become Cu+. Name one crystalline defect that you would expect to form in order to maintain charge neutrality.",
        "answer": "For a Cu2+ O2- compound in which a small fraction of the copper ions exist as Cu+, for each Cu+ formed there is one less positive charge introduced (or one more negative charge). In order to maintain charge neutrality, we must either add an additional positive charge or subtract a negative charge. This may be accomplished be either creating Cu2+ interstitials or O2- vacancies.",
        "type": 4,
        "wrong_answers_1": "For every Al3+ ion that substitutes for Mg2+ in MgO, a single positive charge is added. Thus, in order to maintain charge neutrality, either a positive charge must be removed or a negative charge must be added. Negative charges are added by forming O2- interstitials, which are not likely to form. Positive charges may be removed by forming Mg2+ vacancies.",
        "wrong_answers_2": "For O2- substituting for Cl- in CaCl2, chlorine vacancies would be created. For each O2- substituting for a Cl-, one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to maintain charge neutrality, one O2- ion will lead to the formation of one chlorine vacancy.",
        "wrong_answers_3": "For Ca2+ substituting for Li+ in Li2O, lithium vacancies would be created. For each Ca2+ substituting for Li+, one positive charge is added; in order to maintain charge neutrality, a single positive charge may be removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca2+ ion added, a single lithium vacancy is formed."
    },
    {
        "idx": 1129,
        "question": "Under conditions where cupric oxide (CuO) is exposed to reducing atmospheres at elevated temperatures, some of the Cu2+ ions will become Cu+. How many Cu+ ions are required for the creation of each defect?",
        "answer": "There will be two Cu+ ions required for each of these defects.",
        "type": 1
    },
    {
        "idx": 1130,
        "question": "Under conditions where cupric oxide (CuO) is exposed to reducing atmospheres at elevated temperatures, some of the Cu2+ ions will become Cu+. How would you express the chemical formula for this nonstoichiometric material?",
        "answer": "The chemical formula for this nonstoichiometric material is Cu1+ O or CuO1-x, where x is some small fraction.",
        "type": 4,
        "wrong_answers_1": "Stoichiometric means having exactly the ratio of anions to cations as specified by the chemical formula for the compound.",
        "wrong_answers_2": "Chemical resistance decreases.",
        "wrong_answers_3": "the certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials, and the calculated stress is relatively close to the flexural strength, so there is some chance that fracture will occur."
    },
    {
        "idx": 1131,
        "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for FeO.",
        "answer": "For FeO, the ionic radii of the Mg2+ and Fe2+ are 0.072 nm and 0.077 nm, respectively. The percentage difference in ionic radii is 6.5%, which is within the acceptable range for a high degree of solubility. Additionally, both MgO and FeO have rock salt crystal structures. Therefore, FeO and MgO are expected to form high degrees of solid solubility, and experimentally, they exhibit 100% solubility.",
        "type": 4,
        "wrong_answers_1": "For CoO, the ionic radii of the Mg2+ and Co2+ are 0.072 nm and 0.072 nm, respectively. The percentage difference in ionic radii is 0%, which is within the acceptable range for a high degree of solubility. Therefore, CoO and MgO are expected to form high degrees of solid solubility, likely 100%.",
        "wrong_answers_2": "For MgO, the ionic radii of the Mn2+ and Mg2+ are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.072 nm - 0.067 nm) / 0.072 nm × 100 = 6.9%, which value is within the acceptable range for a high degree of solubility.",
        "wrong_answers_3": "For BaO, the ionic radii of the Mg2+ and Ba2+ are 0.072 nm and 0.136 nm, respectively. The percentage difference in ionic radii is 47%, which is much larger than the acceptable range. Therefore, BaO is not expected to experience any appreciable solubility in MgO. Experimentally, the solubility of BaO in MgO is very small."
    },
    {
        "idx": 1132,
        "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for BaO.",
        "answer": "For BaO, the ionic radii of the Mg2+ and Ba2+ are 0.072 nm and 0.136 nm, respectively. The percentage difference in ionic radii is 47%, which is much larger than the acceptable range. Therefore, BaO is not expected to experience any appreciable solubility in MgO. Experimentally, the solubility of BaO in MgO is very small.",
        "type": 4,
        "wrong_answers_1": "For PbO, the ionic radii of the Mg2+ and Pb2+ are 0.072 nm and 0.120 nm, respectively. The percentage difference in ionic radii is 40%, which is much larger than the acceptable range. Therefore, PbO is not expected to experience any appreciable solubility in MgO.",
        "wrong_answers_2": "For CoO, the ionic radii of the Mg2+ and Co2+ are 0.072 nm and 0.072 nm, respectively. The percentage difference in ionic radii is 0%, which is within the acceptable range for a high degree of solubility. Therefore, CoO and MgO are expected to form high degrees of solid solubility, likely 100%.",
        "wrong_answers_3": "For FeO, the ionic radii of the Mg2+ and Fe2+ are 0.072 nm and 0.077 nm, respectively. The percentage difference in ionic radii is 6.5%, which is within the acceptable range for a high degree of solubility. Additionally, both MgO and FeO have rock salt crystal structures. Therefore, FeO and MgO are expected to form high degrees of solid solubility, and experimentally, they exhibit 100% solubility."
    },
    {
        "idx": 1133,
        "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for PbO.",
        "answer": "For PbO, the ionic radii of the Mg2+ and Pb2+ are 0.072 nm and 0.120 nm, respectively. The percentage difference in ionic radii is 40%, which is much larger than the acceptable range. Therefore, PbO is not expected to experience any appreciable solubility in MgO.",
        "type": 4,
        "wrong_answers_1": "For BaO, the ionic radii of the Mg2+ and Ba2+ are 0.072 nm and 0.136 nm, respectively. The percentage difference in ionic radii is 47%, which is much larger than the acceptable range. Therefore, BaO is not expected to experience any appreciable solubility in MgO. Experimentally, the solubility of BaO in MgO is very small.",
        "wrong_answers_2": "For CoO, the ionic radii of the Mg2+ and Co2+ are 0.072 nm and 0.072 nm, respectively. The percentage difference in ionic radii is 0%, which is within the acceptable range for a high degree of solubility. Therefore, CoO and MgO are expected to form high degrees of solid solubility, likely 100%.",
        "wrong_answers_3": "For MgO, the ionic radii of the Mn2+ and Mg2+ are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.072 nm - 0.067 nm) / 0.072 nm × 100 = 6.9%, which value is within the acceptable range for a high degree of solubility."
    },
    {
        "idx": 1134,
        "question": "Which of the following oxides would you expect to form substitutional solid solutions that have complete (i.e., 100%) solubility with MgO? Explain your answer for CoO.",
        "answer": "For CoO, the ionic radii of the Mg2+ and Co2+ are 0.072 nm and 0.072 nm, respectively. The percentage difference in ionic radii is 0%, which is within the acceptable range for a high degree of solubility. Therefore, CoO and MgO are expected to form high degrees of solid solubility, likely 100%.",
        "type": 4,
        "wrong_answers_1": "For MgO, the ionic radii of the Mn2+ and Mg2+ are 0.067 nm and 0.072 nm, respectively. Therefore the percentage difference in ionic radii, Δr% is determined as follows: Δr% = (0.072 nm - 0.067 nm) / 0.072 nm × 100 = 6.9%, which value is within the acceptable range for a high degree of solubility.",
        "wrong_answers_2": "For FeO, the ionic radii of the Mg2+ and Fe2+ are 0.072 nm and 0.077 nm, respectively. The percentage difference in ionic radii is 6.5%, which is within the acceptable range for a high degree of solubility. Additionally, both MgO and FeO have rock salt crystal structures. Therefore, FeO and MgO are expected to form high degrees of solid solubility, and experimentally, they exhibit 100% solubility.",
        "wrong_answers_3": "For PbO, the ionic radii of the Mg2+ and Pb2+ are 0.072 nm and 0.120 nm, respectively. The percentage difference in ionic radii is 40%, which is much larger than the acceptable range. Therefore, PbO is not expected to experience any appreciable solubility in MgO."
    },
    {
        "idx": 1135,
        "question": "Suppose that CaO is added as an impurity to Li2O. If the Ca2+ substitutes for Li+, what kind of vacancies would you expect to form? How many of these vacancies are created for every Ca2+ added?",
        "answer": "For Ca2+ substituting for Li+ in Li2O, lithium vacancies would be created. For each Ca2+ substituting for Li+, one positive charge is added; in order to maintain charge neutrality, a single positive charge may be removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca2+ ion added, a single lithium vacancy is formed.",
        "type": 4,
        "wrong_answers_1": "For every Al3+ ion that substitutes for Mg2+ in MgO, a single positive charge is added. Thus, in order to maintain charge neutrality, either a positive charge must be removed or a negative charge must be added. Negative charges are added by forming O2- interstitials, which are not likely to form. Positive charges may be removed by forming Mg2+ vacancies.",
        "wrong_answers_2": "For O2- substituting for Cl- in CaCl2, chlorine vacancies would be created. For each O2- substituting for a Cl-, one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to maintain charge neutrality, one O2- ion will lead to the formation of one chlorine vacancy.",
        "wrong_answers_3": "For a Cu2+ O2- compound in which a small fraction of the copper ions exist as Cu+, for each Cu+ formed there is one less positive charge introduced (or one more negative charge). In order to maintain charge neutrality, we must either add an additional positive charge or subtract a negative charge. This may be accomplished be either creating Cu2+ interstitials or O2- vacancies."
    },
    {
        "idx": 1136,
        "question": "Suppose that CaO is added as an impurity to CaCl2. If the O2- substitutes for Cl-, what kind of vacancies would you expect to form? How many of these vacancies are created for every O2- added?",
        "answer": "For O2- substituting for Cl- in CaCl2, chlorine vacancies would be created. For each O2- substituting for a Cl-, one negative charge is added; negative charges are eliminated by creating chlorine vacancies. In order to maintain charge neutrality, one O2- ion will lead to the formation of one chlorine vacancy.",
        "type": 4,
        "wrong_answers_1": "For Ca2+ substituting for Li+ in Li2O, lithium vacancies would be created. For each Ca2+ substituting for Li+, one positive charge is added; in order to maintain charge neutrality, a single positive charge may be removed. Positive charges are eliminated by creating lithium vacancies, and for every Ca2+ ion added, a single lithium vacancy is formed.",
        "wrong_answers_2": "For a Cu2+ O2- compound in which a small fraction of the copper ions exist as Cu+, for each Cu+ formed there is one less positive charge introduced (or one more negative charge). In order to maintain charge neutrality, we must either add an additional positive charge or subtract a negative charge. This may be accomplished be either creating Cu2+ interstitials or O2- vacancies.",
        "wrong_answers_3": "For every Al3+ ion that substitutes for Mg2+ in MgO, a single positive charge is added. Thus, in order to maintain charge neutrality, either a positive charge must be removed or a negative charge must be added. Negative charges are added by forming O2- interstitials, which are not likely to form. Positive charges may be removed by forming Mg2+ vacancies."
    },
    {
        "idx": 1137,
        "question": "Which of these elements would you expect to form a substitutional solid solution having complete solubility with nickel? The four Hume-Rothery rules must be satisfied: (1) the difference in atomic radii between Ni and the other element (ΔR%) must be less than ±15%; (2) the crystal structures must be the same; (3) the electronegativities must be similar; and (4) the valences should be the same.",
        "answer": "Pt is the only element that meets all of the criteria and thus forms a substitutional solid solution having complete solubility. At elevated temperatures Co and Fe experience allotropic transformations to the FCC crystal structure, and thus display complete solid solubility at these temperatures.",
        "type": 2
    },
    {
        "idx": 1138,
        "question": "Which of these elements would you expect to form a substitutional solid solution of incomplete solubility with nickel? The four Hume-Rothery rules must be satisfied: (1) the difference in atomic radii between Ni and the other element (ΔR%) must be less than ±15%; (2) the crystal structures must be the same; (3) the electronegativities must be similar; and (4) the valences should be the same.",
        "answer": "Ag, Al, Co, Cr, Fe, and Zn form substitutional solid solutions of incomplete solubility. All these metals have either BCC or HCP crystal structures, and/or the difference between their atomic radii and that for Ni are greater than ±15%, and/or have a valence different than 2+.",
        "type": 2
    },
    {
        "idx": 1139,
        "question": "Which of these elements would you expect to form an interstitial solid solution with nickel? The four Hume-Rothery rules must be satisfied: (1) the difference in atomic radii between Ni and the other element (ΔR%) must be less than ±15%; (2) the crystal structures must be the same; (3) the electronegativities must be similar; and (4) the valences should be the same.",
        "answer": "C, H, and O form interstitial solid solutions. These elements have atomic radii that are significantly smaller than the atomic radius of Ni.",
        "type": 2
    },
    {
        "idx": 1140,
        "question": "What is the composition, in atom percent, of an alloy that consists of 92.5 wt% \\mathrm{Ag} and 7.5 wt% Cu ?",
        "answer": "the composition of the alloy is 87.9 \\text{ at}% \\text{ ag} and 12.1 \\text{ at}% \\text{ cu}.",
        "type": 1,
        "wrong_answers_1": "the composition of the alloy is 98.7 \\text{ at}% \\text{ ag} and 12.5 \\text{ at}% \\text{ cu}.",
        "wrong_answers_2": "the composition of the alloy is 80.4 \\text{ at}% \\text{ ag} and 11.7 \\text{ at}% \\text{ cu}.",
        "wrong_answers_3": "the composition of the alloy is 76.7 \\text{ at}% \\text{ ag} and 10.5 \\text{ at}% \\text{ cu}."
    },
    {
        "idx": 1141,
        "question": "What is the composition, in atom percent, of an alloy that consists of 5.5 wt% \\mathrm{~Pb} and 94.5 wt% \\mathrm{Sn} ?",
        "answer": "the composition of the alloy is 3.23 \\text{ at% pb} and 96.77 \\text{ at% sn}.",
        "type": 1,
        "wrong_answers_1": "the composition of the alloy is 2.92 \\text{ at% pb} and 107.59 \\text{ at% sn}.",
        "wrong_answers_2": "the composition of the alloy is 3.60 \\text{ at% pb} and 102.03 \\text{ at% sn}.",
        "wrong_answers_3": "the composition of the alloy is 3.64 \\text{ at% pb} and 90.87 \\text{ at% sn}."
    },
    {
        "idx": 1142,
        "question": "What is the composition, in weight percent, of an alloy that consists of 5 at% Cu and 95 at% Pt?",
        "answer": "the composition in weight percent is: 1.68 wt% cu  \n98.32 wt% pt",
        "type": 1,
        "wrong_answers_1": "the composition in weight percent is: 1.45 wt% cu  \n94.22 wt% pt",
        "wrong_answers_2": "the composition in weight percent is: 1.44 wt% cu  \n83.72 wt% pt",
        "wrong_answers_3": "the composition in weight percent is: 1.62 wt% cu  \n104.29 wt% pt"
    },
    {
        "idx": 1143,
        "question": "Calculate the composition, in weight percent, of an alloy that contains 105kg of iron, 0.2 kg of carbon, and 1.0kg of chromium.\n\\title{",
        "answer": "the composition, in weight percent, of the alloy is: c_{\\mathrm{fe}} = 98.87 \\, \\text{wt}% c_{C} = 0.19 \\, \\text{wt}% c_{\\mathrm{cr}} = 0.94 \\, \\text{wt}%",
        "type": 1,
        "wrong_answers_1": "the composition, in weight percent, of the alloy is: c_{\\mathrm{fe}} = 84.99 \\, \\text{wt}% c_{C} = 0.20 \\, \\text{wt}% c_{\\mathrm{cr}} = 0.82 \\, \\text{wt}%",
        "wrong_answers_2": "the composition, in weight percent, of the alloy is: c_{\\mathrm{fe}} = 104.58 \\, \\text{wt}% c_{C} = 0.20 \\, \\text{wt}% c_{\\mathrm{cr}} = 0.90 \\, \\text{wt}%",
        "wrong_answers_3": "the composition, in weight percent, of the alloy is: c_{\\mathrm{fe}} = 86.05 \\, \\text{wt}% c_{C} = 0.18 \\, \\text{wt}% c_{\\mathrm{cr}} = 0.84 \\, \\text{wt}%"
    },
    {
        "idx": 1144,
        "question": "What is the composition, in atom percent, of an alloy that contains 44.5 lb m of \\mathrm{Ag}, 83.7 lb_{m} of \\mathrm{Au}, and 5.3 lb_{m} of Cu ?",
        "answer": "the composition of the alloy in atom percent is:  c_{\\mathrm{ag}} = 44.8 \\, \\text{at}%   c_{\\mathrm{au}} = 46.2 \\, \\text{at}%   c_{\\mathrm{cu}} = 9.0 \\, \\text{at}% ",
        "type": 1,
        "wrong_answers_1": "the composition of the alloy in atom percent is:  c_{\\mathrm{ag}} = 43.3 \\, \\text{at}%   c_{\\mathrm{au}} = 50.4 \\, \\text{at}%   c_{\\mathrm{cu}} = 8.4 \\, \\text{at}% ",
        "wrong_answers_2": "the composition of the alloy in atom percent is:  c_{\\mathrm{ag}} = 50.5 \\, \\text{at}%   c_{\\mathrm{au}} = 50.8 \\, \\text{at}%   c_{\\mathrm{cu}} = 10.3 \\, \\text{at}% ",
        "wrong_answers_3": "the composition of the alloy in atom percent is:  c_{\\mathrm{ag}} = 41.3 \\, \\text{at}%   c_{\\mathrm{au}} = 40.5 \\, \\text{at}%   c_{\\mathrm{cu}} = 9.5 \\, \\text{at}% "
    },
    {
        "idx": 1145,
        "question": "Calculate the number of atoms per cubic meter in \\mathrm{Pb}.",
        "answer": "3.30 × 10^{28} atoms/m^3",
        "type": 1,
        "wrong_answers_1": "2.84 × 10^{28} atoms/m^3",
        "wrong_answers_2": "3.48 × 10^{28} atoms/m^3",
        "wrong_answers_3": "3.74 × 10^{28} atoms/m^3"
    },
    {
        "idx": 1146,
        "question": "Calculate the number of atoms per cubic meter in chromium.",
        "answer": "the number of atoms per cubic meter in chromium is 8.33 × 10^{28} \\ \\text{atoms/m}^3.",
        "type": 1,
        "wrong_answers_1": "the number of atoms per cubic meter in chromium is 9.52 × 10^{28} \\ \\text{atoms/m}^3.",
        "wrong_answers_2": "the number of atoms per cubic meter in chromium is 7.98 × 10^{28} \\ \\text{atoms/m}^3.",
        "wrong_answers_3": "the number of atoms per cubic meter in chromium is 7.65 × 10^{28} \\ \\text{atoms/m}^3."
    },
    {
        "idx": 1147,
        "question": "The concentration of \\mathrm{Si} in an \\mathrm{Fe}-\\mathrm{Si} alloy is 0.25 wt%. What is the concentration in kilograms of Si per cubic meter of alloy?",
        "answer": "the concentration of \\mathrm{si} is 19.6kg / m^{3}.",
        "type": 1,
        "wrong_answers_1": "the concentration of \\mathrm{si} is 20.4kg / m^{3}.",
        "wrong_answers_2": "the concentration of \\mathrm{si} is 21.5kg / m^{3}.",
        "wrong_answers_3": "the concentration of \\mathrm{si} is 20.9kg / m^{3}."
    },
    {
        "idx": 1148,
        "question": "Determine the approximate density of a Ti-6Al-4V titanium alloy that has a composition of 90 wt% \\mathrm{Ti}, 6 wt% \\mathrm{Al}, and 4 wt% V.",
        "answer": "the approximate density of the ti-6al-4v titanium alloy is 4.38 \\mathrm{g/cm}^3.",
        "type": 1,
        "wrong_answers_1": "the approximate density of the ti-6al-3v titanium alloy is 4.12 \\mathrm{g/cm}^3.",
        "wrong_answers_2": "the approximate density of the ti-6al-4v titanium alloy is 3.93 \\mathrm{g/cm}^3.",
        "wrong_answers_3": "the approximate density of the ti-5al-4v titanium alloy is 4.78 \\mathrm{g/cm}^3."
    },
    {
        "idx": 1149,
        "question": "Calculate the unit cell edge length for an 80 wt% \\mathrm{Ag}-20 wt% \\mathrm{Pd} alloy. All of the palladium is in solid solution, the crystal structure for this alloy is FCC, and the roomtemperature density of P d is 12.02g / {cm}^{3}.",
        "answer": "the unit cell edge length for an 80 wt% \\mathrm{ag}-20 wt% \\mathrm{pd} alloy is 0.4050nm.",
        "type": 1,
        "wrong_answers_1": "the unit cell edge length for an 87 wt% \\mathrm{ag}-17 wt% \\mathrm{pd} alloy is 0.3743nm.",
        "wrong_answers_2": "the unit cell edge length for an 91 wt% \\mathrm{ag}-22 wt% \\mathrm{pd} alloy is 0.4389nm.",
        "wrong_answers_3": "the unit cell edge length for an 70 wt% \\mathrm{ag}-22 wt% \\mathrm{pd} alloy is 0.3740nm."
    },
    {
        "idx": 1150,
        "question": "Some hypothetical alloy is composed of 25 wt% of metal A and 75 wt% of metal B. If the densities of metals A and B are 6.17 and 8.00g / {cm}^{3}, respectively, and their respective atomic weights are 171.3 and 162.0g / mol, determine whether the crystal structure for this alloy is simple cubic, face-centered cubic, or body-centered cubic. Assume a unit cell edge length of 0.332nm.",
        "answer": "the crystal structure is simple cubic with 1.00 atom per unit cell.",
        "type": 1,
        "wrong_answers_1": "the crystal structure is simple cubic with 0.91 atom per unit cell.",
        "wrong_answers_2": "the crystal structure is simple cubic with 1.13 atom per unit cell.",
        "wrong_answers_3": "the crystal structure is simple cubic with 1.06 atom per unit cell."
    },
    {
        "idx": 1151,
        "question": "Molybdenum (Mo) forms a substitutional solid solution with tungsten (W). Compute the number of molybdenum atoms per cubic centimeter for a molybdenum-tungsten alloy that contains 16.4 wt% Mo and 83.6 wt% W. The densities of pure molybdenum and tungsten are 10.22 and 19.30g / {cm}^{3}, respectively.",
        "answer": "1.73 × 10^{22} \\text{ atoms/cm}^3",
        "type": 1,
        "wrong_answers_1": "1.87 × 10^{22} \\text{ atoms/cm}^3",
        "wrong_answers_2": "1.60 × 10^{22} \\text{ atoms/cm}^3",
        "wrong_answers_3": "1.95 × 10^{22} \\text{ atoms/cm}^3"
    },
    {
        "idx": 1152,
        "question": "Niobium forms a substitutional solid solution with vanadium. Compute the number of niobium atoms per cubic centimeter for a niobium-vanadium alloy that contains 24 wt% nb and 76 wt% V. The densities of pure niobium and vanadium are 8.57 and 6.10g / {cm}^{3}, respectively.",
        "answer": "the number of niobium atoms per cubic centimeter is 1.02 × 10^{22} \\text{atoms/cm}^3.",
        "type": 1,
        "wrong_answers_1": "the number of niobium atoms per cubic centimeter is 0.92 × 10^{22} \\text{atoms/cm}^3.",
        "wrong_answers_2": "the number of niobium atoms per cubic centimeter is 1.14 × 10^{22} \\text{atoms/cm}^3.",
        "wrong_answers_3": "the number of niobium atoms per cubic centimeter is 0.96 × 10^{22} \\text{atoms/cm}^3."
    },
    {
        "idx": 1153,
        "question": "For a BCC iron-carbon alloy that contains 0.1 wt% C, calculate the fraction of unit cells that contain carbon atoms.\n\\title{",
        "answer": "the fraction of unit cells that contain carbon atoms is 9.31 × 10^{-3} atoms/unit cell. there is one carbon atom per 107.5 unit cells.",
        "type": 1,
        "wrong_answers_1": "the fraction of unit cells that contain carbon atoms is 10.05 × 10^{-3} atoms/unit cell. there is one carbon atom per 104.0 unit cells.",
        "wrong_answers_2": "the fraction of unit cells that contain carbon atoms is 10.08 × 10^{-3} atoms/unit cell. there is one carbon atom per 102.4 unit cells.",
        "wrong_answers_3": "the fraction of unit cells that contain carbon atoms is 10.30 × 10^{-3} atoms/unit cell. there is one carbon atom per 123.4 unit cells."
    },
    {
        "idx": 1154,
        "question": "Gold (Au) forms a substitutional solid solution with silver (Ag). Compute the weight percent of gold that must be added to silver to yield an alloy that contains 5.5 × 10^{21} \\mathrm{Au} atoms per cubic centimeter. The densities of pure \\mathrm{Au} and \\mathrm{Ag} are 19.32 and 10.49g / {cm}^{3}, respectively.",
        "answer": "15.9 wt%",
        "type": 1,
        "wrong_answers_1": "13.6 wt%",
        "wrong_answers_2": "17.3 wt%",
        "wrong_answers_3": "14.0 wt%"
    },
    {
        "idx": 1155,
        "question": "Germanium (Ge) forms a substitutional solid solution with silicon (Si). Compute the weight percent of germanium that must be added to silicon to yield an alloy that contains 2.43 × 10^{21} Ge atoms per cubic centimeter. The densities of pure \\mathrm{Ge} and \\mathrm{Si} are 5.32 and 2.33g / {cm}^{3}, respectively.",
        "answer": "11.7 wt%",
        "type": 1,
        "wrong_answers_1": "13.4 wt%",
        "wrong_answers_2": "11.3 wt%",
        "wrong_answers_3": "11.3 wt%"
    },
    {
        "idx": 1156,
        "question": "For an ASTM grain size of 6, approximately how many grains would there be per square inch at a magnification of 100x?",
        "answer": "32 grains/in.^2",
        "type": 1,
        "wrong_answers_1": "36 grains/in.^2",
        "wrong_answers_2": "35 grains/in.^2",
        "wrong_answers_3": "35 grains/in.^2"
    },
    {
        "idx": 1157,
        "question": "For an ASTM grain size of 6, approximately how many grains would there be per square inch without any magnification?",
        "answer": "320000 grains/in.^2",
        "type": 1,
        "wrong_answers_1": "3640 grains/in.^2",
        "wrong_answers_2": "2820 grains/in.^2",
        "wrong_answers_3": "3540 grains/in.^2"
    },
    {
        "idx": 1158,
        "question": "Determine the ASTM grain size number if 30 grains per square inch are measured at a magnification of 250 ×.",
        "answer": "the astm grain size number is 8.6.",
        "type": 1,
        "wrong_answers_1": "the astm grain size number is 9.6.",
        "wrong_answers_2": "the astm grain size number is 7.6.",
        "wrong_answers_3": "the astm grain size number is 9.3."
    },
    {
        "idx": 1159,
        "question": "Determine the ASTM grain size number if 25 grains per square inch are measured at a magnification of 75 ×.",
        "answer": "the astm grain size number is 4.8.",
        "type": 1,
        "wrong_answers_1": "the astm grain size number is 4.9.",
        "wrong_answers_2": "the astm grain size number is 4.4.",
        "wrong_answers_3": "the astm grain size number is 4.1."
    },
    {
        "idx": 1160,
        "question": "Briefly explain the difference between self-diffusion and interdiffusion.",
        "answer": "Self-diffusion is atomic migration in pure metals--i.e., when all atoms exchanging positions are of the same type. Interdiffusion is diffusion of atoms of one metal into another metal.",
        "type": 4,
        "wrong_answers_1": "Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species--i.e., the diffusion flux is independent of time.",
        "wrong_answers_2": "Self-diffusion may be monitored by using radioactive isotopes of the metal being studied. The motion of these isotopic atoms may be detected by measurement of radioactivity level.",
        "wrong_answers_3": "With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Self-diffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism."
    },
    {
        "idx": 1161,
        "question": "Self-diffusion involves the motion of atoms that are all of the same type; therefore, it is not subject to observation by compositional changes, as with interdiffusion. Suggest one way in which self-diffusion may be monitored.",
        "answer": "Self-diffusion may be monitored by using radioactive isotopes of the metal being studied. The motion of these isotopic atoms may be detected by measurement of radioactivity level.",
        "type": 4,
        "wrong_answers_1": "Self-diffusion is atomic migration in pure metals--i.e., when all atoms exchanging positions are of the same type. Interdiffusion is diffusion of atoms of one metal into another metal.",
        "wrong_answers_2": "Two properties that may be improved by crystallization are (1) a lower coefficient of thermal expansion, and (2) higher strength.",
        "wrong_answers_3": "With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Self-diffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism."
    },
    {
        "idx": 1162,
        "question": "Compare interstitial and vacancy atomic mechanisms for diffusion.",
        "answer": "With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Self-diffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism.",
        "type": 4,
        "wrong_answers_1": "Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller, are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent to a host (or substitutional impurity) atom.",
        "wrong_answers_2": "Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species--i.e., the diffusion flux is independent of time.",
        "wrong_answers_3": "Self-diffusion is atomic migration in pure metals--i.e., when all atoms exchanging positions are of the same type. Interdiffusion is diffusion of atoms of one metal into another metal."
    },
    {
        "idx": 1163,
        "question": "Cite two reasons why interstitial diffusion is normally more rapid than vacancy diffusion.",
        "answer": "Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller, are more mobile; and (2) the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent to a host (or substitutional impurity) atom.",
        "type": 4,
        "wrong_answers_1": "With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Self-diffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism.",
        "wrong_answers_2": "Steady-state diffusion is the situation wherein the rate of diffusion into a given system is just equal to the rate of diffusion out, such that there is no net accumulation or depletion of diffusing species--i.e., the diffusion flux is independent of time.",
        "wrong_answers_3": "Frenkel defects for anions would not exist in appreciable concentrations because the anion is quite large and is highly unlikely to exist as an interstitial."
    },
    {
        "idx": 1164,
        "question": "A sheet of steel 5.0-mm thick has nitrogen atmospheres on both sides at 900^{\\circ} C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 1.85 × 10^{-10}{m}^{2} / s, and the diffusion flux is found to be 1.0 × 10^{-7} kg / m^{2}·s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 2kg / m^{3}. How far into the sheet from this high-pressure side will the concentration be 0.5kg / m^{3} ? Assume a linear concentration profile.",
        "answer": "the concentration will be 0.5 \\mathrm{kg/m}^{3} at a distance of 2.78 mm from the high-pressure side.",
        "type": 1,
        "wrong_answers_1": "the concentration will be 0.4 \\mathrm{kg/m}^{3} at a distance of 2.55 mm from the high-pressure side.",
        "wrong_answers_2": "the concentration will be 0.5 \\mathrm{kg/m}^{3} at a distance of 3.07 mm from the high-pressure side.",
        "wrong_answers_3": "the concentration will be 0.4 \\mathrm{kg/m}^{3} at a distance of 3.12 mm from the high-pressure side."
    },
    {
        "idx": 1165,
        "question": "For a steel alloy it has been determined that a carburizing heat treatment of 15h duration will raise the carbon concentration to 0.35 wt% at a point 2.0mm from the surface. Estimate the time necessary to achieve the same concentration at a 6.0-mm position for an identical steel and at the same carburizing temperature.",
        "answer": "the time necessary to achieve the same concentration at a 6.0-mm position is 135 h.",
        "type": 1,
        "wrong_answers_1": "the time necessary to achieve the same concentration at a 6.7-mm position is 116 h.",
        "wrong_answers_2": "the time necessary to achieve the same concentration at a 6.8-mm position is 148 h.",
        "wrong_answers_3": "the time necessary to achieve the same concentration at a 6.8-mm position is 127 h."
    },
    {
        "idx": 1166,
        "question": "The preexponential and activation energy for the diffusion of chromium in nickel are 1.1 × 10^{-4}{m}^{2} / s and 272,000 J/ mol, respectively. At what temperature will the diffusion coefficient have a value of 1.2 × 10^{-14}{m}^{2} / s ?",
        "answer": "the temperature at which the diffusion coefficient has a value of 1.2 × 10^{-14} m^{2} / s is 1427 k or 1154^{\\circ} C.  alternatively, using the vmse software, the temperature is found to be 1430 k.",
        "type": 1,
        "wrong_answers_1": "the temperature at which the diffusion coefficient has a value of 1.1 × 10^{-14} m^{2} / s is 1287 k or 1023^{\\circ} C.  alternatively, using the vmse software, the temperature is found to be 1300 k.",
        "wrong_answers_2": "the temperature at which the diffusion coefficient has a value of 1.0 × 10^{-14} m^{2} / s is 1266 k or 1264^{\\circ} C.  alternatively, using the vmse software, the temperature is found to be 1550 k.",
        "wrong_answers_3": "the temperature at which the diffusion coefficient has a value of 1.0 × 10^{-14} m^{2} / s is 1248 k or 1105^{\\circ} C.  alternatively, using the vmse software, the temperature is found to be 1600 k."
    },
    {
        "idx": 1167,
        "question": "The activation energy for the diffusion of copper in silver is 193,000 J/ mol. Calculate the diffusion coefficient at 1200 K\\left(927^{\\circ} C\\right), given that D at 1000 K\\left(727^{\\circ} C\\right) is 1.0 × 10^{-14}{m}^{2} / s.",
        "answer": "the diffusion coefficient at 1200k is 4.8 × 10^{-13}{m}^{2} / s.",
        "type": 1,
        "wrong_answers_1": "the diffusion coefficient at 1030k is 5.1 × 10^{-13}{m}^{2} / s.",
        "wrong_answers_2": "the diffusion coefficient at 1300k is 5.3 × 10^{-13}{m}^{2} / s.",
        "wrong_answers_3": "the diffusion coefficient at 1080k is 5.0 × 10^{-13}{m}^{2} / s."
    },
    {
        "idx": 1168,
        "question": "Carbon is allowed to diffuse through a steel plate 10-mm thick. The concentrations of carbon at the two faces are 0.85 and 0.40kg C / {cm}^{3} Fe, which are maintained constant. If the preexponential and activation energy are 5.0 × 10^{-7}{m}^{2} / s and 77,000 J/ mol, respectively, compute the temperature at which the diffusion flux is 6.3 × 10^{-10}kg / m^{2}·s.",
        "answer": "the temperature at which the diffusion flux is 6.3 × 10^{-10} kg/m^2·s is 884 k or 611^{\\circ} C.",
        "type": 1,
        "wrong_answers_1": "the temperature at which the diffusion flux is 5.8 × 10^{-10} kg/m^2·s is 977 k or 592^{\\circ} C.",
        "wrong_answers_2": "the temperature at which the diffusion flux is 6.6 × 10^{-10} kg/m^2·s is 982 k or 580^{\\circ} C.",
        "wrong_answers_3": "the temperature at which the diffusion flux is 5.4 × 10^{-10} kg/m^2·s is 760 k or 681^{\\circ} C."
    },
    {
        "idx": 1169,
        "question": "The steady-state diffusion flux through a metal plate is 7.8 × 10^{-8}kg / m^{2}·s at a temperature of 1200^{\\circ} C(1473 K) and when the concentration gradient is -500kg / m^{4}. Calculate the diffusion flux at 1000^{\\circ} C(1273 K) for the same concentration gradient and assuming an activation energy for diffusion of 145,000 J/ mol.",
        "answer": "the diffusion flux at 1000^{\\circ} C(1273k) is 1.21 × 10^{-8}kg / m^{2}·s.",
        "type": 1,
        "wrong_answers_1": "the diffusion flux at 920^{\\circ} C(1433k) is 1.06 × 10^{-8}kg / m^{2}·s.",
        "wrong_answers_2": "the diffusion flux at 1060^{\\circ} C(1443k) is 1.35 × 10^{-8}kg / m^{2}·s.",
        "wrong_answers_3": "the diffusion flux at 1060^{\\circ} C(1363k) is 1.31 × 10^{-8}kg / m^{2}·s."
    },
    {
        "idx": 1170,
        "question": "Consider the diffusion of some hypothetical metal Y into another hypothetical metal Z at 950^{\\circ} C; after 10h the concentration at the 0.5mm position (in metal Z) is 2.0 wt% Y. At what position will the concentration also be 2.0 wt% Y after a 17.5h heat treatment again at 950^{\\circ} C ? Assume preexponential and activation energy values of 4.3 × 10^{-4}{m}^{2} / s and 180,000 J/ mol, respectively, for this diffusion system.",
        "answer": "the position will be 0.66 mm.",
        "type": 1,
        "wrong_answers_1": "the position will be 0.61 mm.",
        "wrong_answers_2": "the position will be 0.63 mm.",
        "wrong_answers_3": "the position will be 0.60 mm."
    },
    {
        "idx": 1171,
        "question": "Carbon dioxide diffuses through a high-density polyethylene (HDPE) sheet 50mm thick at a rate of 2.2 × 10^{-8}\\left(cm^{3} STP / {cm}^{2}\\right.-s at 325 K. The pressures of carbon dioxide at the two faces are 4000 kPa and 2500 kPa, which are maintained constant. Assuming conditions of steady state, what is the permeability coefficient at 325 K ?",
        "answer": "the permeability coefficient at 325k is 0.73 × 10^{-13} \\frac{{cm}^{3} \\mathrm{stp}·{cm}}{{cm}^{2}·s·\\mathrm{pa}}.",
        "type": 1,
        "wrong_answers_1": "the permeability coefficient at 359k is 0.69 × 10^{-13} \\frac{{cm}^{3} \\mathrm{stp}·{cm}}{{cm}^{2}·s·\\mathrm{pa}}.",
        "wrong_answers_2": "the permeability coefficient at 302k is 0.68 × 10^{-13} \\frac{{cm}^{3} \\mathrm{stp}·{cm}}{{cm}^{2}·s·\\mathrm{pa}}.",
        "wrong_answers_3": "the permeability coefficient at 344k is 0.75 × 10^{-13} \\frac{{cm}^{3} \\mathrm{stp}·{cm}}{{cm}^{2}·s·\\mathrm{pa}}."
    },
    {
        "idx": 1172,
        "question": "A cylindrical specimen of a nickel alloy having an elastic modulus of 207 \\mathrm{GPa}\\left(30 × 10^{6} psi\\right) and an original diameter of 10.2 mm(0.40 in.) experiences only elastic deformation when a tensile load of 8900N\\left(2000 lb f_{\\mathrm{f}}\\right) is applied. Compute the maximum length of the specimen before deformation if the maximum allowable elongation is 0.25 mm(0.010 in.).",
        "answer": "the maximum length of the specimen before deformation is 0.475{m} (475mm or 18.7 in.).",
        "type": 1,
        "wrong_answers_1": "the maximum length of the specimen before deformation is 0.489{m} (413mm or 20.7 in.).",
        "wrong_answers_2": "the maximum length of the specimen before deformation is 0.437{m} (435mm or 20.1 in.).",
        "wrong_answers_3": "the maximum length of the specimen before deformation is 0.542{m} (492mm or 17.2 in.)."
    },
    {
        "idx": 1173,
        "question": "An aluminum bar 125mm (5.0 in.) long and having a square cross section 16.5mm (0.65 in.) on an edge is pulled in tension with a load of 66,700N(15,000 lb) and experiences an elongation of 0.43 mm(1.7 × 10^{-2} in.). Assuming that the deformation is entirely elastic, calculate the modulus of elasticity of the aluminum.",
        "answer": "the modulus of elasticity of the aluminum is 71.2 gpa (10.4 × 10^{6} psi).",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity of the aluminum is 76.6 gpa (9.2 × 10^{6} psi).",
        "wrong_answers_2": "the modulus of elasticity of the aluminum is 64.3 gpa (11.8 × 10^{6} psi).",
        "wrong_answers_3": "the modulus of elasticity of the aluminum is 73.4 gpa (9.1 × 10^{6} psi)."
    },
    {
        "idx": 1174,
        "question": "Consider a cylindrical nickel wire 2.0 mm(0.08 in.) in diameter and 3 × 10^{4}mm (1200 in.) long. Calculate its elongation when a load of 300N\\left(67 lb^{2}\\right) is applied. Assume that the deformation is totally elastic.",
        "answer": "the elongation of the nickel wire is 13.8mm (0.53 in.).",
        "type": 1,
        "wrong_answers_1": "the elongation of the nickel wire is 15.7mm (0.49 in.).",
        "wrong_answers_2": "the elongation of the nickel wire is 12.7mm (0.60 in.).",
        "wrong_answers_3": "the elongation of the nickel wire is 11.9mm (0.49 in.)."
    },
    {
        "idx": 1175,
        "question": "For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0 x 10^6 psi). What is the maximum load that can be applied to a specimen with a cross-sectional area of 130 mm^2 (0.2 in.^2) without plastic deformation?",
        "answer": "the maximum load that can be applied without plastic deformation is 44,850 n (10,000 lb_f).",
        "type": 1,
        "wrong_answers_1": "the maximum load that can be applied without plastic deformation is 41132 n (10552 lb_f).",
        "wrong_answers_2": "the maximum load that can be applied without plastic deformation is 47307 n (8922 lb_f).",
        "wrong_answers_3": "the maximum load that can be applied without plastic deformation is 50303 n (8563 lb_f)."
    },
    {
        "idx": 1176,
        "question": "For a brass alloy, the stress at which plastic deformation begins is 345 MPa (50,000 psi), and the modulus of elasticity is 103 GPa (15.0 x 10^6 psi). If the original specimen length is 76mm (3.0 in.), what is the maximum length to which it can be stretched without causing plastic deformation?",
        "answer": "the maximum length to which the specimen can be stretched without causing plastic deformation is 76.25mm (3.01 in.).",
        "type": 1,
        "wrong_answers_1": "the maximum length to which the specimen can be stretched without causing plastic deformation is 68.78mm (2.78 in.).",
        "wrong_answers_2": "the maximum length to which the specimen can be stretched without causing plastic deformation is 86.96mm (3.20 in.).",
        "wrong_answers_3": "the maximum length to which the specimen can be stretched without causing plastic deformation is 81.14mm (2.91 in.)."
    },
    {
        "idx": 1177,
        "question": "A cylindrical rod of steel (E=207 \\mathrm{GPa}, 30 × 10^{6} psi) having a yield strength of 310 MPa(45,000 psi) is to be subjected to a load of 11,100N(2500 lb). If the length of the rod is 500 mm(20.0 in.), what must be the diameter to allow an elongation of 0.38 mm(0.015 in.)?",
        "answer": "the diameter must be 9.5 mm (0.376 in.).",
        "type": 1,
        "wrong_answers_1": "the diameter must be 8.3 mm (0.395 in.).",
        "wrong_answers_2": "the diameter must be 9.9 mm (0.356 in.).",
        "wrong_answers_3": "the diameter must be 8.1 mm (0.352 in.)."
    },
    {
        "idx": 1178,
        "question": "A cylindrical specimen of a metal alloy 10 mm(0.4 in.) in diameter is stressed elastically in tension. A force of 15,000N(3,370 lb r) produces a reduction in specimen diameter of 7 × 10^{-3} mm(2.8 × 10^{-4} in.). Compute Poisson's ratio for this material if its elastic modulus is 100 \\mathrm{GPa} \\left(14.5 × 10^{6} psi\\right).",
        "answer": "poisson's ratio for the material is  v = 0.367 .",
        "type": 1,
        "wrong_answers_1": "poisson's ratio for the material is  v = 0.395 .",
        "wrong_answers_2": "poisson's ratio for the material is  v = 0.354 .",
        "wrong_answers_3": "poisson's ratio for the material is  v = 0.385 ."
    },
    {
        "idx": 1179,
        "question": "Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 10.0 mm(0.39 in.). A tensile force of 1500N(340 lb) produces an elastic reduction in diameter of 6.7 × 10^{-4} mm(2.64 ×\n10^{-5} in.). Compute the elastic modulus of this alloy, given that Poisson's ratio is 0.35 .",
        "answer": "100 gpa.",
        "type": 1,
        "wrong_answers_1": "112 gpa.",
        "wrong_answers_2": "87 gpa.",
        "wrong_answers_3": "95 gpa."
    },
    {
        "idx": 1180,
        "question": "A cylindrical rod 500mm (20.0 in.) long and having a diameter of 12.7mm (0.50 in.) is to be subjected to a tensile load of 29,000 N (6500 lb). Which of the four metals or alloys listed in the table can experience neither plastic deformation nor an elongation of more than 1.3mm (0.05 in.)?",
        "answer": "Of the four metal alloys listed, only brass and steel satisfy the stipulated criteria. Brass has an elongation of 1.15 mm, and steel has an elongation of 0.56 mm, both of which are less than the maximum allowed elongation of 1.3 mm.",
        "type": 2
    },
    {
        "idx": 1181,
        "question": "Why is aluminum alloy not a candidate for the cylindrical rod subjected to a tensile load of 29,000 N (6500 lb) with a maximum allowed elongation of 1.3mm (0.05 in.)?",
        "answer": "Aluminum alloy is not a candidate because its elongation of 1.64mm exceeds the limit.",
        "type": 4,
        "wrong_answers_1": "Of the alloys listed, the steel alloy is a possible candidate because it meets the criterion regarding diameter reduction. The titanium alloy is not a candidate as it fails this criterion.",
        "wrong_answers_2": "Of the alloys listed, the steel alloy is a possible candidate because it meets the criterion of not experiencing plastic deformation.",
        "wrong_answers_3": "For the aluminum-cast iron couple, corrosion is possible, and aluminum will corrode."
    },
    {
        "idx": 1182,
        "question": "Why is copper not considered for the cylindrical rod subjected to a tensile load of 29,000 N (6500 lb) with a maximum allowed elongation of 1.3mm (0.05 in.)?",
        "answer": "Copper is not considered due to its lower yield strength.",
        "type": 4,
        "wrong_answers_1": "Strength decreases.",
        "wrong_answers_2": "Two properties that may be improved by crystallization are (1) a lower coefficient of thermal expansion, and (2) higher strength.",
        "wrong_answers_3": "diffusion of oxygen is the slower of the two, due to the larger ionic radius of the oxygen."
    },
    {
        "idx": 1183,
        "question": "A cylindrical metal specimen having an original diameter of 12.8mm (0.505 in.) and gauge length of 50.80mm (2.000 in.) is pulled in tension until fracture occurs. The diameter at the point of fracture is 8.13mm (0.320 in.). Calculate the ductility in terms of percent reduction in area.",
        "answer": "the ductility in terms of percent reduction in area is 60%.",
        "type": 1,
        "wrong_answers_1": "the ductility in terms of percent reduction in area is 66%.",
        "wrong_answers_2": "the ductility in terms of percent reduction in area is 52%.",
        "wrong_answers_3": "the ductility in terms of percent reduction in area is 66%."
    },
    {
        "idx": 1184,
        "question": "A cylindrical metal specimen having an original diameter of 12.8mm (0.505 in.) and gauge length of 50.80mm (2.000 in.) is pulled in tension until fracture occurs. The fractured gauge length is 74.17mm (2.920 in.). Calculate the ductility in terms of percent elongation.",
        "answer": "the ductility in terms of percent elongation is 46%.",
        "type": 1,
        "wrong_answers_1": "the ductility in terms of percent elongation is 52%.",
        "wrong_answers_2": "the ductility in terms of percent elongation is 42%.",
        "wrong_answers_3": "the ductility in terms of percent elongation is 42%."
    },
    {
        "idx": 1185,
        "question": "The following true stresses produce the corresponding true plastic strains for a brass alloy:\n\\begin{tabular}{cc}\n\\hline True Stress (psi) & True Strain \\\\\n\\hline 60,000 & 0.15 \\\\\n70,000 & 0.25 \\\\\n\\hline\n\\end{tabular}\nWhat true stress is necessary to produce a true plastic strain of 0.21 ?",
        "answer": "the true stress necessary to produce a true plastic strain of 0.21 is 66,400 \\, \\text{psi} (460 mpa).",
        "type": 1,
        "wrong_answers_1": "the true stress necessary to produce a true plastic strain of 0.23 is 73692 \\, \\text{psi} (429 mpa).",
        "wrong_answers_2": "the true stress necessary to produce a true plastic strain of 0.18 is 69312 \\, \\text{psi} (410 mpa).",
        "wrong_answers_3": "the true stress necessary to produce a true plastic strain of 0.22 is 60635 \\, \\text{psi} (429 mpa)."
    },
    {
        "idx": 1186,
        "question": "A circular specimen of MgO is loaded using a three-point bending mode. Compute the minimum possible radius of the specimen without fracture, given that the applied load is 5560N (1250 lb), the flexural strength is 105 MPa(15,000 psi), and the separation between load points is 45mm (1.75 in.).",
        "answer": "the minimum allowable specimen radius is 9.1mm (0.36 in).",
        "type": 1,
        "wrong_answers_1": "the minimum allowable specimen radius is 8.1mm (0.38 in).",
        "wrong_answers_2": "the minimum allowable specimen radius is 10.2mm (0.33 in).",
        "wrong_answers_3": "the minimum allowable specimen radius is 8.6mm (0.34 in)."
    },
    {
        "idx": 1187,
        "question": "A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.0 mm(0.20 in.); the specimen fractured at a load of 3000N (675 lb f) when the distance between the support points was 40mm (1.6 in.). Another test is to be performed on a specimen of this same material, but one that has a square cross section of 15mm (0.6 in.) length on each edge. At what load would you expect this specimen to fracture if the support point separation is maintained at 40mm (1.6 in.)?",
        "answer": "the specimen with a square cross section would be expected to fracture at a load of 17,200 n (3870 lb f).",
        "type": 1,
        "wrong_answers_1": "the specimen with a square cross section would be expected to fracture at a load of 15429 n (4060 lb f).",
        "wrong_answers_2": "the specimen with a square cross section would be expected to fracture at a load of 16513 n (3490 lb f).",
        "wrong_answers_3": "the specimen with a square cross section would be expected to fracture at a load of 19029 n (3430 lb f)."
    },
    {
        "idx": 1188,
        "question": "A three-point transverse bending test is conducted on a cylindrical specimen of aluminum oxide having a reported flexural strength of 300 MPa (43,500 psi). If the specimen radius is 5.0mm (0.20 in.) and the support point separation distance is 15.0mm (0.61 in.), would you expect the specimen to fracture when a load of 7500 N (1690 lbf) is applied? Justify your answer.",
        "answer": "the flexural stress is 286.5 mpa. since this value is less than the flexural strength of 300 mpa, fracture is not predicted.",
        "type": 1,
        "wrong_answers_1": "the flexural stress is 303.1 mpa. since this value is less than the flexural strength of 327 mpa, fracture is not predicted.",
        "wrong_answers_2": "the flexural stress is 272.5 mpa. since this value is less than the flexural strength of 267 mpa, fracture is not predicted.",
        "wrong_answers_3": "the flexural stress is 275.0 mpa. since this value is less than the flexural strength of 329 mpa, fracture is not predicted."
    },
    {
        "idx": 1189,
        "question": "Would you be 100% certain of the answer in part (a)? Why or why not?",
        "answer": "the certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials, and the calculated stress is relatively close to the flexural strength, so there is some chance that fracture will occur.",
        "type": 4,
        "wrong_answers_1": "the certainty of this prediction is not 100% because there is always some variability in the flexural strength for ceramic materials. since the calculated stress 379 mpa is relatively close to the flexural strength 390 mpa, there is some chance that fracture will occur.",
        "wrong_answers_2": "the applied stress is 61.5 MPa, which is less than the flexural strength of 85 MPa; the polymer is not expected to fracture.",
        "wrong_answers_3": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material."
    },
    {
        "idx": 1190,
        "question": "Compute the modulus of elasticity for the nonporous material given that the modulus of elasticity for spinel (MgAl2O4) having 5 vol% porosity is 240 GPa (35 x 10^6 psi).",
        "answer": "the modulus of elasticity for the nonporous material is 265 gpa (38.6 x 10^6 psi).",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity for the nonporous material is 278 gpa (34.5 x 10^6 psi).",
        "wrong_answers_2": "the modulus of elasticity for the nonporous material is 250 gpa (41.0 x 10^6 psi).",
        "wrong_answers_3": "the modulus of elasticity for the nonporous material is 227 gpa (41.7 x 10^6 psi)."
    },
    {
        "idx": 1191,
        "question": "Compute the modulus of elasticity for 15 vol% porosity given that the modulus of elasticity for spinel (MgAl2O4) having 5 vol% porosity is 240 GPa (35 x 10^6 psi).",
        "answer": "the modulus of elasticity for 15 vol% porosity is 195 gpa (28.4 x 10^6 psi).",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity for 17 vol% porosity is 176 gpa (31.3 x 10^6 psi).",
        "wrong_answers_2": "the modulus of elasticity for 17 vol% porosity is 211 gpa (31.9 x 10^6 psi).",
        "wrong_answers_3": "the modulus of elasticity for 16 vol% porosity is 203 gpa (30.1 x 10^6 psi)."
    },
    {
        "idx": 1192,
        "question": "Compute the modulus of elasticity for the nonporous titanium carbide (TiC) given that the modulus of elasticity for titanium carbide (TiC) having 5 vol% porosity is 310 GPa (45 x 10^6 psi).",
        "answer": "the modulus of elasticity for the nonporous material is 342 gpa (49.6 x 10^6 psi).",
        "type": 1,
        "wrong_answers_1": "the modulus of elasticity for the nonporous material is 305 gpa (43.8 x 10^6 psi).",
        "wrong_answers_2": "the modulus of elasticity for the nonporous material is 358 gpa (54.5 x 10^6 psi).",
        "wrong_answers_3": "the modulus of elasticity for the nonporous material is 314 gpa (54.4 x 10^6 psi)."
    },
    {
        "idx": 1193,
        "question": "At what volume percent porosity will the modulus of elasticity for titanium carbide (TiC) be 240 GPa (35 x 10^6 psi)?",
        "answer": "the volume percent porosity at which the modulus of elasticity is 240 gpa (35 x 10^6 psi) is 17.1 vol%.",
        "type": 1,
        "wrong_answers_1": "the volume percent porosity at which the modulus of elasticity is 250 gpa (32 x 10^6 psi) is 14.6 vol%.",
        "wrong_answers_2": "the volume percent porosity at which the modulus of elasticity is 222 gpa (33 x 10^6 psi) is 17.9 vol%.",
        "wrong_answers_3": "the volume percent porosity at which the modulus of elasticity is 258 gpa (37 x 10^6 psi) is 16.5 vol%."
    },
    {
        "question": "The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows: \begin{tabular}{ll} \\hline (\\sigma_{f_{S}}(\\mathbf{M P a})) & (\boldsymbol{P}) \\ \\hline 70 & 0.10 \\ 60 & 0.15 \\ \\hline \\end{tabular} (a) Compute the flexural strength for a completely nonporous specimen of this material.",
        "answer": "the flexural strength for a completely nonporous specimen of this material is 95.3 mpa.",
        "idx": 1194,
        "type": 1,
        "wrong_answers_1": "the flexural strength for a completely nonporous specimen of this material is 90.8 mpa.",
        "wrong_answers_2": "the flexural strength for a completely nonporous specimen of this material is 102.2 mpa.",
        "wrong_answers_3": "the flexural strength for a completely nonporous specimen of this material is 81.3 mpa."
    },
    {
        "question": "The flexural strength and associated volume fraction porosity for two specimens of the same ceramic material are as follows: \begin{tabular}{ll} \\hline (\\sigma_{f_{S}}(\\mathbf{M P a})) & (\boldsymbol{P}) \\ \\hline 70 & 0.10 \\ 60 & 0.15 \\ \\hline \\end{tabular} (b) Compute the flexural strength for a 0.20 volume fraction porosity.",
        "answer": "the flexural strength for a 0.20 volume fraction porosity is 51.5 mpa.",
        "idx": 1195,
        "type": 1,
        "wrong_answers_1": "the flexural strength for a 0.22 volume fraction porosity is 57.6 mpa.",
        "wrong_answers_2": "the flexural strength for a 0.21 volume fraction porosity is 58.1 mpa.",
        "wrong_answers_3": "the flexural strength for a 0.19 volume fraction porosity is 55.5 mpa."
    },
    {
        "idx": 1196,
        "question": "For some viscoelastic polymers that are subjected to stress relaxation tests, the stress decays with time according to\n\\[\n\\sigma(t)=\\sigma(0) \\exp \\left(-\\frac{t}{\\tau}\\right)\n\\]\nwhere \\sigma(t) and \\sigma(0) represent the time-dependent and initial (i.e., time =0 ) stresses, respectively, and t and \\tau denote elapsed time and the relaxation time, respectively; \\tau is a time-independent constant characteristic of the material. A specimen of a viscoelastic polymer whose stress relaxation obeys Equation was suddenly pulled in tension to a measured strain of 0.5 ; the stress necessary to maintain this constant strain was measured as a function of time. Determine E_{f}(10) for this material if the initial stress level was 3.5 MPa(500 psi), which dropped to 0.5 MPa(70 psi) after 30s.",
        "answer": "e_{f}(10) = 3.66 mpa  (522 \\text{psi})",
        "type": 1,
        "wrong_answers_1": "e_{f}(9) = 3.77 mpa  (478 \\text{psi})",
        "wrong_answers_2": "e_{f}(11) = 3.21 mpa  (444 \\text{psi})",
        "wrong_answers_3": "e_{f}(11) = 3.44 mpa  (565 \\text{psi})"
    },
    {
        "idx": 1197,
        "question": "The following table gives a number of Rockwellg hardness values that were measured on a single steel specimen. Compute average and standard deviation hardness values.\n\\begin{tabular}{rrr}\n47.3 & 48.7 & 47.1 \\\\\n52.1 & 50.0 & 50.4 \\\\\n45.6 & 46.2 & 45.9 \\\\\n49.9 & 48.3 & 46.4 \\\\\n47.6 & 51.1 & 48.5 \\\\\n50.4 & 46.7 & 49.7\n\\end{tabular}",
        "answer": "the average hardness value is 48.4 hrg. the standard deviation of the hardness values is 1.95 hrg.",
        "type": 1,
        "wrong_answers_1": "the average hardness value is 54.4 hrg. the standard deviation of the hardness values is 1.77 hrg.",
        "wrong_answers_2": "the average hardness value is 43.9 hrg. the standard deviation of the hardness values is 1.79 hrg.",
        "wrong_answers_3": "the average hardness value is 52.9 hrg. the standard deviation of the hardness values is 2.23 hrg."
    },
    {
        "idx": 1198,
        "question": "The following table gives a number of yield strength values (in MPa) that were measured on the same aluminum alloy. Compute average and standard deviation yield strength values.\n\\begin{tabular}{rrr}\n274.3 & 277.1 & 263.8 \\\\\n267.5 & 258.6 & 271.2 \\\\\n255.4 & 266.9 & 257.6 \\\\\n270.8 & 260.1 & 264.3 \\\\\n261.7 & 279.4 & 260.5\n\\end{tabular}",
        "answer": "the average yield strength is 265.9 \\text{ mpa}. the standard deviation of the yield strength is 7.34 \\text{ mpa}.",
        "type": 1,
        "wrong_answers_1": "the average yield strength is 301.8 \\text{ mpa}. the standard deviation of the yield strength is 7.72 \\text{ mpa}.",
        "wrong_answers_2": "the average yield strength is 239.6 \\text{ mpa}. the standard deviation of the yield strength is 8.35 \\text{ mpa}.",
        "wrong_answers_3": "the average yield strength is 303.0 \\text{ mpa}. the standard deviation of the yield strength is 6.54 \\text{ mpa}."
    },
    {
        "idx": 1199,
        "question": "What is the first criterion upon which factors of safety are based?",
        "answer": "The first criterion is consequences of failure.",
        "type": 4,
        "wrong_answers_1": "The fourth criterion is economics.",
        "wrong_answers_2": "The second criterion is previous experience.",
        "wrong_answers_3": "Of the alloys listed, the steel alloy is a possible candidate because it meets the criterion regarding diameter reduction. The titanium alloy is not a candidate as it fails this criterion."
    },
    {
        "idx": 1200,
        "question": "What is the second criterion upon which factors of safety are based?",
        "answer": "The second criterion is previous experience.",
        "type": 4,
        "wrong_answers_1": "The fourth criterion is economics.",
        "wrong_answers_2": "The first criterion is consequences of failure.",
        "wrong_answers_3": "Of the alloys listed, the steel alloy is a possible candidate because it meets the criterion regarding diameter reduction. The titanium alloy is not a candidate as it fails this criterion."
    },
    {
        "idx": 1201,
        "question": "What is the third criterion upon which factors of safety are based?",
        "answer": "The third criterion is accuracy of measurement of mechanical forces and/or material properties.",
        "type": 4,
        "wrong_answers_1": "The first criterion is consequences of failure.",
        "wrong_answers_2": "The disadvantages of hot working are: (1) A poor surface finish. (2) A variety of mechanical properties is not possible.",
        "wrong_answers_3": "The fourth criterion is economics."
    },
    {
        "idx": 1202,
        "question": "What is the fourth criterion upon which factors of safety are based?",
        "answer": "The fourth criterion is economics.",
        "type": 4,
        "wrong_answers_1": "The first criterion is consequences of failure.",
        "wrong_answers_2": "The second criterion is previous experience.",
        "wrong_answers_3": "Of the alloys listed, the steel alloy is a possible candidate because it meets the criterion regarding diameter reduction. The titanium alloy is not a candidate as it fails this criterion."
    },
    {
        "idx": 1203,
        "question": "Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60^{\\circ} and 35^{\\circ}, respectively, with the tensile axis. If the critical resolved shear stress is 6.2 MPa(900 psi), will an applied stress of 12 MPa (1750 psi) cause the single crystal to yield? If not, what stress will be necessary?",
        "answer": "the applied stress of 12 MPa will not cause the single crystal to yield. the necessary stress for yielding to occur is 15.1 MPa (2200 psi).",
        "type": 1,
        "wrong_answers_1": "the applied stress of 10 MPa will not cause the single crystal to yield. the necessary stress for yielding to occur is 14.4 MPa (2280 psi).",
        "wrong_answers_2": "the applied stress of 11 MPa will not cause the single crystal to yield. the necessary stress for yielding to occur is 13.1 MPa (1890 psi).",
        "wrong_answers_3": "the applied stress of 11 MPa will not cause the single crystal to yield. the necessary stress for yielding to occur is 13.5 MPa (2460 psi)."
    },
    {
        "idx": 1204,
        "question": "A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of 65 degrees with the tensile axis. Three possible slip directions make angles of 30 degrees, 48 degrees, and 78 degrees with the same tensile axis. Which of these three slip directions is most favored?",
        "answer": "the most favored slip direction is at an angle of 30 degrees with the tensile axis.",
        "type": 2
    },
    {
        "idx": 1205,
        "question": "A single crystal of zinc is oriented for a tensile test such that its slip plane normal makes an angle of 65 degrees with the tensile axis. The most favored slip direction makes an angle of 30 degrees with the same tensile axis. If plastic deformation begins at a tensile stress of 2.5 MPa (355 psi), determine the critical resolved shear stress for zinc.",
        "answer": "the critical resolved shear stress for zinc is 0.91 mpa (130 psi).",
        "type": 1,
        "wrong_answers_1": "the critical resolved shear stress for zinc is 0.84 mpa (148 psi).",
        "wrong_answers_2": "the critical resolved shear stress for zinc is 0.94 mpa (136 psi).",
        "wrong_answers_3": "the critical resolved shear stress for zinc is 0.78 mpa (122 psi)."
    },
    {
        "idx": 1206,
        "question": "The critical resolved shear stress for copper (Cu) is 0.48 MPa(70 psi). Determine the maximum possible yield strength for a single crystal of Cu pulled in tension.",
        "answer": "the maximum possible yield strength for a single crystal of Cu pulled in tension is \\sigma_{y} = 0.96 MPa (140 psi).",
        "type": 1,
        "wrong_answers_1": "the maximum possible yield strength for a single crystal of Cu pulled in tension is \\sigma_{y} = 0.86 MPa (155 psi).",
        "wrong_answers_2": "the maximum possible yield strength for a single crystal of Cu pulled in tension is \\sigma_{y} = 1.09 MPa (146 psi).",
        "wrong_answers_3": "the maximum possible yield strength for a single crystal of Cu pulled in tension is \\sigma_{y} = 0.87 MPa (127 psi)."
    },
    {
        "idx": 1207,
        "question": "The lower yield point for an iron that has an average grain diameter of 1 × 10^{-2}mm is 230 MPa(33,000 psi). At a grain diameter of 6 × 10^{-3} mm, the yield point increases to 275 MPa (40,000 psi). At what grain diameter will the lower yield point be 310 MPa(45,000 psi) ?\n\\underline{\\text {",
        "answer": "the grain diameter at which the lower yield point is 310 MPa is 4.34 × 10^{-3} mm.",
        "type": 1,
        "wrong_answers_1": "the grain diameter at which the lower yield point is 320 MPa is 4.69 × 10^{-3} mm.",
        "wrong_answers_2": "the grain diameter at which the lower yield point is 340 MPa is 4.20 × 10^{-3} mm.",
        "wrong_answers_3": "the grain diameter at which the lower yield point is 322 MPa is 3.71 × 10^{-3} mm."
    },
    {
        "idx": 1208,
        "question": "Two previously undeformed specimens of the same metal are to be plastically deformed by reducing their cross-sectional areas. One has a circular cross section, and the other is rectangular; during deformation the circular cross section is to remain circular, and the rectangular is to remain as rectangular. Their original and deformed dimensions are as follows:\n\\begin{tabular}{lcc}\n\\hline & Circular (diameter, mm ) & Rectangular (mm) \\\\\n\\hline Original dimensions & 18.0 & 20 × 50 \\\\\nDeformed dimensions & 15.9 & 13.7 × 55.1 \\\\\n\\hline\n\\end{tabular}\nWhich of these specimens will be the hardest after plastic deformation, and why?",
        "answer": "the deformed rectangular specimen will be harder since it has the greater% \\text{cw} (24.5% compared to 22.0% for the circular specimen).",
        "type": 2
    },
    {
        "idx": 1209,
        "question": "Experimentally, it has been observed for single crystals of a number of metals that the critical resolved shear stress \\tau_{\\text {CrSS }} is a function of the dislocation density \\rho_{D} as\n\\[\n\\tau_{\\text {crss }}=\\tau_{0}+A \\sqrt{\\rho_{D}}\n\\]\nwhere \\tau_{0} and A are constants. For copper, the critical resolved shear stress is 0.69 MPa(100 psi) at a dislocation density of 10^{4} mm^{-2}. If it is known that the value of \\tau_{0} for copper is 0.069 MPa (10 psi), compute \\tau_{\\text {crss }} at a dislocation density of 10^{6} mm^{-2}.",
        "answer": "the critical resolved shear stress \\tau_{\\text{crss}} at a dislocation density of 10^{6} mm^{-2} is 6.28 MPa (910 psi).",
        "type": 1,
        "wrong_answers_1": "the critical resolved shear stress \\tau_{\\text{crss}} at a dislocation density of 11^{5} mm^{-2} is 5.87 MPa (1006 psi).",
        "wrong_answers_2": "the critical resolved shear stress \\tau_{\\text{crss}} at a dislocation density of 9^{6} mm^{-2} is 6.56 MPa (834 psi).",
        "wrong_answers_3": "the critical resolved shear stress \\tau_{\\text{crss}} at a dislocation density of 11^{6} mm^{-2} is 6.80 MPa (782 psi)."
    },
    {
        "idx": 1210,
        "question": "Briefly cite the differences between the recovery and recrystallization processes.",
        "answer": "For recovery, there is some relief of internal strain energy by dislocation motion; however, there are virtually no changes in either the grain structure or mechanical characteristics. During recrystallization, on the other hand, a new set of strain-free grains forms, and the material becomes softer and more ductile.",
        "type": 4,
        "wrong_answers_1": "During cold-working, the grain structure of the metal has been distorted to accommodate the deformation. Recrystallization produces grains that are equiaxed and smaller than the parent grains.",
        "wrong_answers_2": "The driving force for recrystallization is the difference in internal energy between the strained and unstrained material.",
        "wrong_answers_3": "The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary."
    },
    {
        "idx": 1211,
        "question": "What is the driving force for recrystallization?",
        "answer": "The driving force for recrystallization is the difference in internal energy between the strained and unstrained material.",
        "type": 4,
        "wrong_answers_1": "The driving force for steady-state diffusion is the concentration gradient.",
        "wrong_answers_2": "The driving force for the formation of spheroidite is the net reduction in ferrite-cementite phase boundary area.",
        "wrong_answers_3": "These phenomena occur because right at the phase transition temperature, the driving force is not sufficient to cause the transformation to occur. The driving force is enhanced during superheating or supercooling."
    },
    {
        "idx": 1212,
        "question": "What is the driving force for grain growth?",
        "answer": "The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases.",
        "type": 4,
        "wrong_answers_1": "The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary.",
        "wrong_answers_2": "The driving force for the formation of spheroidite is the net reduction in ferrite-cementite phase boundary area.",
        "wrong_answers_3": "Small-angle grain boundaries are not as effective in interfering with the slip process as are high-angle grain boundaries because there is not as much crystallographic misalignment in the grain boundary region for small-angle, and therefore not as much change in slip direction."
    },
    {
        "idx": 1213,
        "question": "The average grain diameter for a brass material was measured as a function of time at 650 degrees C, which is shown in the following table at two different times: Time (min) 40, Grain Diameter (mm) 5.6 x 10^-2; Time (min) 100, Grain Diameter (mm) 8.0 x 10^-2. What was the original grain diameter?",
        "answer": "the original grain diameter was 0.031 mm.",
        "type": 1,
        "wrong_answers_1": "the original grain diameter was 0.034 mm.",
        "wrong_answers_2": "the original grain diameter was 0.027 mm.",
        "wrong_answers_3": "the original grain diameter was 0.034 mm."
    },
    {
        "idx": 1214,
        "question": "The average grain diameter for a brass material was measured as a function of time at 650 degrees C, which is shown in the following table at two different times: Time (min) 40, Grain Diameter (mm) 5.6 x 10^-2; Time (min) 100, Grain Diameter (mm) 8.0 x 10^-2. What grain diameter would you predict after 200 min at 650 degrees C?",
        "answer": "the predicted grain diameter after 200 min at 650 degrees C is 0.109 mm.",
        "type": 1,
        "wrong_answers_1": "the predicted grain diameter after 184 min at 674 degrees C is 0.105 mm.",
        "wrong_answers_2": "the predicted grain diameter after 215 min at 583 degrees C is 0.116 mm.",
        "wrong_answers_3": "the predicted grain diameter after 216 min at 626 degrees C is 0.115 mm."
    },
    {
        "idx": 1215,
        "question": "An undeformed specimen of some alloy has an average grain diameter of 0.050 mm. You are asked to reduce its average grain diameter to 0.020 mm. Is this possible? If so, explain the procedures you would use and name the processes involved. If it is not possible, explain why.",
        "answer": "Yes, it is possible to reduce the average grain diameter of an undeformed alloy specimen from 0.050mm to 0.020 mm. In order to do this, plastically deform the material at room temperature (i.e., cold work it), and then anneal at an elevated temperature in order to allow recrystallization and some grain growth to occur until the average grain diameter is 0.020 mm.",
        "type": 4,
        "wrong_answers_1": "The driving force for grain growth is the reduction in grain boundary energy as the total grain boundary area decreases.",
        "wrong_answers_2": "The amount of γ2 precipitate at 400°C is 5.4%, and at room temperature, it is 8.5%.",
        "wrong_answers_3": "The surface energy will be greater than the grain boundary energy. For grain boundaries, some atoms on one side of a boundary will bond to atoms on the other side; such is not the case for surface atoms. Therefore, there will be fewer unsatisfied bonds along a grain boundary."
    },
    {
        "idx": 1216,
        "question": "A non-cold-worked brass specimen of average grain size 0.01mm has a yield strength of 150 MPa(21,750 psi). Estimate the yield strength of this alloy after it has been heated to 500^{\\circ} C for 1000s, if it is known that the value of \\sigma_{0} is 25 MPa(3625 psi).",
        "answer": "the yield strength of the brass alloy after heating to 500^{\\circ} C for 1000s is 124 MPa (18,000 psi).",
        "type": 1,
        "wrong_answers_1": "the yield strength of the brass alloy after heating to 452^{\\circ} C for 1110s is 137 MPa (15672 psi).",
        "wrong_answers_2": "the yield strength of the brass alloy after heating to 447^{\\circ} C for 900s is 137 MPa (16847 psi).",
        "wrong_answers_3": "the yield strength of the brass alloy after heating to 556^{\\circ} C for 880s is 106 MPa (16276 psi)."
    },
    {
        "idx": 1217,
        "question": "Normal butane and isobutane have boiling temperatures of -0.5^{\\circ} C and -12.3^{\\circ} C\\left(31.1^{\\circ} F\\right. and 9.9^{\\circ} F ), respectively. Briefly explain this behavior on the basis of their molecular structures.",
        "answer": "Normal butane has a higher melting temperature as a result of its molecular structure . There is more of an opportunity for van der Waals bonds to form between two molecules in close proximity to one another than for isobutane because of the linear nature of each normal butane molecule.",
        "type": 4,
        "wrong_answers_1": "The intermolecular bonding for \\mathrm{HF} is hydrogen, whereas for \\mathrm{HCl}, the intermolecular bonding is van der Waals. Since the hydrogen bond is stronger than van der Waals, HF will have a higher melting temperature.",
        "wrong_answers_2": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will most likely have a higher percent crystallinity, and, therefore, a higher melting temperature than the branched polyethylene. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration.",
        "wrong_answers_3": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The syndiotactic polypropylene will have a higher degree of crystallinity than the atactic material. On the basis of this effect alone, the syndiotactic PP should have the greater Tm, since melting temperature increases with degree of crystallinity. However, the molecular weight for the syndiotactic polypropylene (500,000 g/mol) is less than for the atactic material (750,000 g/mol); a lowering of molecular weight generally results in a reduction of melting temperature."
    },
    {
        "idx": 1218,
        "question": "The tensile strength and number-average molecular weight for two poly(methyl methacrylate) materials are as follows:\n\\begin{tabular}{cc}\n\\hline Tensile Strength (MPa) & Number-Average Molecular Weight (g/mol) \\\\\n\\hline 50 & 30,000 \\\\\n150 & 50,000 \\\\\n\\hline\n\\end{tabular}\nEstimate the tensile strength at a number-average molecular weight of 40,000 g/mol.",
        "answer": "the estimated tensile strength at a number-average molecular weight of 40,000 g/mol is 112.5 mpa.",
        "type": 1,
        "wrong_answers_1": "the estimated tensile strength at a number-average molecular weight of 42885 g/mol is 96.9 mpa.",
        "wrong_answers_2": "the estimated tensile strength at a number-average molecular weight of 38619 g/mol is 126.5 mpa.",
        "wrong_answers_3": "the estimated tensile strength at a number-average molecular weight of 36871 g/mol is 121.4 mpa."
    },
    {
        "idx": 1219,
        "question": "The tensile strength and number-average molecular weight for two polyethylene materials are as follows:\n\\begin{tabular}{cc}\n\\hline Tensile Strength (MPa) & Number-Average Molecular Weight (g/mol) \\\\\n\\hline 90 & 20,000 \\\\\n180 & 40,000 \\\\\n\\hline\n\\end{tabular}\nEstimate the number-average molecular weight that is required to give a tensile strength of 140 MPa.",
        "answer": "the required number-average molecular weight is 27700 g/mol",
        "type": 1,
        "wrong_answers_1": "the required number-average molecular weight is 2970 g/mol",
        "wrong_answers_2": "the required number-average molecular weight is 2970 g/mol",
        "wrong_answers_3": "the required number-average molecular weight is 2580 g/mol"
    },
    {
        "idx": 1220,
        "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Branched and atactic poly(vinyl chloride) with a weight-average molecular weight of 100,000 g/mol; linear and isotactic poly(vinyl chloride) having a weight-average molecular weight of 75,000 g/mol",
        "answer": "Yes, it is possible. The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's crystallinity leads to an increase in its tensile modulus. In addition, tensile modulus is independent of molecular weight - the atactic/branched material has the higher molecular weight.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible. The linear and isotactic polypropylene will have a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. ",
        "wrong_answers_2": "No, it is not possible. Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don't normally crystallize. The atactic polypropylene probably also has a relatively low degree of crystallinity; atactic structures also don't tend to crystallize, and polypropylene has a more complex repeat unit structure than does polyethylene.Tensile modulus increases with degree of crystallinity, and it is not possible to determine which polymer is more crystalline. Furthermore, tensile modulus is independent of molecular weight.",
        "wrong_answers_3": "No, it is not possible. Linear polymers have higher degrees of crystallization (and higher tensile moduli) than branched polymers - on this basis, the PVC material should have the higher value of E. On the other hand, PVC has a more complex repeat unit structure than does polyethylene, which means that, on this basis, the PE would have a higher degree of crystallinity and also a greater tensile modulus. Also, tensile modulus is relatively independent of number-average molecular weight. Therefore, this determination is not possible since it is not possible to determine which of the two materials has the greater degree of crystallinity."
    },
    {
        "idx": 1221,
        "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Random styrene-butadiene copolymer with 5% of possible sites crosslinked; block styrene-butadiene copolymer with 10% of possible sites crosslinked",
        "answer": "Yes, it is possible. The block styrene-butadiene copolymer with 10% of possible sites crosslinked will have the higher modulus. Block copolymers normally have higher degrees of crystallinity than random copolymers of the same material. A higher degree of crystallinity favors larger moduli. In addition, the block copolymer also has a higher degree of crosslinking; increasing the amount of crosslinking also enhances the tensile modulus.",
        "type": 4,
        "wrong_answers_1": "No, it is not possible. The random acrylonitrile-butadiene copolymer will tend to a lower degree of crystallinity than the alternating acrylonitrile-butadiene copolymer inasmuch as random copolymers don't normally crystallize. On this basis only, the alternating material would have a higher modulus inasmuch as tensile modulus increases with degree of crystallinity. On the other hand, the random copolymer has a higher degree of crosslinking (10% versus 5% for the alternating copolymer), and, on this basis only would have the higher tensile modulus-an increase in crosslinking leads to an increase in E. Thus, this determination is not possible; with regard to degree of crystallinity the alternating material has the higher E, whereas the random copolymer would have a higher E value on the basis of degree of crosslinking.",
        "wrong_answers_2": "No it is not possible. The random acrylonitrile-butadiene copolymer has more crosslinking; increased crosslinking leads to an increase in strength. However, the block copolymeric material will most likely have a higher degree of crystallinity; and increasing crystallinity improves the strength.",
        "wrong_answers_3": "No, it is not possible. Linear polymers have higher degrees of crystallization (and higher tensile moduli) than branched polymers - on this basis, the PVC material should have the higher value of E. On the other hand, PVC has a more complex repeat unit structure than does polyethylene, which means that, on this basis, the PE would have a higher degree of crystallinity and also a greater tensile modulus. Also, tensile modulus is relatively independent of number-average molecular weight. Therefore, this determination is not possible since it is not possible to determine which of the two materials has the greater degree of crystallinity."
    },
    {
        "idx": 1222,
        "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile modulus than the other; (2) if this is possible, note which has the higher tensile modulus and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Branched polyethylene with a number-average molecular weight of 100,000 g/mol; atactic polypropylene with a number-average molecular weight of 150,000 g/mol",
        "answer": "No, it is not possible. Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don't normally crystallize. The atactic polypropylene probably also has a relatively low degree of crystallinity; atactic structures also don't tend to crystallize, and polypropylene has a more complex repeat unit structure than does polyethylene.Tensile modulus increases with degree of crystallinity, and it is not possible to determine which polymer is more crystalline. Furthermore, tensile modulus is independent of molecular weight.",
        "type": 4,
        "wrong_answers_1": "No, it is not possible. Linear polymers have higher degrees of crystallization (and higher tensile moduli) than branched polymers - on this basis, the PVC material should have the higher value of E. On the other hand, PVC has a more complex repeat unit structure than does polyethylene, which means that, on this basis, the PE would have a higher degree of crystallinity and also a greater tensile modulus. Also, tensile modulus is relatively independent of number-average molecular weight. Therefore, this determination is not possible since it is not possible to determine which of the two materials has the greater degree of crystallinity.",
        "wrong_answers_2": "Yes, it is possible. The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's crystallinity leads to an increase in its tensile modulus. In addition, tensile modulus is independent of molecular weight - the atactic/branched material has the higher molecular weight.",
        "wrong_answers_3": "Yes, it is possible to decide for these two polymers. The linear polyethylene is more likely to crystallize. The repeat unit structure for polypropylene is chemically more complicated than is the repeat unit structure for polyethylene. Furthermore, branched structures are less likely to crystallize than are linear structures."
    },
    {
        "idx": 1223,
        "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Linear and isotactic poly(vinyl chloride) with a weight-average molecular weight of 100,000 g/mol; branched and atactic poly(vinyl chloride) having a weight-average molecular weight of 75,000 g/mol",
        "answer": "Yes, it is possible. The linear and isotactic material will have the higher tensile strength. Both linearity and isotacticity favor a higher degree of crystallinity than do branching and atacticity; and tensile strength increases with increasing degree of crystallinity. Furthermore, the molecular weight of the linear/isotactic material is higher (100,000 g/mol versus 75,000 g/mol), and tensile strength increases with increasing molecular weight.",
        "type": 4,
        "wrong_answers_1": "Yes it is possible. The syndiotactic polystyrene has the higher tensile strength. Syndiotactic polymers are more likely to crystallize than atactic ones; the greater the crystallinity, the higher the tensile strength. Furthermore, the syndiotactic also has a higher molecular weight; increasing molecular weight also enhances the strength.",
        "wrong_answers_2": "Yes, it is possible. The linear and isotactic poly(vinyl chloride) will display a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. In addition, polymers having isotactic structures will normally have a higher degree of crystallinity that those having atactic structures. Increasing a polymer's crystallinity leads to an increase in its tensile modulus. In addition, tensile modulus is independent of molecular weight - the atactic/branched material has the higher molecular weight.",
        "wrong_answers_3": "No, it is not possible. Alternating copolymers tend to be more crystalline than graft copolymers, and tensile strength increases with degree of crystallinity. However, the graft material has a higher degree of crosslinking, and tensile strength increases with the percentage of crosslinks."
    },
    {
        "idx": 1224,
        "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Graft acrylonitrile-butadiene copolymer with 10% of possible sites crosslinked; alternating acrylonitrile-butadiene copolymer with 5% of possible sites crosslinked",
        "answer": "No, it is not possible. Alternating copolymers tend to be more crystalline than graft copolymers, and tensile strength increases with degree of crystallinity. However, the graft material has a higher degree of crosslinking, and tensile strength increases with the percentage of crosslinks.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible. The linear and isotactic material will have the higher tensile strength. Both linearity and isotacticity favor a higher degree of crystallinity than do branching and atacticity; and tensile strength increases with increasing degree of crystallinity. Furthermore, the molecular weight of the linear/isotactic material is higher (100,000 g/mol versus 75,000 g/mol), and tensile strength increases with increasing molecular weight.",
        "wrong_answers_2": "No, it is not possible. The random acrylonitrile-butadiene copolymer will tend to a lower degree of crystallinity than the alternating acrylonitrile-butadiene copolymer inasmuch as random copolymers don't normally crystallize. On this basis only, the alternating material would have a higher modulus inasmuch as tensile modulus increases with degree of crystallinity. On the other hand, the random copolymer has a higher degree of crosslinking (10% versus 5% for the alternating copolymer), and, on this basis only would have the higher tensile modulus-an increase in crosslinking leads to an increase in E. Thus, this determination is not possible; with regard to degree of crystallinity the alternating material has the higher E, whereas the random copolymer would have a higher E value on the basis of degree of crosslinking.",
        "wrong_answers_3": "No, it is not possible. Branched polyethylene will tend to have a low degree of crystallinity since branched polymers don't normally crystallize. The atactic polypropylene probably also has a relatively low degree of crystallinity; atactic structures also don't tend to crystallize, and polypropylene has a more complex repeat unit structure than does polyethylene.Tensile modulus increases with degree of crystallinity, and it is not possible to determine which polymer is more crystalline. Furthermore, tensile modulus is independent of molecular weight."
    },
    {
        "idx": 1225,
        "question": "For the following pair of polymers, do the following: (1) state whether it is possible to decide whether one polymer has a higher tensile strength than the other; (2) if this is possible, note which has the higher tensile strength and cite the reason(s) for your choice; and (3) if it is not possible to decide, state why. Network polyester; lightly branched polytetrafluoroethylene",
        "answer": "Yes, it is possible. The network polyester will display a greater tensile strength. Relative chain motion is much more restricted than for the lightly branched polytetrafluoroethylene since there are many more of the strong covalent bonds for the network structure.",
        "type": 4,
        "wrong_answers_1": "Yes it is possible. The network polyester will display a greater tensile strength. Relative chain motion is much more restricted than for the lightly branched polypropylene since there are many more of the strong covalent bonds for the network structure.",
        "wrong_answers_2": "Yes, it is possible. The linear and isotactic polypropylene will have a greater tensile modulus. Linear polymers are more likely to crystallize that branched ones. ",
        "wrong_answers_3": "Yes, it is possible to decide for these two polymers. The linear and isotactic polystyrene is more likely to crystallize than network phenol-formaldehyde; network polymers rarely crystallize, whereas isotactic ones crystallize relatively easily."
    },
    {
        "idx": 1226,
        "question": "What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9 × 10^{-4} mm\\left(7.5 × 10^{-6} in.\\right) and a crack length of 3.8 × 10^{-2} mm\\left(1.5 × 10^{-3}\\right. in.) when a tensile stress of 140 MPa(20,000 psi) is applied?",
        "answer": "the magnitude of the maximum stress is 2800 \\text{ mpa} (400,000 psi).",
        "type": 1,
        "wrong_answers_1": "the magnitude of the maximum stress is 2550 \\text{ mpa} (427461 psi).",
        "wrong_answers_2": "the magnitude of the maximum stress is 2460 \\text{ mpa} (351459 psi).",
        "wrong_answers_3": "the magnitude of the maximum stress is 3070 \\text{ mpa} (351866 psi)."
    },
    {
        "idx": 1227,
        "question": "Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an elliptically shaped surface crack of length 0.5 mm(0.02 in.) and a tip radius of curvature of 5 × 10^{-3} mm\\left(2 × 10^{-4} in\\right), when a stress of 1035 MPa(150,000 psi) is applied.",
        "answer": "the theoretical fracture strength of the material is 20.7 gpa (3 × 10^{6} psi).",
        "type": 1,
        "wrong_answers_1": "the theoretical fracture strength of the material is 18.4 gpa (3 × 10^{6} psi).",
        "wrong_answers_2": "the theoretical fracture strength of the material is 23.7 gpa (3 × 10^{6} psi).",
        "wrong_answers_3": "the theoretical fracture strength of the material is 18.7 gpa (3 × 10^{6} psi)."
    },
    {
        "idx": 1228,
        "question": "A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa \\sqrt{m} (50 ksi \\sqrt{m}.) is exposed to a stress of 1030 MPa(150,000 psi). Will this specimen experience fracture if the largest surface crack is 0.5mm ( 0.02 in.) long? Why or why not? Assume that the parameter Y has a value of 1.0 .",
        "answer": "the specimen will not experience fracture because it can tolerate a critical stress of 1380 \\text{ mpa} before fracture, which is greater than the applied stress of 1030 \\text{ mpa}.",
        "type": 1,
        "wrong_answers_1": "the specimen will not experience fracture because it can tolerate a critical stress of 1250 \\text{ mpa} before fracture, which is greater than the applied stress of 1140 \\text{ mpa}.",
        "wrong_answers_2": "the specimen will not experience fracture because it can tolerate a critical stress of 1220 \\text{ mpa} before fracture, which is greater than the applied stress of 1090 \\text{ mpa}.",
        "wrong_answers_3": "the specimen will not experience fracture because it can tolerate a critical stress of 1460 \\text{ mpa} before fracture, which is greater than the applied stress of 960 \\text{ mpa}."
    },
    {
        "idx": 1229,
        "question": "An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 MPa sqrt(m) (36.4 ksi sqrt(in)). It has been determined that fracture results at a stress of 300 MPa (43,500 psi) when the maximum (or critical) internal crack length is 4.0mm (0.16 in.). For this same component and alloy, will fracture occur at a stress level of 260 MPa (38,000 psi) when the maximum internal crack length is 6.0mm (0.24 in.)? Why or why not?",
        "answer": "fracture will occur since the value 42.4 MPa sqrt(m) is greater than the fracture toughness K_IC of the material, 40 MPa sqrt(m).",
        "type": 1,
        "wrong_answers_1": "fracture will occur since the value 44.7 MPa sqrt(m) is greater than the fracture toughness K_IC of the material, 37 MPa sqrt(m).",
        "wrong_answers_2": "fracture will occur since the value 37.0 MPa sqrt(m) is greater than the fracture toughness K_IC of the material, 43 MPa sqrt(m).",
        "wrong_answers_3": "fracture will occur since the value 45.7 MPa sqrt(m) is greater than the fracture toughness K_IC of the material, 42 MPa sqrt(m)."
    },
    {
        "idx": 1230,
        "question": "Suppose that a wing component on an aircraft is fabricated from an aluminum alloy that has a plane-strain fracture toughness of 26.0 MPa \\sqrt{m}(23.7 ksi \\sqrt{m}.). It has been determined that fracture results at a stress of 112 MPa(16,240 psi) when the maximum internal crack length is 8.6 mm(0.34 in.). For this same component and alloy, compute the stress level at which fracture will occur for a critical internal crack length of 6.0 mm(0.24 in.).",
        "answer": "the stress level at which fracture will occur for a critical internal crack length of 6.0mm is 134 MPa (19,300 psi).",
        "type": 1,
        "wrong_answers_1": "the stress level at which fracture will occur for a critical internal crack length of 6.6mm is 129 MPa (17431 psi).",
        "wrong_answers_2": "the stress level at which fracture will occur for a critical internal crack length of 5.2mm is 146 MPa (17232 psi).",
        "wrong_answers_3": "the stress level at which fracture will occur for a critical internal crack length of 5.4mm is 146 MPa (20416 psi)."
    },
    {
        "idx": 1231,
        "question": "A structural component is fabricated from an alloy that has a plane-strain fracture toughness of 62 MPa \\sqrt{m}. It has been determined that this component fails at a stress of 250 MPa when the maximum length of a surface crack is 1.6 mm. What is the maximum allowable surface crack length (in mm ) without fracture for this same component exposed to a stress of 250 MPa and made from another alloy that has a plane strain fracture toughness of 51 MPa \\sqrt{m} ?",
        "answer": "the maximum allowable surface crack length without fracture is 1.08 mm.",
        "type": 1,
        "wrong_answers_1": "the maximum allowable surface crack length without fracture is 1.14 mm.",
        "wrong_answers_2": "the maximum allowable surface crack length without fracture is 1.15 mm.",
        "wrong_answers_3": "the maximum allowable surface crack length without fracture is 1.01 mm."
    },
    {
        "idx": 1232,
        "question": "A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 MPa \\sqrt{m}(75.0 ksi \\sqrt{m}.). If the plate is exposed to a tensile stress of 345 MPa(50,000 psi) during service use, determine the minimum length of a surface crack that will lead to fracture. Assume a value of 1.0 for Y.",
        "answer": "the minimum length of a surface crack that will lead to fracture is 0.0182 \\text{ m} (18.2 mm).",
        "type": 1,
        "wrong_answers_1": "the minimum length of a surface crack that will lead to fracture is 0.0170 \\text{ m} (15.6 mm).",
        "wrong_answers_2": "the minimum length of a surface crack that will lead to fracture is 0.0163 \\text{ m} (19.7 mm).",
        "wrong_answers_3": "the minimum length of a surface crack that will lead to fracture is 0.0171 \\text{ m} (16.4 mm)."
    },
    {
        "idx": 1233,
        "question": "A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane-strain fracture toughness of 98.9 MPa \\sqrt{m} ( 90 ksi \\sqrt{in}.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm(0.12 in.). If the design stress is one-half the yield strength and the value of Y is 1.0 , determine whether a critical flaw for this plate is subject to detection.",
        "answer": "the critical flaw size a_c is 16.8 mm. therefore, the critical flaw is subject to detection since it exceeds the 3.0 mm resolution limit.",
        "type": 1,
        "wrong_answers_1": "the critical flaw size a_c is 17.4 mm. therefore, the critical flaw is subject to detection since it exceeds the 2.8 mm resolution limit.",
        "wrong_answers_2": "the critical flaw size a_c is 14.5 mm. therefore, the critical flaw is subject to detection since it exceeds the 3.3 mm resolution limit.",
        "wrong_answers_3": "the critical flaw size a_c is 16.3 mm. therefore, the critical flaw is subject to detection since it exceeds the 2.6 mm resolution limit."
    },
    {
        "idx": 1234,
        "question": "Why may there be significant scatter in the fracture strength for some given ceramic material?",
        "answer": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material.",
        "type": 4,
        "wrong_answers_1": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material.",
        "wrong_answers_2": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probability of the existence of a flaw that is capable of initiating a crack diminishes.",
        "wrong_answers_3": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probably of the existence of a flaw of that is capable of initiating a crack diminishes."
    },
    {
        "idx": 1235,
        "question": "Why does the fracture strength increase with decreasing specimen size?",
        "answer": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probably of the existence of a flaw of that is capable of initiating a crack diminishes.",
        "type": 4,
        "wrong_answers_1": "The fracture strength increases with decreasing specimen size because as specimen size decreases, the probability of the existence of a flaw that is capable of initiating a crack diminishes.",
        "wrong_answers_2": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material.",
        "wrong_answers_3": "There may be significant scatter in the fracture strength for some given ceramic material because the fracture strength depends on the probability of the existence of a flaw that is capable of initiating a crack; this probability varies from specimen to specimen of the same material."
    },
    {
        "idx": 1236,
        "question": "The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter surface crack geometry (i.e., reduce the crack length and increase the tip radius). Compute the ratio of the etched and original crack-tip radii for a fourfold increase in fracture strength if half of the crack length is removed.",
        "answer": "the ratio of the etched and original crack-tip radii is 8.",
        "type": 1,
        "wrong_answers_1": "the ratio of the etched and original crack-tip radii is 7.",
        "wrong_answers_2": "the ratio of the etched and original crack-tip radii is 8.",
        "wrong_answers_3": "the ratio of the etched and original crack-tip radii is 8."
    },
    {
        "idx": 1237,
        "question": "What is the maximum carbon content possible for a plain carbon steel that must have an impact energy of at least 200 Jat -50^{\\circ} C ?",
        "answer": "the maximum carbon concentration is 0.11% C.",
        "type": 1,
        "wrong_answers_1": "the maximum carbon concentration is 0.12% C.",
        "wrong_answers_2": "the maximum carbon concentration is 0.11% C.",
        "wrong_answers_3": "the maximum carbon concentration is 0.10% C."
    },
    {
        "idx": 1238,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for tin.",
        "answer": "for sn, -71°c or -96°f.",
        "type": 1,
        "wrong_answers_1": "for sn, -63°c or -93°f.",
        "wrong_answers_2": "for sn, -80°c or -101°f.",
        "wrong_answers_3": "for sn, -62°c or -91°f."
    },
    {
        "idx": 1239,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for molybdenum.",
        "answer": "for mo, 776°c or 1429°f.",
        "type": 1,
        "wrong_answers_1": "for mo, 737°c or 1258°f.",
        "wrong_answers_2": "for mo, 717°c or 1509°f.",
        "wrong_answers_3": "for mo, 710°c or 1528°f."
    },
    {
        "idx": 1240,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for iron.",
        "answer": "for fe, 451°c or 845°f.",
        "type": 4,
        "wrong_answers_1": "for fe: 450 degrees c (845 degrees f)",
        "wrong_answers_2": "for zn, 4°c or 39°f.",
        "wrong_answers_3": "In an HCl solution containing dissolved oxygen and Fe2+ ions, possible reactions are Mg -> Mg2+ + 2 e- (oxidation), 4 H+ + O2 + 4 e- -> 2 H2O (reduction), and Fe2+ + 2 e- -> Fe (reduction)."
    },
    {
        "idx": 1241,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for gold.",
        "answer": "for au, 262°c or 504°f.",
        "type": 4,
        "wrong_answers_1": "for zn, 4°c or 39°f.",
        "wrong_answers_2": "for fe, 451°c or 845°f.",
        "wrong_answers_3": "for cr, 586°c or 1087°f."
    },
    {
        "idx": 1242,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for zinc.",
        "answer": "for zn, 4°c or 39°f.",
        "type": 4,
        "wrong_answers_1": "for fe, 451°c or 845°f.",
        "wrong_answers_2": "for cr, 586°c or 1087°f.",
        "wrong_answers_3": "for au, 262°c or 504°f."
    },
    {
        "idx": 1243,
        "question": "Give the approximate temperature at which creep deformation becomes an important consideration for chromium.",
        "answer": "for cr, 586°c or 1087°f.",
        "type": 4,
        "wrong_answers_1": "The alloying elements in tool steels (e.g., \\mathrm{Cr}, V, \\mathrm{W}, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds.",
        "wrong_answers_2": "for zn, 4°c or 39°f.",
        "wrong_answers_3": "for fe, 451°c or 845°f."
    },
    {
        "idx": 1244,
        "question": "A cylindrical specimen 13.2mm in diameter of an S-590 alloy is to be exposed to a tensile load of 27,000N. At approximately what temperature will the steady-state creep be 10^{-3} h^{-1} ?",
        "answer": "the temperature would be approximately 775^\\circ c.",
        "type": 1,
        "wrong_answers_1": "the temperature would be approximately 739^\\circ c.",
        "wrong_answers_2": "the temperature would be approximately 670^\\circ c.",
        "wrong_answers_3": "the temperature would be approximately 876^\\circ c."
    },
    {
        "idx": 1245,
        "question": "Steady-state creep rate data are given in the following table for a nickel alloy at 538^{\\circ} C \\\\ (811 K):\n\\begin{tabular}{lc}\n\\hline \\dot{\\varepsilon}_{s}\\left(h^{-1}\\right) & \\sigma(M P(a) \\\\\n\\hline 10^{-7} & 22.0 \\\\\n10^{-6} & 36.1 \\\\\n\\hline\n\\end{tabular}\nCompute the stress at which the steady-state creep is 10^{-5}h^{-1} (also at 538^{\\circ} C ).",
        "answer": "the stress at which the steady-state creep is 10^{-5}h^{-1} is 59.4 MPa.",
        "type": 1,
        "wrong_answers_1": "the stress at which the steady-state creep is 9^{-5}h^{-1} is 54.9 MPa.",
        "wrong_answers_2": "the stress at which the steady-state creep is 10^{-5}h^{-1} is 68.0 MPa.",
        "wrong_answers_3": "the stress at which the steady-state creep is 11^{-6}h^{-1} is 56.4 MPa."
    },
    {
        "idx": 1246,
        "question": "Steady-state creep rate data are given in the following table for some alloy taken at 200^{\\circ} C(473 K) :\n\\begin{tabular}{ll}\n\\hlines\\left(h^{-1}\\right) & \\sigma[MPa(psi)] \\\\\n\\hline 2.5 × 10^{-3} & 55(8000) \\\\\n2.4 × 10^{-2} & 69(10,000) \\\\\n\\hline\n\\end{tabular}\nIf it is known that the activation energy for creep is 140,000 J/ mol, compute the steady-state creep rate at a temperature of 250^{\\circ} C(523 K) and a stress level of 48 MPa(7000 psi).",
        "answer": "the steady-state creep rate at 250^{\\circ} C and 48 MPa is 1.94 × 10^{-2} h^{-1}.",
        "type": 1,
        "wrong_answers_1": "the steady-state creep rate at 215^{\\circ} C and 49 MPa is 2.20 × 10^{-2} h^{-1}.",
        "wrong_answers_2": "the steady-state creep rate at 259^{\\circ} C and 43 MPa is 1.67 × 10^{-2} h^{-1}.",
        "wrong_answers_3": "the steady-state creep rate at 282^{\\circ} C and 44 MPa is 1.82 × 10^{-2} h^{-1}."
    },
    {
        "idx": 1247,
        "question": "Steady-state creep data taken for an iron at a stress level of 140 MPa(20,000 psi) are given here:\n\\begin{tabular}{cc}\n\\hline\\dot{\\epsilon}_{s}\\left(h^{-1}\\right) & T(K) \\\\\n\\hline 6.6 × 10^{-4} & 1090 \\\\\n8.8 × 10^{-2} & 1200 \\\\\n\\hline\n\\end{tabular}\nIf it is known that the value of the stress exponent n for this alloy is 8.5 , compute the steady-state creep rate at 1300 K and a stress level of 83 MPa(12,000 psi).",
        "answer": "the steady-state creep rate at 1300k and a stress level of 83 MPa is 4.31 × 10^{-2}h^{-1}.",
        "type": 1,
        "wrong_answers_1": "the steady-state creep rate at 1170k and a stress level of 71 MPa is 3.70 × 10^{-2}h^{-1}.",
        "wrong_answers_2": "the steady-state creep rate at 1400k and a stress level of 89 MPa is 3.74 × 10^{-2}h^{-1}.",
        "wrong_answers_3": "the steady-state creep rate at 1480k and a stress level of 80 MPa is 4.67 × 10^{-2}h^{-1}."
    },
    {
        "idx": 1248,
        "question": "If ice homogeneously nucleates at -40^{\\circ} C, calculate the critical radius given values of - 3.1 × 10^{8} J/ m^{3} and 25 × 10^{-3} J/ m^{2}, respectively, for the latent heat of fusion and the surface free energy.",
        "answer": "the critical radius is 1.10 nm.",
        "type": 1,
        "wrong_answers_1": "the critical radius is 1.25 nm.",
        "wrong_answers_2": "the critical radius is 1.05 nm.",
        "wrong_answers_3": "the critical radius is 0.97 nm."
    },
    {
        "idx": 1249,
        "question": "It is known that the kinetics of recrystallization for some alloy obeys the Avrami equation, and that the value of n in the exponential is 5.0. If, at some temperature, the fraction recrystallized is 0.30 after 100 min, determine the rate of recrystallization at this temperature.",
        "answer": "the rate of recrystallization at this temperature is 8.76 × 10^{-3}, \\text{min}^{-1}.",
        "type": 1,
        "wrong_answers_1": "the rate of recrystallization at this temperature is 9.68 × 10^{-3}, \\text{min}^{-1}.",
        "wrong_answers_2": "the rate of recrystallization at this temperature is 9.48 × 10^{-3}, \\text{min}^{-1}.",
        "wrong_answers_3": "the rate of recrystallization at this temperature is 9.57 × 10^{-3}, \\text{min}^{-1}."
    },
    {
        "idx": 1250,
        "question": "The kinetics of the austenite-to-pearlite transformation obeys the Avrami relationship. Using the fraction transformed-time data given here, determine the total time required for 95% of the austenite to transform to pearlite:\n\\begin{tabular}{cc}\n\\hline Fraction Transformed & Time (s) \\\\\n\\hline 0.2 & 280 \\\\\n0.6 & 425 \\\\\n\\hline\n\\end{tabular}",
        "answer": "the total time required for 95% of the austenite to transform to pearlite is 603s.",
        "type": 1,
        "wrong_answers_1": "the total time required for 82% of the austenite to transform to pearlite is 642s.",
        "wrong_answers_2": "the total time required for 98% of the austenite to transform to pearlite is 664s.",
        "wrong_answers_3": "the total time required for 104% of the austenite to transform to pearlite is 625s."
    },
    {
        "idx": 1251,
        "question": "The fraction recrystallized-time data for the recrystallization at 350^{\\circ} C of a previously deformed aluminum are tabulated here. Assuming that the kinetics of this process obey the Avrami relationship, determine the fraction recrystallized after a total time of 116.8min.\n\\begin{tabular}{cc}\n\\hline Fraction Recrystallized & Time (min) \\\\\n\\hline 0.30 & 95.2 \\\\\n0.80 & 126.6 \\\\\n\\hline\n\\end{tabular}",
        "answer": "the fraction recrystallized after a total time of 116.8min is 0.65.",
        "type": 1,
        "wrong_answers_1": "the fraction recrystallized after a total time of 125.1min is 0.60.",
        "wrong_answers_2": "the fraction recrystallized after a total time of 104.1min is 0.58.",
        "wrong_answers_3": "the fraction recrystallized after a total time of 130.8min is 0.56."
    },
    {
        "idx": 1252,
        "question": "Briefly cite the differences among pearlite, bainite, and spheroidite relative to microstructure.",
        "answer": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers, which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded by an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.",
        "type": 4,
        "wrong_answers_1": "The microstructures of pearlite, bainite, and spheroidite all consist of α-ferrite and cementite phases. For pearlite, the two phases exist as layers which alternate with one another. Bainite consists of very fine and parallel needle-shaped particles of cementite that are surrounded an α-ferrite matrix. For spheroidite, the matrix is ferrite, and the cementite phase is in the shape of sphere-shaped particles.",
        "wrong_answers_2": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_3": "The diagram provides no indication as to the time-temperature relationships for the formation of pearlite, bainite, and spheroidite, all of which are composed of the equilibrium ferrite and cementite phases."
    },
    {
        "idx": 1253,
        "question": "Briefly cite the differences among pearlite, bainite, and spheroidite relative to mechanical properties.",
        "answer": "Bainite is harder and stronger than pearlite, which, in turn, is harder and stronger than spheroidite.",
        "type": 4,
        "wrong_answers_1": "Bainite is harder and stronger than pearlite, which, in turn, is harder and stronger than spheroidite.",
        "wrong_answers_2": "The hardness and strength of iron-carbon alloys that have microstructures consisting of \\alpha-ferrite and cementite phases depend on the boundary area between the two phases. The greater this area, the harder and stronger the alloy inasmuch as (1) these boundaries impede the motion of dislocations, and (2) the cementite phase restricts the deformation of the ferrite phase in regions adjacent to the phase boundaries. Fine pearlite is harder and stronger than coarse pearlite because the alternating ferrite-cementite layers are thinner for fine, and therefore, there is more phase boundary area. The phase boundary area between the sphere-like cementite particles and the ferrite matrix is less in spheroidite than for the alternating layered microstructure found in coarse pearlite.",
        "wrong_answers_3": "Tempered martensite is harder and stronger inasmuch as there is much more ferrite-cementite phase boundary area for the smaller particles; thus, there is greater reinforcement of the ferrite phase, and more phase boundary barriers to dislocation motion."
    },
    {
        "idx": 1254,
        "question": "What is the principal difference between natural and artificial aging processes?",
        "answer": "For precipitation hardening, natural aging is allowing the precipitation process to occur at the ambient temperature; artificial aging is carried out at an elevated temperature.",
        "type": 4,
        "wrong_answers_1": "The two differences are: (1) the hardening/strengthening effect is not retained at elevated temperatures for precipitation hardening--however, it is retained for dispersion strengthening; and (2) the strength is developed by a heat treatment for precipitation hardening--such is not the case for dispersion strengthening.",
        "wrong_answers_2": "Rivets of a 2017 aluminum alloy must be refrigerated before they are used because, after being solution heat treated, they precipitation harden at room temperature. Once precipitation hardened, they are too strong and brittle to be driven.",
        "wrong_answers_3": "The glass transition temperature is, for a noncrystalline ceramic, that temperature at which there is a change of slope for the specific volume versus temperature curve."
    },
    {
        "idx": 1255,
        "question": "Which of the following polymers would be suitable for the fabrication of cups to contain hot coffee: polyethylene, polypropylene, poly(vinyl chloride), PET polyester, and polycarbonate. Why?",
        "answer": "This question asks us to name, which, of five polymers, would be suitable for the fabrication of cups to contain hot coffee. At its glass transition temperature, an amorphous polymer begins to soften. The maximum temperature of hot coffee is probably slightly below 100^{\\circ} C\\left(212^{\\circ} F\\right). Of the polymers listed, only polystyrene and polycarbonate have glass transition temperatures of 100^{\\circ} C or above, and would be suitable for this application.",
        "type": 2
    },
    {
        "idx": 1256,
        "question": "For the following pair of polymers, do the following: (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Branched polyethylene having a number-average molecular weight of 850,000 g/mol; linear polyethylene having a number-average molecular weight of 850,000 g/mol",
        "answer": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will most likely have a higher percent crystallinity, and, therefore, a higher melting temperature than the branched polyethylene. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to determine which polymer has the higher melting temperature. Of these two polytetrafluoroethylene polymers, the PTFE with the higher density (2.20 g/cm^3) will have the higher percent crystallinity, and, therefore, a higher melting temperature than the lower density PTFE. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration.",
        "wrong_answers_2": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration.",
        "wrong_answers_3": "Yes, it is possible to determine which of the two polystyrenes has the higher Tm. The isotactic polystyrene will have the higher melting temperature because it has a higher density (i.e., less branching) and also the greater weight-average molecular weight."
    },
    {
        "idx": 1257,
        "question": "For the following pair of polymers, do the following: (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Polytetrafluoroethylene having a density of 2.14 g/cm^3 and a weight-average molecular weight of 600,000 g/mol; PTFE having a density of 2.20 g/cm^3 and a weight-average molecular weight of 600,000 g/mol",
        "answer": "Yes, it is possible to determine which polymer has the higher melting temperature. Of these two polytetrafluoroethylene polymers, the PTFE with the higher density (2.20 g/cm^3) will have the higher percent crystallinity, and, therefore, a higher melting temperature than the lower density PTFE. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will most likely have a higher percent crystallinity, and, therefore, a higher melting temperature than the branched polyethylene. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration.",
        "wrong_answers_2": "Yes, it is possible to determine which of the two polystyrenes has the higher Tm. The isotactic polystyrene will have the higher melting temperature because it has a higher density (i.e., less branching) and also the greater weight-average molecular weight.",
        "wrong_answers_3": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration."
    },
    {
        "idx": 1258,
        "question": "For the following pair of polymers, do the following: (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Linear and syndiotactic poly(vinyl chloride) having a number-average molecular weight of 500,000 g/mol; linear polyethylene having a number-average molecular weight of 225,000 g/mol",
        "answer": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration.",
        "type": 4,
        "wrong_answers_1": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will most likely have a higher percent crystallinity, and, therefore, a higher melting temperature than the branched polyethylene. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration.",
        "wrong_answers_2": "Yes, it is possible to determine which polymer has the higher melting temperature. Of these two polytetrafluoroethylene polymers, the PTFE with the higher density (2.20 g/cm^3) will have the higher percent crystallinity, and, therefore, a higher melting temperature than the lower density PTFE. The molecular weights of both materials are the same and, thus, molecular weight is not a consideration.",
        "wrong_answers_3": "Yes, it is possible to determine which polymer has the higher melting temperature. The polypropylene will have the higher Tm because it has a bulky phenyl side group in its repeat unit structure, which is absent in the polyethylene. Furthermore, the polypropylene has a higher degree of polymerization."
    },
    {
        "idx": 1259,
        "question": "For the following pair of polymers, do the following: (1) state whether it is possible to determine whether one polymer has a higher melting temperature than the other; (2) if it is possible, note which has the higher melting temperature and then cite reason(s) for your choice; and (3) if it is not possible to decide, then state why. Linear and syndiotactic polypropylene having a weight-average molecular weight of 500,000 g/mol; linear and atactic polypropylene having a weight-average molecular weight of 750,000 g/mol",
        "answer": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The syndiotactic polypropylene will have a higher degree of crystallinity than the atactic material. On the basis of this effect alone, the syndiotactic PP should have the greater Tm, since melting temperature increases with degree of crystallinity. However, the molecular weight for the syndiotactic polypropylene (500,000 g/mol) is less than for the atactic material (750,000 g/mol); a lowering of molecular weight generally results in a reduction of melting temperature.",
        "type": 4,
        "wrong_answers_1": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The polystyrene has a bulkier side group than the polypropylene; on the basis of this effect alone, the polystyrene should have the greater Tm. However, the polystyrene has more branching and a lower degree of polymerization; both of these factors lead to a lowering of the melting temperature.",
        "wrong_answers_2": "Yes it is possible. The syndiotactic polystyrene has the higher tensile strength. Syndiotactic polymers are more likely to crystallize than atactic ones; the greater the crystallinity, the higher the tensile strength. Furthermore, the syndiotactic also has a higher molecular weight; increasing molecular weight also enhances the strength.",
        "wrong_answers_3": "Yes, it is possible to determine which polymer has the higher melting temperature. The linear polyethylene will have the greater melting temperature inasmuch as it will have a higher degree of crystallinity; polymers having a syndiotactic structure do not crystallize as easily as those polymers having identical single-atom side groups. With regard to molecular weight, or rather, degree of polymerization, it is about the same for both materials (8000), and therefore, is not a consideration."
    },
    {
        "question": "Compute the electrical conductivity of a cylindrical silicon specimen 7.0mm (0.28 in.) diameter and 57mm (2.25 in.) in length in which a current of 0.25 A passes in an axial direction. A voltage of 24V is measured across two probes that are separated by 45mm (1.75 in.).",
        "answer": "12.2 (omega·m)^{-1}.",
        "idx": 1260,
        "type": 1,
        "wrong_answers_1": "11.4 (omega·m)^{-1}.",
        "wrong_answers_2": "13.4 (omega·m)^{-1}.",
        "wrong_answers_3": "10.9 (omega·m)^{-1}."
    },
    {
        "question": "Compute the resistance over the entire 57mm (2.25 in.) of the specimen.",
        "answer": "the resistance over the entire specimen length is 121.4 omega.",
        "idx": 1261,
        "type": 1,
        "wrong_answers_1": "the resistance over the entire specimen length is 131.1 omega.",
        "wrong_answers_2": "the resistance over the entire specimen length is 108.0 omega.",
        "wrong_answers_3": "the resistance over the entire specimen length is 133.8 omega."
    },
    {
        "idx": 1262,
        "question": "What is the distinction between electronic and ionic conduction? \\\\",
        "answer": "When a current arises from a flow of electrons, the conduction is termed electronic; for \\\\ ionic conduction, the current results from the net motion of charged ions.",
        "type": 4,
        "wrong_answers_1": "For a small anode-to-cathode area ratio, the corrosion rate will be higher than for A large ratio. The reason for this is that for some given current flow associated with the corrosion reaction, for a small area ratio the current density at the anode will be greater than for a large ratio. The corrosion rate is proportional to the current density (i) according to Equation.",
        "wrong_answers_2": "In ionic materials, the larger the size of the component ions the greater the degree of electronic polarization.",
        "wrong_answers_3": "An impressed current from an external dc power source provides excess electrons to the metallic structure to be protected."
    },
    {
        "idx": 1263,
        "question": "In terms of electron energy band structure, discuss reasons for the difference in electrical \\\\ conductivity among metals, semiconductors, and insulators.",
        "answer": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities.",
        "type": 4,
        "wrong_answers_1": "For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid.",
        "wrong_answers_2": "Metals are typically better thermal conductors than are ceramic materials because, for metals, most of the heat is transported by free electrons (of which there are relatively large numbers). In ceramic materials, the primary mode of thermal conduction is via phonons, and phonons are more easily scattered than are free electrons.",
        "wrong_answers_3": "No free electron is generated by an electron excitation involving an acceptor impurity atom because the electron is excited from the valence band into the impurity level within the band gap; no free electron is introduced into the conduction band."
    },
    {
        "idx": 1264,
        "question": "How does the electron structure of an isolated atom differ from that of a solid material?",
        "answer": "For an isolated atom, there exist discrete electron energy states (arranged into shells and subshells); each state may be occupied by, at most, two electrons, which must have opposite spins. On the other hand, an electron band structure is found for solid materials; within each band exist closely spaced yet discrete electron states, each of which may be occupied by, at most, two electrons, having opposite spins. The number of electron states in each band will equal the total number of corresponding states contributed by all of the atoms in the solid.",
        "type": 4,
        "wrong_answers_1": "The Pauli exclusion principle states that each electron state can hold no more than two electrons, which must have opposite spins.",
        "wrong_answers_2": "The mj quantum number designates the number of electron states in each electron subshell.",
        "wrong_answers_3": "For metallic materials, there are vacant electron energy states adjacent to the highest filled state; thus, very little energy is required to excite large numbers of electrons into conducting states. These electrons are those that participate in the conduction process, and, because there are so many of them, metals are good electrical conductors.\nThere are no empty electron states adjacent to and above filled states for semiconductors and insulators, but rather, an energy band gap across which electrons must be excited in order to participate in the conduction process. Thermal excitation of electrons occurs, and the number of electrons excited will be less than for metals, and will depend on the band gap energy. For semiconductors, the band gap is narrower than for insulators; consequently, at a specific temperature more electrons will be excited for semiconductors, giving rise to higher conductivities."
    },
    {
        "idx": 1265,
        "question": "Briefly state what is meant by the drift velocity of a free electron.",
        "answer": "The drift velocity of a free electron is the average electron velocity in the direction of the force imposed by an electric field.",
        "type": 4,
        "wrong_answers_1": "The mobility is the proportionality constant between the drift velocity and the electric field. It is also a measure of the frequency of scattering events (and is inversely proportional to the frequency of scattering).",
        "wrong_answers_2": "For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field.",
        "wrong_answers_3": "(a) For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field."
    },
    {
        "idx": 1266,
        "question": "Briefly state what is meant by the mobility of a free electron.",
        "answer": "The mobility is the proportionality constant between the drift velocity and the electric field. It is also a measure of the frequency of scattering events (and is inversely proportional to the frequency of scattering).",
        "type": 4,
        "wrong_answers_1": "The drift velocity of a free electron is the average electron velocity in the direction of the force imposed by an electric field.",
        "wrong_answers_2": "For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field.",
        "wrong_answers_3": "(a) For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field."
    },
    {
        "idx": 1267,
        "question": "(a) Calculate the drift velocity of electrons in silicon at room temperature and when the magnitude of the electric field is 500 V/m.",
        "answer": "the drift velocity of electrons in silicon at room temperature and when the magnitude of the electric field is 500 V/m is 72.5 m/s.",
        "type": 1,
        "wrong_answers_1": "the drift velocity of electrons in silicon at room temperature and when the magnitude of the electric field is 427 V/m is 75.9 m/s.",
        "wrong_answers_2": "the drift velocity of electrons in silicon at room temperature and when the magnitude of the electric field is 535 V/m is 67.8 m/s.",
        "wrong_answers_3": "the drift velocity of electrons in silicon at room temperature and when the magnitude of the electric field is 558 V/m is 65.4 m/s."
    },
    {
        "idx": 1268,
        "question": "(b) Under these circumstances, how long does it take an electron to traverse a 25-mm length of crystal?",
        "answer": "under these circumstances, it takes an electron 3.45 x 10^-4 s to traverse a 25-mm length of crystal.",
        "type": 1,
        "wrong_answers_1": "under these circumstances, it takes an electron 3.22 x 10^-4 s to traverse a 28-mm length of crystal.",
        "wrong_answers_2": "under these circumstances, it takes an electron 3.61 x 10^-4 s to traverse a 26-mm length of crystal.",
        "wrong_answers_3": "under these circumstances, it takes an electron 3.68 x 10^-4 s to traverse a 24-mm length of crystal."
    },
    {
        "idx": 1269,
        "question": "At room temperature the electrical conductivity and the electron mobility for aluminum are 3.8 x 10^7 (Ω·m)^-1 and 0.0012 m^2/V·s, respectively. Compute the number of free electrons per cubic meter for aluminum at room temperature.",
        "answer": "the number of free electrons per cubic meter for aluminum at room temperature is 1.98 x 10^29 m^-3.",
        "type": 1,
        "wrong_answers_1": "the number of free electrons per cubic meter for aluminum at room temperature is 1.87 x 10^29 m^-3.",
        "wrong_answers_2": "the number of free electrons per cubic meter for aluminum at room temperature is 2.19 x 10^29 m^-3.",
        "wrong_answers_3": "the number of free electrons per cubic meter for aluminum at room temperature is 2.26 x 10^29 m^-3."
    },
    {
        "idx": 1270,
        "question": "At room temperature the electrical conductivity and the electron mobility for aluminum are 3.8 x 10^7 (Ω·m)^-1 and 0.0012 m^2/V·s, respectively. What is the number of free electrons per aluminum atom? Assume a density of 2.7 g/cm^3.",
        "answer": "the number of free electrons per aluminum atom is 3.28 electrons/al atom.",
        "type": 1,
        "wrong_answers_1": "the number of free electrons per aluminum atom is 3.72 electrons/al atom.",
        "wrong_answers_2": "the number of free electrons per aluminum atom is 3.51 electrons/al atom.",
        "wrong_answers_3": "the number of free electrons per aluminum atom is 2.84 electrons/al atom."
    },
    {
        "idx": 1271,
        "question": "At room temperature the electrical conductivity of P b S is 25\\left(\\Omega·m\\right)^{-1}, whereas the electron and hole mobilities are 0.06 and 0.02{m}^{2} / V·s, respectively. Compute the intrinsic carrier concentration for PbS at room temperature.",
        "answer": "the intrinsic carrier concentration for pbs at room temperature is 1.95 × 10^{21}{m}^{-3}.",
        "type": 1,
        "wrong_answers_1": "the intrinsic carrier concentration for pbs at room temperature is 1.83 × 10^{21}{m}^{-3}.",
        "wrong_answers_2": "the intrinsic carrier concentration for pbs at room temperature is 2.05 × 10^{21}{m}^{-3}.",
        "wrong_answers_3": "the intrinsic carrier concentration for pbs at room temperature is 2.13 × 10^{21}{m}^{-3}."
    },
    {
        "idx": 1272,
        "question": "An n-type semiconductor is known to have an electron concentration of 5 × 10^{17}{m}^{-3}. If the electron drift velocity is 350{m} / s in an electric field of 1000V / m, calculate the conductivity of this material.",
        "answer": "the conductivity of the material is 0.028, (\\omega·m)^{-1}.",
        "type": 1,
        "wrong_answers_1": "the conductivity of the material is 0.032, (\\omega·m)^{-1}.",
        "wrong_answers_2": "the conductivity of the material is 0.027, (\\omega·m)^{-1}.",
        "wrong_answers_3": "the conductivity of the material is 0.027, (\\omega·m)^{-1}."
    },
    {
        "idx": 1273,
        "question": "Predict whether Nitrogen (N) will act as a donor or an acceptor when added to Silicon (Si). Assume that the impurity elements are substitutional.",
        "answer": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom.",
        "type": 4,
        "wrong_answers_1": "Sulfur will act as a donor in InSb. Since S is from group VIA of the periodic table, it will substitute for Sb; an S atom has one more valence electron than an Sb atom.",
        "wrong_answers_2": "Arsenic will act as an acceptor in ZnTe. Since As is from group VA of the periodic table, it will substitute for Te; furthermore, an As atom has one less valence electron than a Te atom.",
        "wrong_answers_3": "Indium will act as a donor in CdS. Since In is from group IIIA of the periodic table, it will substitute for Cd; an In atom has one more valence electron than a Cd atom."
    },
    {
        "idx": 1274,
        "question": "Predict whether Boron (B) will act as a donor or an acceptor when added to Germanium (Ge). Assume that the impurity elements are substitutional.",
        "answer": "Boron will act as an acceptor in Ge. Since it (B) is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom.",
        "type": 4,
        "wrong_answers_1": "Arsenic will act as an acceptor in ZnTe. Since As is from group VA of the periodic table, it will substitute for Te; furthermore, an As atom has one less valence electron than a Te atom.",
        "wrong_answers_2": "Indium will act as a donor in CdS. Since In is from group IIIA of the periodic table, it will substitute for Cd; an In atom has one more valence electron than a Cd atom.",
        "wrong_answers_3": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom."
    },
    {
        "idx": 1275,
        "question": "Predict whether Sulfur (S) will act as a donor or an acceptor when added to Indium Antimonide (InSb). Assume that the impurity elements are substitutional.",
        "answer": "Sulfur will act as a donor in InSb. Since S is from group VIA of the periodic table, it will substitute for Sb; an S atom has one more valence electron than an Sb atom.",
        "type": 4,
        "wrong_answers_1": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom.",
        "wrong_answers_2": "Indium will act as a donor in CdS. Since In is from group IIIA of the periodic table, it will substitute for Cd; an In atom has one more valence electron than a Cd atom.",
        "wrong_answers_3": "Arsenic will act as an acceptor in ZnTe. Since As is from group VA of the periodic table, it will substitute for Te; furthermore, an As atom has one less valence electron than a Te atom."
    },
    {
        "idx": 1276,
        "question": "Predict whether Indium (In) will act as a donor or an acceptor when added to Cadmium Sulfide (CdS). Assume that the impurity elements are substitutional.",
        "answer": "Indium will act as a donor in CdS. Since In is from group IIIA of the periodic table, it will substitute for Cd; an In atom has one more valence electron than a Cd atom.",
        "type": 4,
        "wrong_answers_1": "Sulfur will act as a donor in InSb. Since S is from group VIA of the periodic table, it will substitute for Sb; an S atom has one more valence electron than an Sb atom.",
        "wrong_answers_2": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom.",
        "wrong_answers_3": "Boron will act as an acceptor in Ge. Since it (B) is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom."
    },
    {
        "idx": 1277,
        "question": "Predict whether Arsenic (As) will act as a donor or an acceptor when added to Zinc Telluride (ZnTe). Assume that the impurity elements are substitutional.",
        "answer": "Arsenic will act as an acceptor in ZnTe. Since As is from group VA of the periodic table, it will substitute for Te; furthermore, an As atom has one less valence electron than a Te atom.",
        "type": 4,
        "wrong_answers_1": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom.",
        "wrong_answers_2": "Boron will act as an acceptor in Ge. Since it (B) is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom.",
        "wrong_answers_3": "Sulfur will act as a donor in InSb. Since S is from group VIA of the periodic table, it will substitute for Sb; an S atom has one more valence electron than an Sb atom."
    },
    {
        "idx": 1278,
        "question": "Germanium to which 10^24 m^-3 As atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the As atoms may be thought of as being ionized (i.e., one charge carrier exists for each As atom). Is this material n-type or p-type?",
        "answer": "this germanium material to which has been added 10^24 m^-3 as atoms is n-type since as is a donor in ge.",
        "type": 4,
        "wrong_answers_1": "this germanium material to which has been added 5 × 10^22 m^-3 sb atoms is n-type since sb is a donor in ge.",
        "wrong_answers_2": "Boron will act as an acceptor in Ge. Since it (B) is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom.",
        "wrong_answers_3": "Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table, and an N atom has one more valence electron than an Si atom."
    },
    {
        "idx": 1279,
        "question": "Germanium to which 10^24 m^-3 As atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the As atoms may be thought of as being ionized (i.e., one charge carrier exists for each As atom). Calculate the electrical conductivity of this material, assuming electron and hole mobilities of 0.1 and 0.05 m^2/V-s, respectively.",
        "answer": "the conductivity is equal to 1.60 × 10^4 (ω·m)^-1.",
        "type": 1,
        "wrong_answers_1": "the conductivity is equal to 1.48 × 10^4 (ω·m)^-1.",
        "wrong_answers_2": "the conductivity is equal to 1.73 × 10^4 (ω·m)^-1.",
        "wrong_answers_3": "the conductivity is equal to 1.76 × 10^4 (ω·m)^-1."
    },
    {
        "idx": 1280,
        "question": "A metal alloy is known to have electrical conductivity and electron mobility values of 1.2 × 10^{7}(\\Omega·m)^{-1} and 0.0050{m}^{2} / V·s, respectively. A current of 40 \\mathrm{~A} is passed through a specimen of this alloy that is 35mm thick. What magnetic field would need to be imposed to yield a Hall voltage of -3.5 × 10^{-7}V ?",
        "answer": "the required magnetic field is 0.735, \\text{tesla}.",
        "type": 1,
        "wrong_answers_1": "the required magnetic field is 0.787, \\text{tesla}.",
        "wrong_answers_2": "the required magnetic field is 0.767, \\text{tesla}.",
        "wrong_answers_3": "the required magnetic field is 0.784, \\text{tesla}."
    },
    {
        "idx": 1281,
        "question": "At temperatures between 540^{\\circ} C(813 K) and 727^{\\circ} C(1000 K), the activation energy and preexponential for the diffusion coefficient of \\mathrm{Na}^{+}in \\mathrm{NaCl} are 173,000 J/ mol and 4.0 × 10^{-4} m^{2} / s, respectively. Compute the mobility for an \\mathrm{Na}^{+}ion at 600^{\\circ} C(873 K).",
        "answer": "the mobility for a \\mathrm{na}^{+} ion at 600^{\\circ} C (873 k) is 2.34 × 10^{-13} m^{2}/\\mathrm{v}-s.",
        "type": 1,
        "wrong_answers_1": "the mobility for a \\mathrm{na}^{+} ion at 527^{\\circ} C (809 k) is 2.25 × 10^{-13} m^{2}/\\mathrm{v}-s.",
        "wrong_answers_2": "the mobility for a \\mathrm{na}^{+} ion at 554^{\\circ} C (970 k) is 2.05 × 10^{-13} m^{2}/\\mathrm{v}-s.",
        "wrong_answers_3": "the mobility for a \\mathrm{na}^{+} ion at 644^{\\circ} C (764 k) is 2.59 × 10^{-13} m^{2}/\\mathrm{v}-s."
    },
    {
        "idx": 1282,
        "question": "A parallel-plate capacitor using a dielectric material having an \\varepsilon_{\\mathrm{r}} of 2.2 has a plate spacing of 2mm ( 0.08 in.). If another material having a dielectric constant of 3.7 is used and the capacitance is to be unchanged, what must be the new spacing between the plates?",
        "answer": "the new spacing between the plates must be 3.36 mm.",
        "type": 1,
        "wrong_answers_1": "the new spacing between the plates must be 3.83 mm.",
        "wrong_answers_2": "the new spacing between the plates must be 3.82 mm.",
        "wrong_answers_3": "the new spacing between the plates must be 3.86 mm."
    },
    {
        "idx": 1283,
        "question": "Consider a parallel-plate capacitor having an area of 3225 mm2 (5 in.2), a plate separation of 1mm (0.04 in.), and a material having a dielectric constant of 3.5 positioned between the plates. What is the capacitance of this capacitor?",
        "answer": "the capacitance of this capacitor is 100 pf.",
        "type": 1,
        "wrong_answers_1": "the capacitance of this capacitor is 96 pf.",
        "wrong_answers_2": "the capacitance of this capacitor is 106 pf.",
        "wrong_answers_3": "the capacitance of this capacitor is 93 pf."
    },
    {
        "idx": 1284,
        "question": "Consider a parallel-plate capacitor having an area of 3225 mm2 (5 in.2), a plate separation of 1mm (0.04 in.), and a material having a dielectric constant of 3.5 positioned between the plates. Compute the electric field that must be applied for 2 × 10−8 C to be stored on each plate.",
        "answer": "the electric field that must be applied is 2.0 × 105 v/m.",
        "type": 1,
        "wrong_answers_1": "the electric field that must be applied is 1.9 × 105 v/m.",
        "wrong_answers_2": "the electric field that must be applied is 1.9 × 105 v/m.",
        "wrong_answers_3": "the electric field that must be applied is 2.3 × 105 v/m."
    },
    {
        "idx": 1285,
        "question": "The polarization P of a dielectric material positioned within a parallel-plate capacitor is to be 4.0 × 10^-6 C/m^2. What must be the dielectric constant if an electric field of 10^5 V/m is applied?",
        "answer": "the dielectric constant εr must be 5.52.",
        "type": 1,
        "wrong_answers_1": "the dielectric constant εr must be 4.79.",
        "wrong_answers_2": "the dielectric constant εr must be 5.79.",
        "wrong_answers_3": "the dielectric constant εr must be 5.20."
    },
    {
        "idx": 1286,
        "question": "The polarization P of a dielectric material positioned within a parallel-plate capacitor is to be 4.0 × 10^-6 C/m^2. What will be the dielectric displacement D?",
        "answer": "the dielectric displacement D will be 4.89 × 10^-6 c/m^2.",
        "type": 1,
        "wrong_answers_1": "the dielectric displacement D will be 5.52 × 10^-6 c/m^2.",
        "wrong_answers_2": "the dielectric displacement D will be 4.58 × 10^-6 c/m^2.",
        "wrong_answers_3": "the dielectric displacement D will be 5.31 × 10^-6 c/m^2."
    },
    {
        "idx": 1287,
        "question": "For each of the three types of polarization, briefly describe the mechanism by which dipoles are induced and/or oriented by the action of an applied electric field.",
        "answer": "For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field.",
        "type": 4,
        "wrong_answers_1": "(a) For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field.",
        "wrong_answers_2": "(b) Electronic, ionic, and orientation polarizations would be observed in lead titanate. The lead, titanium, and oxygen would undoubtedly be largely ionic in character. Furthermore, orientation polarization is also possible inasmuch as permanent dipole moments may be induced in the same manner as for BaTiO3. Only electronic polarization is to be found in gaseous neon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Only electronic polarization is to be found in solid diamond; this material does not have molecules with permanent dipole moments, nor is it an ionic material. Both electronic and ionic polarizations will be found in solid KCl, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid NH3. The NH3 molecules have permanent dipole moments that are easily oriented in the liquid state.",
        "wrong_answers_3": "Only electronic polarization is found in gaseous argon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Both electronic and ionic polarizations are found in solid LiF, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid H2O. The H2O molecules have permanent dipole moments that are easily oriented in the liquid state. Only electronic polarization is to be found in solid Si; this material does not have molecules with permanent dipole moments, nor is it an ionic material."
    },
    {
        "idx": 1288,
        "question": "For gaseous argon, solid LiF, liquid H2O, and solid Si, what kind(s) of polarization is (are) possible? Why?",
        "answer": "Only electronic polarization is found in gaseous argon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Both electronic and ionic polarizations are found in solid LiF, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid H2O. The H2O molecules have permanent dipole moments that are easily oriented in the liquid state. Only electronic polarization is to be found in solid Si; this material does not have molecules with permanent dipole moments, nor is it an ionic material.",
        "type": 4,
        "wrong_answers_1": "(b) Electronic, ionic, and orientation polarizations would be observed in lead titanate. The lead, titanium, and oxygen would undoubtedly be largely ionic in character. Furthermore, orientation polarization is also possible inasmuch as permanent dipole moments may be induced in the same manner as for BaTiO3. Only electronic polarization is to be found in gaseous neon; being an inert gas, its atoms will not be ionized nor possess permanent dipole moments. Only electronic polarization is to be found in solid diamond; this material does not have molecules with permanent dipole moments, nor is it an ionic material. Both electronic and ionic polarizations will be found in solid KCl, since it is strongly ionic. In all probability, no permanent dipole moments will be found in this material. Both electronic and orientation polarizations are found in liquid NH3. The NH3 molecules have permanent dipole moments that are easily oriented in the liquid state.",
        "wrong_answers_2": "For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field.",
        "wrong_answers_3": "(a) For electronic polarization, the electric field causes a net displacement of the center of the negatively charged electron cloud relative to the positive nucleus. With ionic polarization, the cations and anions are displaced in opposite directions as a result of the application of an electric field. Orientation polarization is found in substances that possess permanent dipole moments; these dipole moments become aligned in the direction of the electric field."
    },
    {
        "idx": 1289,
        "question": "List the four classifications of steels.",
        "answer": "The four classifications of steels are: Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, and High Alloy Steels (Stainless and Tool).",
        "type": 4,
        "wrong_answers_1": "Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, High Alloy Steels (Stainless and Tool)",
        "wrong_answers_2": "It is normally used on medium- and high-carbon steels, which, by virtue of carbon content, are relatively hard and strong.",
        "wrong_answers_3": "Stainless steels and aluminum alloys often passivate."
    },
    {
        "idx": 1290,
        "question": "For Low Carbon Steels, briefly describe the properties and typical applications.",
        "answer": "Properties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable. Typical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans.",
        "type": 4,
        "wrong_answers_1": "Properties: nonresponsive to heat treatments; relatively soft and weak; machinable and weldable. Typical applications: automobile bodies, structural shapes, pipelines, buildings, bridges, and tin cans.",
        "wrong_answers_2": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades.",
        "wrong_answers_3": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades."
    },
    {
        "idx": 1291,
        "question": "For Medium Carbon Steels, briefly describe the properties and typical applications.",
        "answer": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.",
        "type": 4,
        "wrong_answers_1": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.",
        "wrong_answers_2": "The chief difference between heat-treatable and nonheat-treatable alloys is that heat-treatable alloys may be strengthened by a heat treatment wherein a precipitate phase is formed (precipitation hardening) or a martensitic transformation occurs. Nonheat-treatable alloys are not amenable to strengthening by such treatments.",
        "wrong_answers_3": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades."
    },
    {
        "idx": 1292,
        "question": "For High Carbon Steels, briefly describe the properties and typical applications.",
        "answer": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades.",
        "type": 4,
        "wrong_answers_1": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades.",
        "wrong_answers_2": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts.",
        "wrong_answers_3": "Properties: heat treatable, relatively large combinations of mechanical characteristics. Typical applications: railway wheels and tracks, gears, crankshafts, and machine parts."
    },
    {
        "idx": 1293,
        "question": "For High Alloy Steels (Stainless and Tool), briefly describe the properties and typical applications.",
        "answer": "Properties: hard and wear resistant; resistant to corrosion in a large variety of environments. Typical applications: cutting tools, drills, cutlery, food processing, and surgical tools.",
        "type": 4,
        "wrong_answers_1": "Properties: hard and wear resistant; resistant to corrosion in a large variety of environments. Typical applications: cutting tools, drills, cutlery, food processing, and surgical tools.",
        "wrong_answers_2": "The alloying elements in tool steels (e.g., \\mathrm{Cr}, V, \\mathrm{W}, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds.",
        "wrong_answers_3": "Properties: hard, strong, and relatively brittle. Typical applications: chisels, hammers, knives, and hacksaw blades."
    },
    {
        "idx": 1294,
        "question": "Compare gray and malleable cast irons with respect to composition and heat treatment",
        "answer": "Gray iron--2.5 to 4.0 wt% C and 1.0 to 3.0 wt% Si. For most gray irons there is no heat treatment after solidification. Malleable iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. White iron is heated in a nonoxidizing atmosphere and at a temperature between 800 and 900°C for an extended time period.",
        "type": 4,
        "wrong_answers_1": "White iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. No heat treatment; however, cooling is rapid during solidification. Nodular cast iron--2.5 to 4.0 wt% C, 1.0 to 3.0 wt% Si, and a small amount of Mg or Ce. A heat treatment at about 700°C may be necessary to produce a ferritic matrix.",
        "wrong_answers_2": "Gray iron--Relatively weak and brittle in tension; good capacity for damping vibrations. Malleable iron--Moderate strength and ductility.",
        "wrong_answers_3": "White iron--Extremely hard and brittle. Nodular cast iron--Moderate strength and ductility."
    },
    {
        "idx": 1295,
        "question": "Compare gray and malleable cast irons with respect to microstructure",
        "answer": "Gray iron--Graphite flakes are embedded in a ferrite or pearlite matrix. Malleable iron--Graphite clusters are embedded in a ferrite or pearlite matrix.",
        "type": 4,
        "wrong_answers_1": "White iron--There are regions of cementite interspersed within pearlite. Nodular cast iron--Nodules of graphite are embedded in a ferrite or pearlite matrix.",
        "wrong_answers_2": "Gray iron is weak and brittle in tension because the tips of the graphite flakes act as points of stress concentration.",
        "wrong_answers_3": "Gray iron--Relatively weak and brittle in tension; good capacity for damping vibrations. Malleable iron--Moderate strength and ductility."
    },
    {
        "idx": 1296,
        "question": "Compare gray and malleable cast irons with respect to mechanical characteristics",
        "answer": "Gray iron--Relatively weak and brittle in tension; good capacity for damping vibrations. Malleable iron--Moderate strength and ductility.",
        "type": 4,
        "wrong_answers_1": "White iron--Extremely hard and brittle. Nodular cast iron--Moderate strength and ductility.",
        "wrong_answers_2": "Gray iron is weak and brittle in tension because the tips of the graphite flakes act as points of stress concentration.",
        "wrong_answers_3": "Gray iron--2.5 to 4.0 wt% C and 1.0 to 3.0 wt% Si. For most gray irons there is no heat treatment after solidification. Malleable iron--2.5 to 4.0 wt% C and less than 1.0 wt% Si. White iron is heated in a nonoxidizing atmosphere and at a temperature between 800 and 900°C for an extended time period."
    },
    {
        "idx": 1297,
        "question": "What are the distinctive features, limitations, and applications of titanium alloys?",
        "answer": "Distinctive features: relatively low density, high melting temperatures, and high strengths are possible. Limitation: because of chemical reactivity with other materials at elevated temperatures, these alloys are expensive to refine. Applications: aircraft structures, space vehicles, and in chemical and petroleum industries.",
        "type": 4,
        "wrong_answers_1": "Distinctive features: extremely high melting temperatures; large elastic moduli, hardnesses, and strengths. Limitation: some experience rapid oxidation at elevated temperatures. Applications: extrusion dies, structural parts in space vehicles, incandescent light filaments, x-ray tubes, and welding electrodes.",
        "wrong_answers_2": "Distinctive features: highly resistant to oxidation, especially at elevated temperatures; soft and ductile. Limitation: expensive. Applications: jewelry, dental restoration materials, coins, catalysts, and thermocouples.",
        "wrong_answers_3": "Five important characteristics for polymers that are to be used in thin-film applications are: (1) low density; (2) high flexibility; (3) high tensile and tear strengths; (4) resistance to moisture/chemical attack; and (5) low gas permeability.\n}"
    },
    {
        "idx": 1298,
        "question": "What are the distinctive features, limitations, and applications of refractory metals?",
        "answer": "Distinctive features: extremely high melting temperatures; large elastic moduli, hardnesses, and strengths. Limitation: some experience rapid oxidation at elevated temperatures. Applications: extrusion dies, structural parts in space vehicles, incandescent light filaments, x-ray tubes, and welding electrodes.",
        "type": 4,
        "wrong_answers_1": "Distinctive features: relatively low density, high melting temperatures, and high strengths are possible. Limitation: because of chemical reactivity with other materials at elevated temperatures, these alloys are expensive to refine. Applications: aircraft structures, space vehicles, and in chemical and petroleum industries.",
        "wrong_answers_2": "Distinctive features: highly resistant to oxidation, especially at elevated temperatures; soft and ductile. Limitation: expensive. Applications: jewelry, dental restoration materials, coins, catalysts, and thermocouples.",
        "wrong_answers_3": "Distinctive features: able to withstand high temperatures and oxidizing atmospheres for long time periods. Applications: aircraft turbines, nuclear reactors, and petrochemical equipment."
    },
    {
        "idx": 1299,
        "question": "What are the distinctive features, limitations, and applications of superalloys?",
        "answer": "Distinctive features: able to withstand high temperatures and oxidizing atmospheres for long time periods. Applications: aircraft turbines, nuclear reactors, and petrochemical equipment.",
        "type": 4,
        "wrong_answers_1": "Distinctive features: relatively low density, high melting temperatures, and high strengths are possible. Limitation: because of chemical reactivity with other materials at elevated temperatures, these alloys are expensive to refine. Applications: aircraft structures, space vehicles, and in chemical and petroleum industries.",
        "wrong_answers_2": "Distinctive features: highly resistant to oxidation, especially at elevated temperatures; soft and ductile. Limitation: expensive. Applications: jewelry, dental restoration materials, coins, catalysts, and thermocouples.",
        "wrong_answers_3": "Distinctive features: extremely high melting temperatures; large elastic moduli, hardnesses, and strengths. Limitation: some experience rapid oxidation at elevated temperatures. Applications: extrusion dies, structural parts in space vehicles, incandescent light filaments, x-ray tubes, and welding electrodes."
    },
    {
        "idx": 1300,
        "question": "What are the distinctive features, limitations, and applications of noble metals?",
        "answer": "Distinctive features: highly resistant to oxidation, especially at elevated temperatures; soft and ductile. Limitation: expensive. Applications: jewelry, dental restoration materials, coins, catalysts, and thermocouples.",
        "type": 4,
        "wrong_answers_1": "Distinctive features: relatively low density, high melting temperatures, and high strengths are possible. Limitation: because of chemical reactivity with other materials at elevated temperatures, these alloys are expensive to refine. Applications: aircraft structures, space vehicles, and in chemical and petroleum industries.",
        "wrong_answers_2": "Distinctive features: extremely high melting temperatures; large elastic moduli, hardnesses, and strengths. Limitation: some experience rapid oxidation at elevated temperatures. Applications: extrusion dies, structural parts in space vehicles, incandescent light filaments, x-ray tubes, and welding electrodes.",
        "wrong_answers_3": "Distinctive features: able to withstand high temperatures and oxidizing atmospheres for long time periods. Applications: aircraft turbines, nuclear reactors, and petrochemical equipment."
    },
    {
        "idx": 1301,
        "question": "For the MgO-Al2O3 system, what is the maximum temperature that is possible without the formation of a liquid phase?",
        "answer": "This problem asks that we specify, for the MgO-Al2O3 system, the maximum temperature without the formation of a liquid phase; it is approximately 2800 degrees Celsius.",
        "type": 1,
        "wrong_answers_1": "This problem asks that we specify, for the MgO-Al2O3 system, the maximum temperature without the formation of a liquid phase; it is approximately 2540 degrees Celsius.",
        "wrong_answers_2": "This problem asks that we specify, for the MgO-Al2O3 system, the maximum temperature without the formation of a liquid phase; it is approximately 3210 degrees Celsius.",
        "wrong_answers_3": "This problem asks that we specify, for the MgO-Al2O3 system, the maximum temperature without the formation of a liquid phase; it is approximately 2990 degrees Celsius."
    },
    {
        "idx": 1302,
        "question": "For the MgO-Al2O3 system, at what composition or over what range of compositions will this maximum temperature be achieved?",
        "answer": "This maximum temperature is possible for pure MgO.",
        "type": 4,
        "wrong_answers_1": "( MgO has ionic bonds, which are strong compared to the metallic bonds in Mg. A higher force will be required to cause the same separation between the ions in MgO compared to the atoms in Mg. Therefore, MgO should have the higher modulus of elasticity. In Mg, E=6 × 10^{6} psi; in MgO, E=30 × 10^{6} psi.",
        "wrong_answers_2": "The glass transition temperature is, for a noncrystalline ceramic, that temperature at which there is a change of slope for the specific volume versus temperature curve.",
        "wrong_answers_3": "The melting temperature is, for a crystalline material and upon cooling, that temperature at which there is a sudden and discontinuous decrease in the specific-volume-versus-temperature curve."
    },
    {
        "idx": 1303,
        "question": "What are the characteristics of sand casting technique?",
        "answer": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast.",
        "type": 4,
        "wrong_answers_1": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast.",
        "wrong_answers_2": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "wrong_answers_3": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast."
    },
    {
        "idx": 1304,
        "question": "What are the characteristics of die casting technique?",
        "answer": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "type": 4,
        "wrong_answers_1": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "wrong_answers_2": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast.",
        "wrong_answers_3": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast."
    },
    {
        "idx": 1305,
        "question": "What are the characteristics of investment casting technique?",
        "answer": "For investment casting, a single-piece mold is used, which is not reusable; it results in high dimensional accuracy, good reproduction of detail, and a fine surface finish; and casting rates are low.",
        "type": 4,
        "wrong_answers_1": "For investment casting, a single-piece mold is used, which is not reusable; it results in high dimensional accuracy, good reproduction of detail, and a fine surface finish; and casting rates are low.",
        "wrong_answers_2": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "wrong_answers_3": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast."
    },
    {
        "idx": 1306,
        "question": "What are the characteristics of lost-foam casting technique?",
        "answer": "For lost foam casting, the pattern is polystyrene foam, whereas the mold material is sand. Complex geometries and tight tolerances are possible. Casting rates are higher than for investment, and there are few environmental wastes.",
        "type": 4,
        "wrong_answers_1": "For lost foam casting, the pattern is polystyrene foam, whereas the mold material is sand. Complex geometries and tight tolerances are possible. Casting rates are higher than for investment, and there are few environmental wastes.",
        "wrong_answers_2": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast.",
        "wrong_answers_3": "For sand casting, sand is the mold material, a two-piece mold is used, ordinarily the surface finish is not an important consideration, the sand may be reused (but the mold may not), casting rates are low, and large pieces are usually cast."
    },
    {
        "idx": 1307,
        "question": "What are the characteristics of continuous casting technique?",
        "answer": "For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and mechanical properties are relatively uniform throughout the cross-section.",
        "type": 4,
        "wrong_answers_1": "For continuous casting, at the conclusion of the extraction process, the molten metal is cast into a continuous strand having either a rectangular or circular cross-section; these shapes are desirable for subsequent secondary metal-forming operations. The chemical composition and mechanical properties are relatively uniform throughout the cross-section.",
        "wrong_answers_2": "The disadvantages of extrusion over rolling are as follows: (1) Nonuniform deformation over the cross-section. (2) A variation in properties may result over a cross-section of an extruded piece.",
        "wrong_answers_3": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast."
    },
    {
        "idx": 1308,
        "question": "What is the distinction between dye and pigment colorants?",
        "answer": "The distinction between dye and pigment colorants is that a dye dissolves within and becomes a part of the polymer structure, whereas a pigment does not dissolve, but remains as a separate phase.",
        "type": 4,
        "wrong_answers_1": "For clay-based aggregates, a liquid phase forms during firing, which infiltrates the pores between the unmelted particles; upon cooling, this liquid becomes a glass, that serves as the bonding phase.",
        "wrong_answers_2": "A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time.",
        "wrong_answers_3": "For polymer-matrix fiber-reinforced composites, three functions of the polymer-matrix phase are: (1) to bind the fibers together so that the applied stress is distributed among the fibers; (2) to protect the surface of the fibers from being damaged; and (3) to separate the fibers and inhibit crack propagation."
    },
    {
        "idx": 1309,
        "question": "Cite four factors that determine what fabrication technique is used to form polymeric materials.",
        "answer": "Four factors that determine what fabrication technique is used to form polymeric materials are: (1) whether the polymer is thermoplastic or thermosetting; (2) if thermoplastic, the softening temperature; (3) atmospheric stability; and (4) the geometry and size of the finished product.",
        "type": 4,
        "wrong_answers_1": "Four situations in which casting is the preferred fabrication technique are:\n(1) For large pieces and/or complicated shapes.\n(2) When mechanical strength is not an important consideration.\n(3) For alloys having low ductilities.\n(4) When it is the most economical fabrication technique.",
        "wrong_answers_2": "For thermoplastic polymers, five factors that favor brittle fracture are as follows: (1) a reduction in temperature, (2) an increase in strain rate, (3) the presence of a sharp notch, (4) increased specimen thickness, and (5) modifications of the polymer structure.",
        "wrong_answers_3": "A high frequency will cause heating of the polymer. As the temperature of the polymer increases, the fatigue strength will decrease. If the applied stress is not reduced, then the polymer will fail in a shorter time."
    },
    {
        "idx": 1310,
        "question": "What is the process of compression molding used to form plastic materials?",
        "answer": "For compression molding, both heat and pressure are applied after the polymer and necessary additives are situated between the mold members.",
        "type": 4,
        "wrong_answers_1": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "wrong_answers_2": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "wrong_answers_3": "The principal disadvantage of hot-isostatic pressing is that it is expensive. The pressure is applied on a pre-formed green piece by a gas. Thus, the process is slow, and the equipment required to supply the gas and withstand the elevated temperature and pressure is costly."
    },
    {
        "idx": 1311,
        "question": "What is the process of transfer molding used to form plastic materials?",
        "answer": "For transfer molding, the solid materials (normally thermosetting in nature) are first melted in the transfer chamber prior to being forced into the die.",
        "type": 4,
        "wrong_answers_1": "For injection molding (normally used for thermoplastic materials), the raw materials are impelled by a ram through a heating chamber, and finally into the die cavity.",
        "wrong_answers_2": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "wrong_answers_3": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast."
    },
    {
        "idx": 1312,
        "question": "What is the process of injection molding used to form plastic materials?",
        "answer": "For injection molding (normally used for thermoplastic materials), the raw materials are impelled by a ram through a heating chamber, and finally into the die cavity.",
        "type": 4,
        "wrong_answers_1": "For transfer molding, the solid materials (normally thermosetting in nature) are first melted in the transfer chamber prior to being forced into the die.",
        "wrong_answers_2": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast.",
        "wrong_answers_3": "For die casting, a permanent mold is used, casting rates are high, the molten metal is forced into the mold under pressure, a two-piece mold is used, and small pieces are normally cast."
    },
    {
        "idx": 1313,
        "question": "Why must fiber materials that are melt-spun be thermoplastic?",
        "answer": "Fiber materials that are melt spun must be thermoplastic because they must be capable of forming a viscous liquid when heated, which is not possible for thermosets.",
        "type": 4,
        "wrong_answers_1": "Fiber materials that are drawn must be thermoplastic because during drawing, mechanical elongation must be possible; inasmuch as thermosetting materials are, in general, hard and relatively brittle, they are not easily elongated.",
        "wrong_answers_2": "Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of crystallinity.",
        "wrong_answers_3": "The matrix phase must be ductile and is usually relatively soft, whereas the fiber phase must be stiff and strong."
    },
    {
        "idx": 1314,
        "question": "Why must fiber materials that are drawn be thermoplastic?",
        "answer": "Fiber materials that are drawn must be thermoplastic because during drawing, mechanical elongation must be possible; inasmuch as thermosetting materials are, in general, hard and relatively brittle, they are not easily elongated.",
        "type": 4,
        "wrong_answers_1": "Fiber materials that are melt spun must be thermoplastic because they must be capable of forming a viscous liquid when heated, which is not possible for thermosets.",
        "wrong_answers_2": "Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of crystallinity.",
        "wrong_answers_3": "The matrix phase must be ductile and is usually relatively soft, whereas the fiber phase must be stiff and strong."
    },
    {
        "idx": 1315,
        "question": "Estimate the maximum thermal conductivity value for a cermet that contains 90 vol% titanium carbide (TiC) particles in a nickel matrix. Assume thermal conductivities of 27 and 67 W/m·K for TiC and Ni, respectively.",
        "answer": "the maximum thermal conductivity k_max is 31.0 W/m·K.",
        "type": 1,
        "wrong_answers_1": "the maximum thermal conductivity k_max is 34.9 W/m·K.",
        "wrong_answers_2": "the maximum thermal conductivity k_max is 27.1 W/m·K.",
        "wrong_answers_3": "the maximum thermal conductivity k_max is 27.6 W/m·K."
    },
    {
        "idx": 1316,
        "question": "Estimate the minimum thermal conductivity value for a cermet that contains 90 vol% titanium carbide (TiC) particles in a nickel matrix. Assume thermal conductivities of 27 and 67 W/m·K for TiC and Ni, respectively.",
        "answer": "the minimum thermal conductivity k_min is 28.7 W/m·K.",
        "type": 1,
        "wrong_answers_1": "the minimum thermal conductivity k_min is 24.6 W/m·K.",
        "wrong_answers_2": "the minimum thermal conductivity k_min is 32.1 W/m·K.",
        "wrong_answers_3": "the minimum thermal conductivity k_min is 27.0 W/m·K."
    },
    {
        "idx": 1317,
        "question": "A large-particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of tungsten and copper are 0.70 and 0.30 , respectively, estimate the upper limit for the specific stiffness of this composite, given the data that follow.\n\\begin{tabular}{lcc}\n\\hline & Specific Gravity & Modulus of Elasticity (GPa) \\\\\n\\hline Copper & 8.9 & 110 \\\\\nTungsten & 19.3 & 407 \\\\\n\\hline\n\\end{tabular}",
        "answer": "the upper limit for the specific stiffness of this composite is 19.65 gpa. alternatively, the specific stiffness can be estimated as 18.47 gpa.",
        "type": 1,
        "wrong_answers_1": "the upper limit for the specific stiffness of this composite is 17.91 gpa. alternatively, the specific stiffness can be estimated as 17.05 gpa.",
        "wrong_answers_2": "the upper limit for the specific stiffness of this composite is 22.42 gpa. alternatively, the specific stiffness can be estimated as 16.72 gpa.",
        "wrong_answers_3": "the upper limit for the specific stiffness of this composite is 21.17 gpa. alternatively, the specific stiffness can be estimated as 17.44 gpa."
    },
    {
        "idx": 1318,
        "question": "For a continuous and oriented fiber-reinforced composite, the moduli of elasticity in the longitudinal and transverse directions are 33.1 and 3.66 \\mathrm{GPa}\\left(4.8 × 10^{6}\\right. and \\left.5.3 × 10^{5} psi\\right), respectively. If the volume fraction of fibers is 0.30 , determine the moduli of elasticity of fiber and matrix phases.",
        "answer": "the moduli of elasticity of the matrix and fiber phases are: \\[\ne_{m} = 2.6 \\, \\text{gpa}  (3.77 × 10^{5} \\, \\text{psi})\n\\] \\[\ne_{f} = 104 \\, \\text{gpa}  (15 × 10^{6} \\, \\text{psi})\n\\]",
        "type": 1,
        "wrong_answers_1": "the moduli of elasticity of the matrix and fiber phases are: \\[\ne_{m} = 2.5 \\, \\text{gpa}  (3.49 × 10^{5} \\, \\text{psi})\n\\] \\[\ne_{f} = 110 \\, \\text{gpa}  (13 × 10^{6} \\, \\text{psi})\n\\]",
        "wrong_answers_2": "the moduli of elasticity of the matrix and fiber phases are: \\[\ne_{m} = 2.2 \\, \\text{gpa}  (4.25 × 10^{5} \\, \\text{psi})\n\\] \\[\ne_{f} = 112 \\, \\text{gpa}  (17 × 10^{6} \\, \\text{psi})\n\\]",
        "wrong_answers_3": "the moduli of elasticity of the matrix and fiber phases are: \\[\ne_{m} = 3.0 \\, \\text{gpa}  (3.36 × 10^{5} \\, \\text{psi})\n\\] \\[\ne_{f} = 98 \\, \\text{gpa}  (14 × 10^{6} \\, \\text{psi})\n\\]"
    },
    {
        "idx": 1319,
        "question": "In an aligned and continuous carbon fiber-reinforced nylon 6,6 composite, the fibers are to carry 97% of a load applied in the longitudinal direction. Using the data provided, determine the volume fraction of fibers required.",
        "answer": "the volume fraction of fibers required is 0.258.",
        "type": 1,
        "wrong_answers_1": "the volume fraction of fibers required is 0.293.",
        "wrong_answers_2": "the volume fraction of fibers required is 0.238.",
        "wrong_answers_3": "the volume fraction of fibers required is 0.233."
    },
    {
        "idx": 1320,
        "question": "In an aligned and continuous carbon fiber-reinforced nylon 6,6 composite, the fibers are to carry 97% of a load applied in the longitudinal direction. What will be the tensile strength of this composite? Assume that the matrix stress at fiber failure is 50 MPa (7250 psi).",
        "answer": "the tensile strength of this composite is 1070 mpa (155,000 psi).",
        "type": 1,
        "wrong_answers_1": "the tensile strength of this composite is 1220 mpa (144039 psi).",
        "wrong_answers_2": "the tensile strength of this composite is 960 mpa (167313 psi).",
        "wrong_answers_3": "the tensile strength of this composite is 940 mpa (162345 psi)."
    },
    {
        "idx": 1321,
        "question": "Cite several reasons why fiberglass-reinforced composites are used extensively.",
        "answer": "Reasons why fiberglass-reinforced composites are utilized extensively are: (1) glass fibers are very inexpensive to produce; (2) these composites have relatively high specific strengths; and (3) they are chemically inert in a wide variety of environments.",
        "type": 4,
        "wrong_answers_1": "Several limitations of these composites are: (1) care must be exercised in handling the fibers inasmuch as they are susceptible to surface damage; (2) they are lacking in stiffness in comparison to other fibrous composites; and (3) they are limited as to maximum temperature use.",
        "wrong_answers_2": "These composites are constructed in order to have a relatively high strength in virtually all directions within the plane of the laminate.",
        "wrong_answers_3": "Disadvantages of ferrous alloys are: (1) They are susceptible to corrosion. (2) They have a relatively high density. (3) They have relatively low electrical conductivities."
    },
    {
        "idx": 1322,
        "question": "Cite several limitations of fiberglass-reinforced composites.",
        "answer": "Several limitations of these composites are: (1) care must be exercised in handling the fibers inasmuch as they are susceptible to surface damage; (2) they are lacking in stiffness in comparison to other fibrous composites; and (3) they are limited as to maximum temperature use.",
        "type": 4,
        "wrong_answers_1": "Disadvantages of ferrous alloys are: (1) They are susceptible to corrosion. (2) They have a relatively high density. (3) They have relatively low electrical conductivities.",
        "wrong_answers_2": "Reasons why fiberglass-reinforced composites are utilized extensively are: (1) glass fibers are very inexpensive to produce; (2) these composites have relatively high specific strengths; and (3) they are chemically inert in a wide variety of environments.",
        "wrong_answers_3": "Two important characteristics for polymers that are to be used in fiber applications are: (1) they must have high molecular weights, and (2) they must have chain configurations/structures that will allow for high degrees of crystallinity."
    },
    {
        "idx": 1323,
        "question": "What is a hybrid composite?",
        "answer": "A hybrid composite is a composite that is reinforced with two or more different fiber materials in a single matrix.",
        "type": 4,
        "wrong_answers_1": "Two advantages of hybrid composites are: (1) better overall property combinations, and (2) failure is not as catastrophic as with single-fiber composites.",
        "wrong_answers_2": "Bimetallic materials are produced by bonding two materials having different coefficients of thermal expansion to one another, forming a laminar composite. When the temperature changes, one of the materials will expand or contract more than the other material. This difference in expansion or contraction causes the bimetallic material to change shape; if the original shape is that of a coil, then the device will coil or uncoil, depending on the direction of the temperature change.",
        "wrong_answers_3": "There must be a strong interfacial bond between fiber and matrix in order to: (1) maximize the stress transmittance between matrix and fiber phases; and (2) minimize fiber pull-out, and the probability of failure."
    },
    {
        "idx": 1324,
        "question": "List two important advantages of hybrid composites over normal fiber composites.",
        "answer": "Two advantages of hybrid composites are: (1) better overall property combinations, and (2) failure is not as catastrophic as with single-fiber composites.",
        "type": 4,
        "wrong_answers_1": "A hybrid composite is a composite that is reinforced with two or more different fiber materials in a single matrix.",
        "wrong_answers_2": "Several limitations of these composites are: (1) care must be exercised in handling the fibers inasmuch as they are susceptible to surface damage; (2) they are lacking in stiffness in comparison to other fibrous composites; and (3) they are limited as to maximum temperature use.",
        "wrong_answers_3": "Reasons why fiberglass-reinforced composites are utilized extensively are: (1) glass fibers are very inexpensive to produce; (2) these composites have relatively high specific strengths; and (3) they are chemically inert in a wide variety of environments."
    },
    {
        "idx": 1325,
        "question": "(a) Write an expression for the modulus of elasticity for a hybrid composite in which all fibers of both types are oriented in the same direction.",
        "answer": "the expression for the modulus of elasticity for a hybrid composite with all fibers aligned in the same direction is: e_cl = e_m v_m + e_f1 v_f1 + e_f2 v_f2",
        "type": 1,
        "wrong_answers_1": "the expression for the modulus of elasticity for a hybrid composite with all fibers aligned in the same direction is: e_cl = e_m v_m + e_f1 v_f1 + e_f2 v_f2",
        "wrong_answers_2": "the expression for the modulus of elasticity for a hybrid composite with all fibers aligned in the same direction is: e_cl = e_m v_m + e_f1 v_f1 + e_f2 v_f2",
        "wrong_answers_3": "the expression for the modulus of elasticity for a hybrid composite with all fibers aligned in the same direction is: e_cl = e_m v_m + e_f1 v_f1 + e_f2 v_f2"
    },
    {
        "idx": 1326,
        "question": "(b) Using this expression, compute the longitudinal modulus of elasticity of a hybrid composite consisting of aramid and glass fibers in volume fractions of 0.25 and 0.35 , respectively, within a polyester resin matrix [E_m=4.0 GPa (6x10^5 psi)].",
        "answer": "the longitudinal modulus of elasticity of the hybrid composite is: e_cl = 59.7 gpa (8.67x10^6 psi)",
        "type": 1,
        "wrong_answers_1": "the longitudinal modulus of elasticity of the hybrid composite is: e_cl = 63.6 gpa (8.40x10^6 psi)",
        "wrong_answers_2": "the longitudinal modulus of elasticity of the hybrid composite is: e_cl = 66.3 gpa (7.98x10^6 psi)",
        "wrong_answers_3": "the longitudinal modulus of elasticity of the hybrid composite is: e_cl = 68.1 gpa (7.99x10^6 psi)"
    },
    {
        "idx": 1327,
        "question": "Briefly describe laminar composites.",
        "answer": "Laminar composites are a series of sheets or panels, each of which has a preferred high-strength direction. These sheets are stacked and then cemented together such that the orientation of the high-strength direction varies from layer to layer.",
        "type": 4,
        "wrong_answers_1": "Increase in strength in the direction of drawing.",
        "wrong_answers_2": "Decrease in strength perpendicular to the direction of drawing.",
        "wrong_answers_3": "These composites are constructed in order to have a relatively high strength in virtually all directions within the plane of the laminate."
    },
    {
        "idx": 1328,
        "question": "What is the prime reason for fabricating laminar composites?",
        "answer": "These composites are constructed in order to have a relatively high strength in virtually all directions within the plane of the laminate.",
        "type": 4,
        "wrong_answers_1": "Reasons why fiberglass-reinforced composites are utilized extensively are: (1) glass fibers are very inexpensive to produce; (2) these composites have relatively high specific strengths; and (3) they are chemically inert in a wide variety of environments.",
        "wrong_answers_2": "Laminar composites are a series of sheets or panels, each of which has a preferred high-strength direction. These sheets are stacked and then cemented together such that the orientation of the high-strength direction varies from layer to layer.",
        "wrong_answers_3": "All metals do not have the same slip system. The reason for this is that for most metals, the slip system will consist of the most densely packed crystallographic plane, and within that plane the most closely packed direction. This plane and direction will vary from crystal structure to crystal structure."
    },
    {
        "idx": 1329,
        "question": "Is a voltage generated between the two cell halves of an Fe/Fe2+ concentration cell where both electrodes are pure iron, with Fe2+ concentrations of 0.5 M and 2 × 10^-2 M?",
        "answer": "Yes, a voltage is generated.",
        "type": 3
    },
    {
        "idx": 1330,
        "question": "What is the magnitude of the voltage generated in an Fe/Fe2+ concentration cell with Fe2+ concentrations of 0.5 M and 2 × 10^-2 M?",
        "answer": "The magnitude of the voltage is 0.0414 V.",
        "type": 1,
        "wrong_answers_1": "The magnitude of the voltage is 0.0448 V.",
        "wrong_answers_2": "The magnitude of the voltage is 0.0447 V.",
        "wrong_answers_3": "The magnitude of the voltage is 0.0470 V."
    },
    {
        "idx": 1331,
        "question": "Which electrode will be oxidized in an Fe/Fe2+ concentration cell with Fe2+ concentrations of 0.5 M and 2 × 10^-2 M?",
        "answer": "Oxidation occurs in the cell half with the lower Fe2+ concentration (2 × 10^-2 M).",
        "type": 4,
        "wrong_answers_1": "The electrode in the cell half with the Zn2+ concentration of 10^-2 M will be oxidized.",
        "wrong_answers_2": "Oxidation occurs at the anode; reduction at the cathode.",
        "wrong_answers_3": "In an HCl solution containing dissolved oxygen and Fe2+ ions, possible reactions are Mg -> Mg2+ + 2 e- (oxidation), 4 H+ + O2 + 4 e- -> 2 H2O (reduction), and Fe2+ + 2 e- -> Fe (reduction)."
    },
    {
        "idx": 1332,
        "question": "For the aluminum and cast iron pair that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.",
        "answer": "For the aluminum-cast iron couple, corrosion is possible, and aluminum will corrode.",
        "type": 4,
        "wrong_answers_1": "For the cadmium-zinc couple, corrosion is possible, and zinc will corrode.",
        "wrong_answers_2": "For the low-carbon steel-copper couple, corrosion is possible, and the low-carbon steel will corrode.",
        "wrong_answers_3": "For the brass-titanium pair, corrosion is possible, and brass will corrode."
    },
    {
        "idx": 1333,
        "question": "For the Inconel and nickel pair that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.",
        "answer": "For the Inconel-nickel couple, corrosion is unlikely inasmuch as both alloys appear within the same set of brackets (in both active and passive states).",
        "type": 4,
        "wrong_answers_1": "both martensite and bainite form.",
        "wrong_answers_2": "For the aluminum-cast iron couple, corrosion is possible, and aluminum will corrode.",
        "wrong_answers_3": "For the cadmium-zinc couple, corrosion is possible, and zinc will corrode."
    },
    {
        "idx": 1334,
        "question": "For the cadmium and zinc pair that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.",
        "answer": "For the cadmium-zinc couple, corrosion is possible, and zinc will corrode.",
        "type": 4,
        "wrong_answers_1": "For the aluminum-cast iron couple, corrosion is possible, and aluminum will corrode.",
        "wrong_answers_2": "BeO has the zinc blende structure.",
        "wrong_answers_3": "For the low-carbon steel-copper couple, corrosion is possible, and the low-carbon steel will corrode."
    },
    {
        "idx": 1335,
        "question": "For the brass and titanium pair that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.",
        "answer": "For the brass-titanium pair, corrosion is possible, and brass will corrode.",
        "type": 4,
        "wrong_answers_1": "For brass, the bonding is metallic since it is a metal alloy.",
        "wrong_answers_2": "For the aluminum-cast iron couple, corrosion is possible, and aluminum will corrode.",
        "wrong_answers_3": "For the cadmium-zinc couple, corrosion is possible, and zinc will corrode."
    },
    {
        "idx": 1336,
        "question": "For the low-carbon steel and copper pair that are coupled in seawater, predict the possibility of corrosion; if corrosion is probable, note which metal/alloy will corrode.",
        "answer": "For the low-carbon steel-copper couple, corrosion is possible, and the low-carbon steel will corrode.",
        "type": 4,
        "wrong_answers_1": "A \"hypoeutectoid\" steel has a carbon concentration less than the eutectoid; on the other hand, a \"hypereutectoid\" steel has a carbon content greater than the eutectoid.",
        "wrong_answers_2": "For the aluminum-cast iron couple, corrosion is possible, and aluminum will corrode.",
        "wrong_answers_3": "For the cadmium-zinc couple, corrosion is possible, and zinc will corrode."
    },
    {
        "idx": 1337,
        "question": "A piece of corroded metal alloy plate was found in a submerged ocean vessel. It was estimated that the original area of the plate was 800 cm^2 and that approximately 7.6kg had corroded away during the submersion. Assuming a corrosion penetration rate of 4mm / yr for this alloy in seawater, estimate the time of submersion in years. The density of the alloy is 4.5g/cm^3.",
        "answer": "the time of submersion is 5.27 yr.",
        "type": 1,
        "wrong_answers_1": "the time of submersion is 5.96 yr.",
        "wrong_answers_2": "the time of submersion is 4.74 yr.",
        "wrong_answers_3": "the time of submersion is 5.11 yr."
    },
    {
        "idx": 1338,
        "question": "Assuming that activation polarization controls both oxidation and reduction reactions, determine the rate of corrosion of metal M (in mol/cm^2-s). The following corrosion data are known about the metal and solution: For Metal M: V_(MM^2+) = -0.90 V, i_0 = 10^-12 A/cm^2, beta = +0.10. For Hydrogen: V_(H^-/H_2) = 0 V, i_0 = 10^-10 A/cm^2.",
        "answer": "the rate of corrosion of metal M is 3.27 x 10^-13 mol/cm^2-s.",
        "type": 1,
        "wrong_answers_1": "the rate of corrosion of metal M is 2.83 x 10^-13 mol/cm^2-s.",
        "wrong_answers_2": "the rate of corrosion of metal M is 3.63 x 10^-13 mol/cm^2-s.",
        "wrong_answers_3": "the rate of corrosion of metal M is 3.07 x 10^-13 mol/cm^2-s."
    },
    {
        "idx": 1339,
        "question": "Compute the corrosion potential for this reaction. The following corrosion data are known about the metal and solution: For Metal M: V_(MM^2+) = -0.90 V, i_0 = 10^-12 A/cm^2, beta = +0.10. For Hydrogen: V_(H^-/H_2) = 0 V, i_0 = 10^-10 A/cm^2.",
        "answer": "the corrosion potential for this reaction is -0.420 V.",
        "type": 1,
        "wrong_answers_1": "the corrosion potential for this reaction is -0.436 V.",
        "wrong_answers_2": "the corrosion potential for this reaction is -0.388 V.",
        "wrong_answers_3": "the corrosion potential for this reaction is -0.365 V."
    },
    {
        "idx": 1340,
        "question": "Why does chromium in stainless steels make them more corrosion resistant than plain carbon steels in many environments?",
        "answer": "The chromium in stainless steels causes a very thin and highly adherent surface coating to form over the surface of the alloy, which protects it from further corrosion. For plain carbon steels, rust, instead of this adherent coating, forms.",
        "type": 4,
        "wrong_answers_1": "The chromium in stainless steels causes a very thin and highly adherent surface coating to form over the surface of the alloy, which protects it from further corrosion. For plain carbon steels, rust, instead of this adherent coating, forms.",
        "wrong_answers_2": "The four classifications of steels are: Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, and High Alloy Steels (Stainless and Tool).",
        "wrong_answers_3": "Low Carbon Steels, Medium Carbon Steels, High Carbon Steels, High Alloy Steels (Stainless and Tool)"
    },
    {
        "idx": 1341,
        "question": "For a concentration cell, briefly explain why corrosion occurs at the region having the lower concentration.",
        "answer": "Inasmuch as \\left[\\mathrm{M}_{\\mathrm{L}}^{2+}\\right] \\prec\\left[\\mathrm{M}_{H}^{2+}\\right] then the natural logarithm of the \\left[\\mathrm{M}^{2+}\\right] ratio is negative, which yields a positive value for \\Delta V. This means that the electrochemical reaction is spontaneous as written, or that oxidation occurs at the electrode having the lower \\mathrm{M}^{2+} concentration.",
        "type": 4,
        "wrong_answers_1": "The softening point of a glass is that temperature at which the viscosity is 4 × 10^{6} \\mathrm{~Pa}·s; from Figure 13.7, these temperatures for the 96% silica, borosilicate, and soda-lime glasses are 1540^{\\circ} C\\left(2800^{\\circ} F\\right), 830^{\\circ} C\\left(1525^{\\circ} F\\right), and 700^{\\circ} C\\left(1290^{\\circ} F\\right), respectively.",
        "wrong_answers_2": "Chemical tempering will be accomplished by substitution, for \\mathrm{Na}^{+}, another monovalent cation with a slightly larger diameter. Both K^{+}and \\mathrm{Cs}^{+}fill these criteria, having ionic radii of 0.138 and 0.170 nm, respectively, which are larger than the ionic radius of \\mathrm{Na}^{+}(0.102nm). In fact, soda-lime glasses are tempered by a K^{+}-\\mathrm{Na}^{+}ion exchange.",
        "wrong_answers_3": "The alloying elements in tool steels (e.g., \\mathrm{Cr}, V, \\mathrm{W}, and Mo) combine with the carbon to form very hard and wear-resistant carbide compounds."
    },
    {
        "idx": 1342,
        "question": "For copper, the heat capacity at constant volume C_V at 20 K is 0.38 J/mol-K and the Debye temperature is 340 K. Estimate the specific heat at 40 K.",
        "answer": "the specific heat for copper at 40 K is 47.8 J/kg-K.",
        "type": 1,
        "wrong_answers_1": "the specific heat for copper at 42 K is 44.1 J/kg-K.",
        "wrong_answers_2": "the specific heat for copper at 45 K is 52.5 J/kg-K.",
        "wrong_answers_3": "the specific heat for copper at 46 K is 43.8 J/kg-K."
    },
    {
        "idx": 1343,
        "question": "For copper, the heat capacity at constant volume C_V at 20 K is 0.38 J/mol-K and the Debye temperature is 340 K. Estimate the specific heat at 400 K.",
        "answer": "the specific heat for copper at 400 K is 392 J/kg-K.",
        "type": 1,
        "wrong_answers_1": "the specific heat for copper at 340 K is 334 J/kg-K.",
        "wrong_answers_2": "the specific heat for copper at 358 K is 421 J/kg-K.",
        "wrong_answers_3": "the specific heat for copper at 445 K is 405 J/kg-K."
    },
    {
        "idx": 1344,
        "question": "A 0.4-m (15.7-in.) rod of a metal elongates 0.48mm (0.019 in.) on heating from 20^{\\circ} C to 100^{\\circ} C\\left(68^{\\circ} F\\right. to \\left.212^{\\circ} F\\right). Determine the value of the linear coefficient of thermal expansion for this material.",
        "answer": "the linear coefficient of thermal expansion for this material is 15.0 × 10^{-6} \\, ({}^{\\circ} C)^{-1} [8.40 × 10^{-6} \\, ({}^{\\circ} \\mathrm{f})^{-1}].",
        "type": 1,
        "wrong_answers_1": "the linear coefficient of thermal expansion for this material is 13.9 × 10^{-6} \\, ({}^{\\circ} C)^{-1} [7.78 × 10^{-6} \\, ({}^{\\circ} \\mathrm{f})^{-1}].",
        "wrong_answers_2": "the linear coefficient of thermal expansion for this material is 14.3 × 10^{-6} \\, ({}^{\\circ} C)^{-1} [9.48 × 10^{-6} \\, ({}^{\\circ} \\mathrm{f})^{-1}].",
        "wrong_answers_3": "the linear coefficient of thermal expansion for this material is 17.0 × 10^{-6} \\, ({}^{\\circ} C)^{-1} [7.72 × 10^{-6} \\, ({}^{\\circ} \\mathrm{f})^{-1}]."
    },
    {
        "idx": 1345,
        "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Pure silver; sterling silver (92.5 wt% Ag-7.5 wt% Cu)",
        "answer": "Pure silver will have a larger conductivity than sterling silver because the impurity atoms in the latter will lead to a greater degree of free electron scattering.",
        "type": 4,
        "wrong_answers_1": "Pure copper will have a larger conductivity than aluminum bronze because the impurity atoms in the latter will lead to a greater degree of free electron scattering.",
        "wrong_answers_2": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "wrong_answers_3": "The linear polyethylene will have the larger conductivity than the branched polyethylene because the former will have the higher degree of crystallinity. Linear polymers have higher degrees of crystallinity than branched polymers. Since heat transfer is accomplished by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity."
    },
    {
        "idx": 1346,
        "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Fused silica; polycrystalline silica",
        "answer": "Polycrystalline silica will have a larger conductivity than fused silica because fused silica is noncrystalline and lattice vibrations are more effectively scattered in noncrystalline materials.",
        "type": 4,
        "wrong_answers_1": "Quartz will have a larger conductivity than fused silica because fused silica is noncrystalline (whereas quartz is crystalline) and lattice vibrations are more effectively scattered in noncrystalline materials.",
        "wrong_answers_2": "Borosilicate glasses and fused silica are resistant to thermal shock because they have relatively low coefficients of thermal expansion; therefore, upon heating or cooling, the difference in the degree of expansion or contraction across a cross-section of a ware that is constructed from these materials will be relatively low.",
        "wrong_answers_3": "Thermal conductivities are higher for crystalline than for noncrystalline ceramics because, for noncrystalline, phonon scattering, and thus the resistance to heat transport, is much more effective due to the highly disordered and irregular atomic structure."
    },
    {
        "idx": 1347,
        "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Linear and syndiotactic poly(vinyl chloride) (DP =1000); linear and syndiotactic polystyrene (DP=1000)",
        "answer": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "type": 4,
        "wrong_answers_1": "The linear polyethylene will have the larger conductivity than the branched polyethylene because the former will have the higher degree of crystallinity. Linear polymers have higher degrees of crystallinity than branched polymers. Since heat transfer is accomplished by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "wrong_answers_2": "The isotactic polypropylene will have a larger thermal conductivity than the atactic polypropylene because isotactic polymers have higher degrees of crystallinity. The influence of crystallinity on conductivity is explained in part (c).",
        "wrong_answers_3": "No, it is not possible to determine which of the two polymers has the higher melting temperature. The syndiotactic polypropylene will have a higher degree of crystallinity than the atactic material. On the basis of this effect alone, the syndiotactic PP should have the greater Tm, since melting temperature increases with degree of crystallinity. However, the molecular weight for the syndiotactic polypropylene (500,000 g/mol) is less than for the atactic material (750,000 g/mol); a lowering of molecular weight generally results in a reduction of melting temperature."
    },
    {
        "idx": 1348,
        "question": "For the following pair of materials, decide which has the larger thermal conductivity. Justify your choice. Atactic polypropylene (Mw=10^6 g/mol); isotactic polypropylene (Mw=10^5 g/mol)",
        "answer": "The isotactic polypropylene will have a larger thermal conductivity than the atactic polypropylene because isotactic polymers have higher degrees of crystallinity. The influence of crystallinity on conductivity is explained in part (c).",
        "type": 4,
        "wrong_answers_1": "The linear polyethylene will have the larger conductivity than the branched polyethylene because the former will have the higher degree of crystallinity. Linear polymers have higher degrees of crystallinity than branched polymers. Since heat transfer is accomplished by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity.",
        "wrong_answers_2": "The alternating poly(styrene-butadiene) copolymer will have a higher crystallinity than the random copolymer; alternating copolymers crystallize more easily than random ones. The influence of crystallinity on conductivity is explained in part (c).",
        "wrong_answers_3": "The poly(vinyl chloride) will have the larger conductivity than the polystyrene because the former will have the higher degree of crystallinity. Both polymers are syndiotactic and have the same degree of polymerization. However, with regard to side-group bulkiness, the PVC is more likely to crystallize. Since heat transfer is by molecular chain vibrations, and the coordination of these vibrations increases with percent crystallinity, the higher the crystallinity, the greater the thermal conductivity."
    },
    {
        "idx": 1349,
        "question": "Briefly explain why thermal stresses may be introduced into a structure by rapid heating or cooling.",
        "answer": "Thermal stresses may be introduced into a structure by rapid heating or cooling because temperature gradients are established across the cross section due to more rapid temperature changes at the surface than within the interior; thus, the surface will expand or contract at a different rate than the interior and since this surface expansion or contraction will be restrained by the interior, stresses will be introduced.",
        "type": 4,
        "wrong_answers_1": "For cooling, the surface stresses will be tensile in nature since the interior contracts to a lesser degree than the cooler surface.",
        "wrong_answers_2": "Residual thermal stresses are introduced into a glass piece when it is cooled because surface and interior regions cool at different rates, and, therefore, contract different amounts; since the material will experience very little, if any deformation, stresses are established.",
        "wrong_answers_3": "For heating, the surface stresses will be compressive in nature since the interior expands to a lesser degree than the hotter surface."
    },
    {
        "idx": 1350,
        "question": "For cooling, what is the nature of the surface stresses?",
        "answer": "For cooling, the surface stresses will be tensile in nature since the interior contracts to a lesser degree than the cooler surface.",
        "type": 4,
        "wrong_answers_1": "For heating, the surface stresses will be compressive in nature since the interior expands to a lesser degree than the hotter surface.",
        "wrong_answers_2": "Thermal stresses may be introduced into a structure by rapid heating or cooling because temperature gradients are established across the cross section due to more rapid temperature changes at the surface than within the interior; thus, the surface will expand or contract at a different rate than the interior and since this surface expansion or contraction will be restrained by the interior, stresses will be introduced.",
        "wrong_answers_3": "Residual thermal stresses are introduced into a glass piece when it is cooled because surface and interior regions cool at different rates, and, therefore, contract different amounts; since the material will experience very little, if any deformation, stresses are established."
    },
    {
        "idx": 1351,
        "question": "For heating, what is the nature of the surface stresses?",
        "answer": "For heating, the surface stresses will be compressive in nature since the interior expands to a lesser degree than the hotter surface.",
        "type": 4,
        "wrong_answers_1": "For cooling, the surface stresses will be tensile in nature since the interior contracts to a lesser degree than the cooler surface.",
        "wrong_answers_2": "Thermal stresses may be introduced into a structure by rapid heating or cooling because temperature gradients are established across the cross section due to more rapid temperature changes at the surface than within the interior; thus, the surface will expand or contract at a different rate than the interior and since this surface expansion or contraction will be restrained by the interior, stresses will be introduced.",
        "wrong_answers_3": "Residual thermal stresses are introduced into a glass piece when it is cooled because surface and interior regions cool at different rates, and, therefore, contract different amounts; since the material will experience very little, if any deformation, stresses are established."
    },
    {
        "idx": 1352,
        "question": "The magnetic flux density within a bar of some material is 0.630 tesla at an H field of 5 x 10^5 A/m. Compute the magnetic permeability for this material.",
        "answer": "1.260 x 10^-6 h/m.",
        "type": 1,
        "wrong_answers_1": "1.178 x 10^-6 h/m.",
        "wrong_answers_2": "1.345 x 10^-6 h/m.",
        "wrong_answers_3": "1.157 x 10^-6 h/m."
    },
    {
        "idx": 1353,
        "question": "The magnetic flux density within a bar of some material is 0.630 tesla at an H field of 5 x 10^5 A/m. Compute the magnetic susceptibility for this material.",
        "answer": "2.387 x 10^-3.",
        "type": 1,
        "wrong_answers_1": "2.569 x 10^-3.",
        "wrong_answers_2": "2.602 x 10^-3.",
        "wrong_answers_3": "2.060 x 10^-3."
    },
    {
        "idx": 1354,
        "question": "The magnetic flux density within a bar of some material is 0.630 tesla at an H field of 5 x 10^5 A/m. What type(s) of magnetism would you suggest is (are) being displayed by this material? Why?",
        "answer": "This material would display both diamagnetic and paramagnetic behavior. All materials are diamagnetic, and since χm is positive and on the order of 10^-3, there would also be a paramagnetic contribution.",
        "type": 4,
        "wrong_answers_1": "No, it is not possible. Linear polymers have higher degrees of crystallization (and higher tensile moduli) than branched polymers - on this basis, the PVC material should have the higher value of E. On the other hand, PVC has a more complex repeat unit structure than does polyethylene, which means that, on this basis, the PE would have a higher degree of crystallinity and also a greater tensile modulus. Also, tensile modulus is relatively independent of number-average molecular weight. Therefore, this determination is not possible since it is not possible to determine which of the two materials has the greater degree of crystallinity.",
        "wrong_answers_2": "In order for a material to have an index of refraction less than unity, the velocity of light in the material (v) would necessarily have to be greater than the velocity of light in a vacuum . This is not possible.",
        "wrong_answers_3": "In order for a material to have an index of refraction less than unity, the velocity of light in the material (v) would necessarily have to be greater than the velocity of light in a vacuum. This is not possible."
    },
    {
        "idx": 1355,
        "question": "A net magnetic moment is associated with each atom in paramagnetic and ferromagnetic materials. Explain why ferromagnetic materials can be permanently magnetized whereas paramagnetic ones cannot.",
        "answer": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions - domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.For paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains.",
        "type": 4,
        "wrong_answers_1": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions--domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.\nFor paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains.",
        "wrong_answers_2": "All atoms do not have a net magnetic moment. If an atom has completely filled electron shells or subshells, there will be a cancellation of both orbital and spin magnetic moments.",
        "wrong_answers_3": "For type I superconductors, with increasing magnetic field the material is completely diamagnetic and superconductive below H_{C}, while at H_{C} conduction becomes normal and complete magnetic flux penetration takes place. On the other hand, for type II superconductors upon increasing the magnitude of the magnetic field, the transition from the superconducting to normal conducting states is gradual between lower-critical and upper-critical fields; so also is magnetic flux penetration gradual. Furthermore, type II superconductors generally have higher critical temperatures and critical magnetic fields."
    },
    {
        "idx": 1356,
        "question": "The chemical formula for copper ferrite may be written as \\left(\\mathrm{CuFe}_{2} \\mathrm{O}_{4}\\right)_{8} because there are eight formula units per unit cell. If this material has a saturation magnetization of 1.35 × 10^{5} A/m and a density of 5.40g /cm^3, estimate the number of Bohr magnetons associated with each Cu^{2+} ion.",
        "answer": "1.07 bohr magnetons/cu^{2+} ion",
        "type": 1,
        "wrong_answers_1": "1.22 bohr magnetons/cu^{2+} ion",
        "wrong_answers_2": "1.20 bohr magnetons/cu^{2+} ion",
        "wrong_answers_3": "1.22 bohr magnetons/cu^{2+} ion"
    },
    {
        "idx": 1357,
        "question": "The formula for samarium iron garnet (Sm3Fe5O12) may be written in the form Sm3^aFe2^cFe3^dO12, where the superscripts a, c, and d represent different sites on which the Sm^3+ and Fe^3+ ions are located. The spin magnetic moments for the Sm^3+ and Fe^3+ ions positioned in a and c sites are oriented parallel to one another and antiparallel to the Fe^3+ ions in d sites. Compute the number of Bohr magnetons associated with each Sm^3+ ion, given the following information: (1) each unit cell consists of eight formula (Sm3Fe5O12) units; (2) the unit cell is cubic with an edge length of 1.2529 nm; (3) the saturation magnetization for this material is 1.35 × 10^5 A/m; and (4) there are 5 Bohr magnetons associated with each Fe^3+ ion.",
        "answer": "2.86 bm.",
        "type": 1,
        "wrong_answers_1": "2.59 bm.",
        "wrong_answers_2": "3.18 bm.",
        "wrong_answers_3": "3.01 bm."
    },
    {
        "idx": 1358,
        "question": "An iron bar magnet having a coercivity of 7000 A/m is to be demagnetized. If the bar is inserted within a cylindrical wire coil 0.25m long and having 150 turns, what electric current is required to generate the necessary magnetic field?",
        "answer": "11.7A",
        "type": 1,
        "wrong_answers_1": "13.3A",
        "wrong_answers_2": "13.4A",
        "wrong_answers_3": "10.8A"
    },
    {
        "idx": 1359,
        "question": "Cite the differences between type I and type II superconductors.",
        "answer": "For type I superconductors, with increasing magnetic field the material is completely diamagnetic and superconductive below H_{C}, while at H_{C} conduction becomes normal and complete magnetic flux penetration takes place. On the other hand, for type II superconductors upon increasing the magnitude of the magnetic field, the transition from the superconducting to normal conducting states is gradual between lower-critical and upper-critical fields; so also is magnetic flux penetration gradual. Furthermore, type II superconductors generally have higher critical temperatures and critical magnetic fields.",
        "type": 4,
        "wrong_answers_1": "The Meissner effect is a phenomenon found in superconductors wherein, in the superconducting state, the material is diamagnetic and completely excludes any external magnetic field from its interior. In the normal conducting state complete magnetic flux penetration of the material occurs.",
        "wrong_answers_2": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions--domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.\nFor paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains.",
        "wrong_answers_3": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions - domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.For paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains."
    },
    {
        "idx": 1360,
        "question": "Briefly describe the Meissner effect.",
        "answer": "The Meissner effect is a phenomenon found in superconductors wherein, in the superconducting state, the material is diamagnetic and completely excludes any external magnetic field from its interior. In the normal conducting state complete magnetic flux penetration of the material occurs.",
        "type": 4,
        "wrong_answers_1": "For type I superconductors, with increasing magnetic field the material is completely diamagnetic and superconductive below H_{C}, while at H_{C} conduction becomes normal and complete magnetic flux penetration takes place. On the other hand, for type II superconductors upon increasing the magnitude of the magnetic field, the transition from the superconducting to normal conducting states is gradual between lower-critical and upper-critical fields; so also is magnetic flux penetration gradual. Furthermore, type II superconductors generally have higher critical temperatures and critical magnetic fields.",
        "wrong_answers_2": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions - domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.For paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains.",
        "wrong_answers_3": "Ferromagnetic materials may be permanently magnetized (whereas paramagnetic ones may not) because of the ability of net spin magnetic moments of adjacent atoms to align with one another. This mutual magnetic moment alignment in the same direction exists within small volume regions--domains. When a magnetic field is applied, favorably oriented domains grow at the expense of unfavorably oriented ones, by the motion of domain walls. When the magnetic field is removed, there remains a net magnetization by virtue of the resistance to movement of domain walls; even after total removal of the magnetic field, the magnetization of some net domain volume will be aligned near the direction that the external field was oriented.\nFor paramagnetic materials, there is no magnetic dipole coupling, and, consequently, domains do not form. When a magnetic field is removed, the atomic dipoles assume random orientations, and no magnetic moment remains."
    },
    {
        "idx": 1361,
        "question": "Visible light having a wavelength of 5 x 10^-7 m appears green. Compute the frequency of a photon of this light.",
        "answer": "the frequency of the photon is 6 x 10^14 s^-1.",
        "type": 1,
        "wrong_answers_1": "the frequency of the photon is 6 x 10^14 s^-1.",
        "wrong_answers_2": "the frequency of the photon is 7 x 10^14 s^-1.",
        "wrong_answers_3": "the frequency of the photon is 7 x 10^14 s^-1."
    },
    {
        "idx": 1362,
        "question": "Visible light having a wavelength of 5 x 10^-7 m appears green. Compute the energy of a photon of this light.",
        "answer": "the energy of the photon is 3.98 x 10^-19 j (2.48 ev).",
        "type": 1,
        "wrong_answers_1": "the energy of the photon is 3.77 x 10^-19 j (2.68 ev).",
        "wrong_answers_2": "the energy of the photon is 4.25 x 10^-19 j (2.21 ev).",
        "wrong_answers_3": "the energy of the photon is 4.48 x 10^-19 j (2.80 ev)."
    },
    {
        "idx": 1363,
        "question": "What are the characteristics of opaque materials in terms of their appearance and light transmittance?",
        "answer": "Opaque materials are impervious to light transmission; it is not possible to see through them.",
        "type": 4,
        "wrong_answers_1": "Opaque materials are impervious to light transmission; it is not possible to see through them.",
        "wrong_answers_2": "An opaque material is impervious to light transmission.",
        "wrong_answers_3": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them."
    },
    {
        "idx": 1364,
        "question": "What are the characteristics of translucent materials in terms of their appearance and light transmittance?",
        "answer": "Light is transmitted diffusely through translucent materials (there is some internal light scattering). Objects are not clearly distinguishable when viewed through a translucent material.",
        "type": 4,
        "wrong_answers_1": "Light is transmitted diffusely through translucent materials (there is some internal light scattering). Objects are not clearly distinguishable when viewed through a translucent material.",
        "wrong_answers_2": "A translucent material transmits light diffusely.",
        "wrong_answers_3": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them."
    },
    {
        "idx": 1365,
        "question": "What are the characteristics of transparent materials in terms of their appearance and light transmittance?",
        "answer": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them.",
        "type": 4,
        "wrong_answers_1": "Virtually all of the incident light is transmitted through transparent materials, and one can see clearly through them.",
        "wrong_answers_2": "Opaque materials are impervious to light transmission; it is not possible to see through them.",
        "wrong_answers_3": "Opaque materials are impervious to light transmission; it is not possible to see through them."
    },
    {
        "idx": 1366,
        "question": "Can a material have a positive index of refraction less than unity? Why or why not?",
        "answer": "In order for a material to have an index of refraction less than unity, the velocity of light in the material (v) would necessarily have to be greater than the velocity of light in a vacuum . This is not possible.",
        "type": 4,
        "wrong_answers_1": "In order for a material to have an index of refraction less than unity, the velocity of light in the material (v) would necessarily have to be greater than the velocity of light in a vacuum. This is not possible.",
        "wrong_answers_2": "The drift velocity of a free electron is the average electron velocity in the direction of the force imposed by an electric field.",
        "wrong_answers_3": "No, it is not possible. Linear polymers have higher degrees of crystallization (and higher tensile moduli) than branched polymers - on this basis, the PVC material should have the higher value of E. On the other hand, PVC has a more complex repeat unit structure than does polyethylene, which means that, on this basis, the PE would have a higher degree of crystallinity and also a greater tensile modulus. Also, tensile modulus is relatively independent of number-average molecular weight. Therefore, this determination is not possible since it is not possible to determine which of the two materials has the greater degree of crystallinity."
    },
    {
        "idx": 1367,
        "question": "The index of refraction of quartz is anisotropic. Suppose that visible light is passing from one grain to another of different crystallographic orientation and at normal incidence to the grain boundary. Calculate the reflectivity at the boundary if the indices of refraction for the two grains are 1.544 and 1.553 in the direction of light propagation.",
        "answer": "the reflectivity at the boundary is 8.45 × 10^{-6}.",
        "type": 1,
        "wrong_answers_1": "the reflectivity at the boundary is 8.81 × 10^{-6}.",
        "wrong_answers_2": "the reflectivity at the boundary is 8.77 × 10^{-6}.",
        "wrong_answers_3": "the reflectivity at the boundary is 7.70 × 10^{-6}."
    },
    {
        "idx": 1368,
        "question": "Zinc selenide has a band gap of 2.58 eV. Over what range of wavelengths of visible light is it transparent?",
        "answer": "pure znse is transparent to visible light having wavelengths between 0.48 and 0.7 \\mu m.",
        "type": 1,
        "wrong_answers_1": "pure znse is transparent to visible light having wavelengths between 0.45 and 0.7 \\mu m.",
        "wrong_answers_2": "pure znse is transparent to visible light having wavelengths between 0.42 and 0.6 \\mu m.",
        "wrong_answers_3": "pure znse is transparent to visible light having wavelengths between 0.51 and 0.6 \\mu m."
    }
]