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rather than the final length itself."},{"text":"final length","correct":false,"explanation":"The final length includes the original length as well as the extension. The correct term is increase in length, which is the change from the original length."},{"text":"decrease in length","correct":false,"explanation":"A decrease in length describes compression rather than tensile extension. The correct term is increase in length, measured relative to the original length."},{"text":"original length","correct":false,"explanation":"The original length is the reference length, not the change produced by the force. The correct term is increase in length compared with that original value."}]},{"id":"ae8933d4-8c6c-4213-a661-e362ea93098f","cardType":"cloze","points":null,"front":"Hooke’s law is [ ] and applies before the [ ] is exceeded.","back":"Hooke’s law is $F = kx$ and applies before the limit of proportionality is exceeded.","mcqOptions":[{"text":"Hooke’s law is $F = kx$ and applies before the limit of proportionality is exceeded.","correct":true,"explanation":"Hooke’s law states that force equals spring constant multiplied by extension, and this proportional relationship holds only up to the limit of proportionality."},{"text":"Hooke’s law is $F = kx$ and applies before the elastic limit is exceeded.","correct":false,"explanation":"The elastic limit concerns whether an object returns to its original shape, whereas proportionality can fail earlier. The correct boundary for Hooke’s law is the limit of proportionality."},{"text":"Hooke’s law is $F = rac{x}{k}$ and applies before the limit of proportionality is exceeded.","correct":false,"explanation":"Rearranging the relationship incorrectly puts the spring constant in the denominator. The correct equation is $F = kx$, valid before the limit of proportionality."},{"text":"Hooke’s law is $F = kx$ and applies before the breaking point is exceeded.","correct":false,"explanation":"Breaking occurs much later than the point where force and extension stop being proportional. The correct term is limit of proportionality, where the straight-line relationship ends."}]},{"id":"ad89b4ee-2ac4-4fd3-be0c-04b89028185f","cardType":"basic","points":1,"front":"On a Hooke’s law force-extension graph, what does the gradient give?","back":"The spring constant, $k$.","mcqOptions":[{"text":"The spring constant, $k$.","correct":true,"explanation":"On a force-extension graph, the gradient is force divided by extension, $F/x$. From $F = kx$, this ratio equals the spring constant, $k$."},{"text":"The extension, $x$.","correct":false,"explanation":"Extension is read from the horizontal axis, not found from the gradient. The gradient is $F/x$, which equals the spring constant, $k$."},{"text":"The Young modulus, $E$.","correct":false,"explanation":"Young modulus is obtained from stress divided by strain for a material, not directly from a force-extension graph. The gradient here gives the spring constant, $k$."},{"text":"The elastic limit, $F$.","correct":false,"explanation":"The elastic limit is identified from where permanent deformation begins, not from the gradient. The gradient of the force-extension line gives the spring constant, $k$."}]},{"id":"b238ea04-6aab-4716-9442-11955b04507c","cardType":"basic","points":1,"front":"If extension is used in millimetres in $F = kx$, what happens to $k$?","back":"$$k$ is wrong by a factor of 1000.","mcqOptions":[{"text":"$$k$ is wrong by a factor of 1000.","correct":true,"explanation":"The extension must be converted from millimetres to metres before using $F = kx$. Failing to divide by 1000 makes the calculated spring constant wrong by a factor of 1000."},{"text":"$$k$ is wrong by a factor of 100.","correct":false,"explanation":"There are 1000 millimetres in a metre, not 100. Using millimetres without conversion therefore makes $k$ wrong by a factor of 1000."},{"text":"$$k$ is wrong by a factor of 10.","correct":false,"explanation":"The factor is not based on centimetre conversion; it comes from 1000 millimetres per metre. Thus using millimetres directly makes $k$ wrong by a factor of 1000."},{"text":"$$k$ is wrong by a factor of 0.001.","correct":false,"explanation":"This reverses the conversion factor. Since one metre equals 1000 millimetres, using millimetres directly produces a value 1000 times too large, so the error factor is 1000."}]},{"id":"648c9d2d-71dc-4425-9da3-ee1672610b98","cardType":"basic","points":2,"front":"A spring extends by 24 mm under 3.6 N. What is $k$?","back":"$$k = \\dfrac{3.6}{2.4 \\times 10^{-2}} = \\boldsymbol{150\\text{ N m}^{-1}}$.","mcqOptions":[{"text":"$$k = \\dfrac{3.6}{2.4 \\times 10^{-2}} = \\boldsymbol{150\\text{ N m}^{-1}}$.","correct":true,"explanation":"The extension is $24\\text{ mm} = 2.4 \\times 10^{-2}\\text{ m}$. Substituting into $k = F/x$ gives $3.6/(2.4 \\times 10^{-2}) = 150\\text{ N m}^{-1}$."},{"text":"$$k = \\dfrac{3.6}{2.4 \\times 10^{-3}} = \\boldsymbol{1500\\text{ N m}^{-1}}$.","correct":false,"explanation":"The millimetre-to-metre conversion has an extra factor of ten: $24\\text{ mm} = 2.4 \\times 10^{-2}\\text{ m}$, not $2.4 \\times 10^{-3}\\text{ m}$. The correct result is $150\\text{ N m}^{-1}$."},{"text":"$$k = \\dfrac{3.6}{2.4 \\times 10^{-2}} = \\boldsymbol{15\\text{ N m}^{-1}}$.","correct":false,"explanation":"This loses a factor of ten when dividing by the converted extension. Using $k = F/x$ correctly gives $3.6/(2.4 \\times 10^{-2}) = 150\\text{ N m}^{-1}$."},{"text":"$$k = \\dfrac{2.4 \\times 10^{-2}}{3.6} = \\boldsymbol{6.7 \\times 10^{-3}\\text{ N m}^{-1}}$.","correct":false,"explanation":"This reverses $F/x$ and calculates extension per force instead of force per extension. The spring constant is $k = F/x = 150\\text{ N m}^{-1}$."}]},{"id":"72233657-e9fc-49e3-a235-e81ab487c391","cardType":"basic","points":1,"front":"What equation gives tensile stress from force and area?","back":"$\\sigma = \\dfrac{F}{A}$.","mcqOptions":[{"text":"$\\sigma = \\dfrac{F}{A}$.","correct":true,"explanation":"Tensile stress is defined as the tensile force divided by the cross-sectional area over which it acts, so $\\sigma = F/A$."},{"text":"$\\sigma = \\dfrac{A}{F}$.","correct":false,"explanation":"This reverses the definition and gives area per force. Tensile stress is force distributed over area, so the correct equation is $\\sigma = F/A$."},{"text":"$\\sigma = \\dfrac{\\Delta l}{l}$.","correct":false,"explanation":"This expression defines tensile strain, not stress. Tensile stress depends on force and cross-sectional area, using $\\sigma = F/A$."},{"text":"$\\sigma = F A$.","correct":false,"explanation":"Multiplying force by area gives the wrong dependence and units. Stress is force divided by area, so the correct equation is $\\sigma = F/A$."}]},{"id":"e46444b9-dac9-412e-acd3-e92a377e205f","cardType":"cloze","points":null,"front":"Tensile strain is [ ] and has [ ].","back":"Tensile strain is $\\varepsilon = \\dfrac{\\Delta l}{l}$ and has no unit.","mcqOptions":[{"text":"Tensile strain is $\\varepsilon = \\dfrac{\\Delta l}{l}$ and has no unit.","correct":true,"explanation":"Tensile strain is the extension divided by the original length. Both quantities are lengths, so their ratio is dimensionless and has no unit."},{"text":"Tensile strain is $\\varepsilon = \\dfrac{l}{\\Delta l}$ and has no unit.","correct":false,"explanation":"This ratio is the inverse of strain. Strain is extension divided by original length, $\\Delta l/l$, and remains dimensionless with no unit."},{"text":"Tensile strain is $\\varepsilon = \\dfrac{\\Delta l}{l}$ and has units of metres.","correct":false,"explanation":"Although the equation is correct, metres cancel because both terms are lengths. The correct description is $\\Delta l/l$ with no unit."},{"text":"Tensile strain is $\\varepsilon = \\dfrac{F}{A}$ and has units of pascals.","correct":false,"explanation":"$$F/A$ defines tensile stress, whose unit is the pascal. Tensile strain is $\\Delta l/l$, a dimensionless ratio with no unit."}]},{"id":"b8705d04-3bbe-4359-93e8-3b4412b1925d","cardType":"basic","points":1,"front":"What does a large Young modulus mean about a material?","back":"It is stiff: large stress gives small strain.","mcqOptions":[{"text":"It is stiff: large stress gives small strain.","correct":true,"explanation":"A stiff material undergoes little strain for a given stress. Since Young modulus is stress divided by strain, a large Young modulus indicates stiffness."},{"text":"It is flexible: large stress gives small strain.","correct":false,"explanation":"Small strain under a given stress indicates stiffness, not flexibility. A large Young modulus therefore means the material is stiff."},{"text":"It is stiff: small stress gives large strain.","correct":false,"explanation":"Large strain from small stress describes a compliant or flexible material. A large Young modulus means large stress is needed to produce a small strain."},{"text":"It is brittle: large stress gives small strain.","correct":false,"explanation":"Brittleness describes fracture behaviour and is not defined by Young modulus. A large Young modulus specifically indicates stiffness: large stress produces small strain."}]},{"id":"81a9626f-1d91-4712-a363-79228220b79b","cardType":"cloze","points":null,"front":"On a stress-strain graph, the gradient of the straight elastic region is the [ ].","back":"On a stress-strain graph, the gradient of the straight elastic region is the Young modulus.","mcqOptions":[{"text":"Young modulus","correct":true,"explanation":"Young modulus is the ratio of tensile stress to tensile strain within the linear elastic region, so it equals the gradient of a stress-strain graph there."},{"text":"spring constant","correct":false,"explanation":"The spring constant is the gradient of a force-extension graph, not a stress-strain graph. The straight elastic region of a stress-strain graph has gradient equal to Young modulus."},{"text":"elastic limit","correct":false,"explanation":"The elastic limit is a point beyond which deformation may become permanent, rather than a graph gradient. The gradient in the straight elastic region is Young modulus."},{"text":"tensile strength","correct":false,"explanation":"Tensile strength is the maximum tensile stress a material withstands, not the gradient of its graph. The straight elastic gradient represents Young modulus."}]},{"id":"735f476e-d91c-407a-9d3e-475fd139c46f","cardType":"basic","points":1,"front":"What equation gives the cross-sectional area of a wire from diameter?","back":"$A = \\dfrac{\\pi d^2}{4}$.","mcqOptions":[{"text":"$A = \\dfrac{\\pi d^2}{4}$.","correct":true,"explanation":"A wire has a circular cross-section with radius $d/2$. Substituting this into $A=\\pi r^2$ gives $A=\\pi(d/2)^2=\\pi d^2/4$."},{"text":"$A = \\pi d^2$.","correct":false,"explanation":"This uses the diameter directly as the radius, making the calculated area four times too large. The radius is $d/2$, so $A=\\pi d^2/4$."},{"text":"$A = \\dfrac{\\pi d^2}{2}$.","correct":false,"explanation":"This squares the factor of two incorrectly when substituting $r=d/2$. Since $(d/2)^2=d^2/4$, the correct area is $A=\\pi d^2/4$."},{"text":"$A = \\dfrac{\\pi d}{2}$.","correct":false,"explanation":"This expression resembles a circumference calculation and does not square the diameter, so it cannot represent area. A circular wire has $A=\\pi d^2/4$."}]},{"id":"0bdf8023-aa9a-4e91-aab9-80d99c5efc44","cardType":"basic","points":1,"front":"What graph area equals the work done deforming a solid?","back":"Area under the force-extension graph.","mcqOptions":[{"text":"Area under the force-extension graph.","correct":true,"explanation":"Work done is force multiplied by each small extension, so summing these contributions gives the area under a force-extension graph."},{"text":"Gradient of the force-extension graph.","correct":false,"explanation":"The gradient of a force-extension graph gives the spring constant in its linear region, not work done. Work done is the area under the graph."},{"text":"Area under the stress-strain graph.","correct":false,"explanation":"A stress-strain graph’s area represents energy transferred per unit volume, not the total work done. Total work is the area under the force-extension graph."},{"text":"Gradient of the stress-strain graph.","correct":false,"explanation":"The gradient of a stress-strain graph gives Young modulus in the linear elastic region. Work done deforming the solid is the area under the force-extension graph."}]},{"id":"f52eb9b5-b148-43fb-88a0-59708896cbb4","cardType":"cloze","points":null,"front":"For Hooke’s law stretching, strain energy is [ ] because the graph area is a [ ].","back":"For Hooke’s law stretching, strain energy is $W = \\dfrac{1}{2}Fx$ because the graph area is a triangle.","mcqOptions":[{"text":"For Hooke’s law stretching, strain energy is $W = \\dfrac{1}{2}Fx$ because the graph area is a triangle.","correct":true,"explanation":"For Hooke’s law, force rises linearly from zero to $F$, making a triangular force-extension area. Therefore the stored strain energy is $W=\\frac{1}{2}Fx$."},{"text":"For Hooke’s law stretching, strain energy is $W = Fx$ because the graph area is a rectangle.","correct":false,"explanation":"Using $Fx$ treats the force as constant at its final value and gives a rectangular area. Hooke’s law produces a triangle, so $W=\\frac{1}{2}Fx$."},{"text":"For Hooke’s law stretching, strain energy is $W = \\dfrac{1}{2}Fx$ because the graph area is a rectangle.","correct":false,"explanation":"The energy formula is correct, but a linear force-extension graph encloses a triangle, not a rectangle. The triangular area gives $W=\\frac{1}{2}Fx$."},{"text":"For Hooke’s law stretching, strain energy is $W = 2Fx$ because the graph area is a triangle.","correct":false,"explanation":"A triangular graph area requires the factor of one-half, not two. Because force increases linearly from zero to $F$, the strain energy is $W=\\frac{1}{2}Fx$."}]},{"id":"c9c5e747-d833-4483-9fb1-4f5f4e75f518","cardType":"basic","points":1,"front":"For a curved force-extension graph, how should strain energy be found?","back":"Estimate the area under the curve.","mcqOptions":[{"text":"Estimate the area under the curve.","correct":true,"explanation":"For a changing force, strain energy is the area under the force-extension curve. Because the boundary is curved, that area must be estimated graphically or integrated."},{"text":"Multiply final force by the extension.","correct":false,"explanation":"Multiplying final force by extension assumes that force remained at its final value, producing a rectangular area. Strain energy must instead be estimated from the area under the curve."},{"text":"Estimate the gradient along the curve.","correct":false,"explanation":"Using the gradient would determine how stiffness varies, rather than the transferred energy. Strain energy is found by estimating the area under the force-extension curve."},{"text":"Use $\\dfrac{1}{2}Fx$ for the curve.","correct":false,"explanation":"The triangular formula applies only when force is proportional to extension. For a curved relationship, strain energy must be found by estimating the actual area under the curve."}]},{"id":"7ddf9f02-dde6-4b74-8852-71fae6ac902f","cardType":"cloze","points":null,"front":"[ ] returns to original shape after unloading; [ ] leaves permanent deformation.","back":"Elastic deformation returns to original shape after unloading; plastic deformation leaves permanent deformation.","mcqOptions":[{"text":"Elastic deformation returns to original shape after unloading; plastic deformation leaves permanent deformation.","correct":true,"explanation":"Elastic deformation is reversible when the load is removed, whereas plastic deformation involves irreversible structural changes and therefore leaves the material permanently deformed."},{"text":"Plastic deformation returns to original shape after unloading; elastic deformation leaves permanent deformation.","correct":false,"explanation":"The two behaviours have been reversed. Elastic deformation disappears after unloading, while plastic deformation remains and produces a permanent change in shape."},{"text":"Brittle fracture returns to original shape after unloading; plastic deformation leaves permanent deformation.","correct":false,"explanation":"Brittle fracture means breaking with little plastic deformation, not returning to the original shape. Elastic deformation is reversible, while plastic deformation leaves a permanent change."},{"text":"Elastic deformation returns to original shape after unloading; ductile deformation leaves permanent deformation.","correct":false,"explanation":"Elastic deformation is correctly identified as reversible, but ductile deformation describes substantial plastic deformation before fracture rather than permanent deformation itself. The required term is plastic deformation."}]},{"id":"03da252c-a7cd-4a67-94c5-abc2bc21a507","cardType":"basic","points":1,"front":"On a ductile metal graph, what happens in the yield region?","back":"A small force increase causes a large extension increase.","mcqOptions":[{"text":"A small force increase causes a large extension increase.","correct":true,"explanation":"In the yield region, dislocations move readily through the metal lattice, allowing a large increase in extension for only a small increase in applied force."},{"text":"A large force increase causes a small extension increase.","correct":false,"explanation":"A small extension increase for a large force increase describes a stiff response, not yielding. During yield, a small force increase produces a large extension increase."},{"text":"A proportional force increase causes a proportional extension increase.","correct":false,"explanation":"A proportional force-extension increase characterises the linear Hooke’s law region before yielding. In the yield region, extension increases greatly for only a small force increase."},{"text":"A small force increase causes a large extension decrease.","correct":false,"explanation":"A decrease in extension during increasing load would imply contraction rather than tensile yielding. Yielding produces a large extension increase for a comparatively small force increase."}]},{"id":"393755bd-5c66-4f1f-a97e-c435c709c0dd","cardType":"basic","points":1,"front":"What microscopic process causes most plastic deformation in metals?","back":"Dislocations move through the crystal lattice.","mcqOptions":[{"text":"Dislocations move through the crystal lattice.","correct":true,"explanation":"Plastic deformation in metals occurs mainly when dislocations move through the crystal lattice, allowing atomic planes to slip without requiring all bonds across a plane to break simultaneously."},{"text":"Atoms vibrate within the crystal lattice.","correct":false,"explanation":"Atomic vibration increases with temperature but does not usually produce permanent deformation by itself. Most plastic deformation occurs because dislocations move through the crystal lattice."},{"text":"Atomic bonds stretch within the crystal lattice.","correct":false,"explanation":"Elastic bond stretching is reversible and accounts for elastic deformation, not lasting shape change. Plastic deformation mainly occurs when dislocations move through the crystal lattice."},{"text":"Grain boundaries move through the crystal lattice.","correct":false,"explanation":"Grain boundaries can impede dislocation motion and strengthen a metal rather than provide the principal deformation mechanism. Plastic deformation mainly results from dislocations moving through the lattice."}]},{"id":"85ebe000-03f5-4fcd-be5e-a24cfc1f3d81","cardType":"basic","points":2,"front":"Why do barriers to dislocation movement strengthen metals?","back":"A larger stress is needed for plastic deformation.","mcqOptions":[{"text":"A larger stress is needed for plastic deformation.","correct":true,"explanation":"Obstacles impede dislocations, so a greater applied stress is required to move them through the lattice and produce permanent plastic deformation."},{"text":"A larger strain is needed for elastic deformation.","correct":false,"explanation":"A larger strain is an outcome of deformation, not the cause of strengthening. Barriers strengthen metals because a larger stress is needed to move dislocations and initiate plastic deformation."},{"text":"A smaller stress is needed for plastic deformation.","correct":false,"explanation":"Reducing the stress needed for plastic deformation would make a metal weaker. Dislocation barriers increase the stress required for dislocations to move through the lattice."},{"text":"A larger stress is needed for elastic deformation.","correct":false,"explanation":"Dislocation barriers affect permanent deformation rather than requiring greater stress for reversible deformation. They strengthen metals because a larger stress is needed for plastic deformation."}]},{"id":"9ec6a27a-34fb-42e7-8771-310d2addf97c","cardType":"cloze","points":null,"front":"A crystalline solid has a [ ]; an amorphous solid has no [ ].","back":"A crystalline solid has a repeating lattice; an amorphous solid has no long-range repeating structure.","mcqOptions":[{"text":"A crystalline solid has a repeating lattice; an amorphous solid has no long-range repeating structure.","correct":true,"explanation":"Crystalline solids possess an ordered arrangement that repeats over long distances. Amorphous solids have only short-range order and lack a long-range repeating structure."},{"text":"A crystalline solid has a random arrangement; an amorphous solid has no long-range repeating structure.","correct":false,"explanation":"Crystalline solids are not randomly arranged; their particles occupy a repeating lattice. Amorphous solids are the materials that lack a long-range repeating structure."},{"text":"A crystalline solid has a repeating lattice; an amorphous solid has no short-range local order.","correct":false,"explanation":"Amorphous solids can possess short-range order between neighbouring particles. What they lack is long-range repeating structure, while a crystalline solid has a repeating lattice."},{"text":"A crystalline solid has a disordered structure; an amorphous solid has no random particle arrangement.","correct":false,"explanation":"The descriptions have been reversed. A crystalline solid has a repeating lattice, whereas an amorphous solid has no long-range repeating structure."}]},{"id":"acfcbf5b-2ea3-407b-a38e-298a2622eb8d","cardType":"basic","points":2,"front":"What mechanism often causes brittle fracture in glass?","back":"Crack propagation from stress concentration at crack tips.","mcqOptions":[{"text":"Crack propagation from stress concentration at crack tips.","correct":true,"explanation":"Stress is strongly concentrated at a crack tip, allowing the crack to extend rapidly through glass with little plastic deformation and produce brittle fracture."},{"text":"Dislocation movement from stress concentration at crack tips.","correct":false,"explanation":"Dislocation movement produces plastic deformation and is limited in glass. Brittle fracture instead commonly occurs through crack propagation driven by stress concentration at crack tips."},{"text":"Elastic deformation from stress concentration at crack tips.","correct":false,"explanation":"Elastic deformation is reversible and does not itself describe fracture. Glass usually fails when concentrated stress at a crack tip causes rapid crack propagation."},{"text":"Work hardening from stress concentration at crack tips.","correct":false,"explanation":"Work hardening is associated with dislocation interactions in metals, not brittle glass fracture. In glass, stress concentration at crack tips promotes crack propagation."}]},{"id":"68b65024-21ac-4fba-ba98-90d04f11429a","cardType":"basic","points":1,"front":"How does toughened glass increase breaking stress?","back":"Its surface is put under compression.","mcqOptions":[{"text":"Its surface is put under compression.","correct":true,"explanation":"Surface compression means an applied tensile stress must first overcome the compressive stress before surface cracks can open and propagate, increasing the breaking stress."},{"text":"Its surface is put under tension.","correct":false,"explanation":"Surface tension would help cracks open and therefore reduce resistance to fracture. Toughened glass is strengthened by placing its surface under compression."},{"text":"Its interior is put under compression.","correct":false,"explanation":"Compressing only the interior does not directly suppress the opening of surface flaws. Toughened glass increases breaking stress by putting its surface under compression."},{"text":"Its interior is put under tension.","correct":false,"explanation":"Tension in the interior accompanies the toughening process but is not the feature that inhibits surface cracks. The surface is placed under compression."}]},{"id":"77ac4a1f-ff6c-4f8d-b54e-ef34ceab95ae","cardType":"basic","points":1,"front":"What happens to rubber’s long-chain molecules when stretched?","back":"They uncoil and straighten.","mcqOptions":[{"text":"They uncoil and straighten.","correct":true,"explanation":"Rubber contains coiled, tangled polymer chains. Stretching makes these long-chain molecules uncoil and become straighter and more aligned with the applied force."},{"text":"They break and shorten.","correct":false,"explanation":"Stretching rubber primarily changes the arrangement of polymer chains rather than breaking them. The long-chain molecules uncoil and straighten, allowing substantial extension."},{"text":"They coil and contract.","correct":false,"explanation":"Further coiling would shorten the rubber rather than extend it. When rubber is stretched, its long-chain molecules uncoil and straighten along the applied force."},{"text":"They tangle and randomise.","correct":false,"explanation":"The chains become more ordered rather than randomly arranged during stretching. Rubber extends because its long-chain molecules uncoil and straighten."}]},{"id":"5860e4d2-ddc8-41c8-970e-2eee683c4ab2","cardType":"basic","points":1,"front":"What does the area inside a rubber hysteresis loop represent?","back":"Energy dissipated to surroundings, mostly thermal.","mcqOptions":[{"text":"Energy dissipated to surroundings, mostly thermal.","correct":true,"explanation":"The loop area is the difference between energy supplied during loading and energy returned during unloading. This energy is dissipated mainly as thermal energy."},{"text":"Energy stored in the rubber, mostly elastic.","correct":false,"explanation":"Stored elastic energy corresponds to an area under a loading or unloading curve, not the enclosed loop. The loop represents energy dissipated to the surroundings, mainly thermally."},{"text":"Energy gained from surroundings, mostly mechanical.","correct":false,"explanation":"The rubber does not gain mechanical energy over a complete loading-unloading cycle. The loop area represents energy dissipated to the surroundings, mostly as thermal energy."},{"text":"Force retained after unloading, mostly elastic.","correct":false,"explanation":"The hysteresis loop measures an energy difference, not the force remaining after unloading. Its area represents energy dissipated to the surroundings, mainly as thermal energy."}]},{"id":"007b50f1-d5ab-42c3-bd20-ba9f2f8e8f4e","cardType":"basic","points":1,"front":"If a wire’s force-extension gradient is $G$, what equation gives $E$?","back":"$E = \\dfrac{Gl}{A}$.","mcqOptions":[{"text":"$E = \\dfrac{Gl}{A}$.","correct":true,"explanation":"For a wire, $G=F/\\Delta l$. Combining this with $E=(F/A)/(\\Delta l/l)$ gives $E=Gl/A$."},{"text":"$E = \\dfrac{GA}{l}$.","correct":false,"explanation":"This inverts the required length factor. Substituting $G=F/\\Delta l$ into the Young modulus definition gives $E=Gl/A$, not $GA/l$."},{"text":"$E = GlA$.","correct":false,"explanation":"Area must divide rather than multiply because stress is $F/A$. Combining stress and strain with the force-extension gradient gives $E=Gl/A$."},{"text":"$E = \\dfrac{A}{Gl}$.","correct":false,"explanation":"This is the reciprocal of the correct relationship. Using $E=(F/A)/(\\Delta l/l)$ and $G=F/\\Delta l$ gives $E=Gl/A$."}]},{"id":"f3d74fa2-9238-4f00-ace4-10f827e95329","cardType":"basic","points":2,"front":"Why is wire diameter measurement especially important for Young modulus?","back":"$$A \\propto d^2$, so area % uncertainty is about twice diameter % uncertainty.","mcqOptions":[{"text":"$$A \\propto d^2$, so area % uncertainty is about twice diameter % uncertainty.","correct":true,"explanation":"A wire’s cross-sectional area is $A=\\pi d^2/4$. For a squared quantity, the percentage uncertainty is approximately doubled, so diameter uncertainty strongly affects Young modulus."},{"text":"$$A \\propto d$, so area % uncertainty is about equal to diameter % uncertainty.","correct":false,"explanation":"Cross-sectional area does not vary linearly with diameter; it is proportional to $d^2$. Consequently, its percentage uncertainty is about twice the diameter percentage uncertainty."},{"text":"$$A \\propto d^2$, so area % uncertainty is about equal to diameter % uncertainty.","correct":false,"explanation":"Although the proportionality is correct, the uncertainty factor is not. Squaring a measured diameter approximately doubles its percentage uncertainty rather than leaving it unchanged."},{"text":"$$A \\propto d^2$, so area % uncertainty is about half diameter % uncertainty.","correct":false,"explanation":"The uncertainty relationship has been reversed. Because area depends on the square of diameter, area percentage uncertainty is about twice, not half, diameter percentage uncertainty."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Component 1"},"heading":"Solids Under Stress","title":"Solids Under Stress Flashcards","topbar":{"backHref":"/a-level/physics/eduqas/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"5b8debf5-c737-5bac-b0f4-d675fe6da998","walled":"$L39"}],"$L3a","$L3b"]