A Level Physics 6.2 Elastic and Plastic Behaviour MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice MCQs for Cambridge International AS & A Level Physics (9702), Section 6.2 Elastic and Plastic Behaviour, written for the 2025-2027 syllabus. Work through all 30 questions and open the answer panels to check the worked explanations.
Formula recap: Hooke’s law: F = kx. Spring constant k = F / x = gradient of the force-extension graph. Elastic potential energy: E = 0.5 F x = 0.5 k x2 = area under the force-extension graph. Strain energy per unit volume = 0.5 × stress × strain = area under the stress-strain graph. Keep every equation in one place with the A Level Physics Formula Sheet.
Elastic & Plastic Behaviour
Q1. Hooke’s law states that the extension of a spring or wire is
- (A) proportional to its original length
- (B) proportional to the applied force, provided the limit of proportionality is not exceeded
- (C) equal to the applied force divided by the cross-sectional area
- (D) proportional to the applied force for all values of the force
Show Answer
Answer: B – Hooke’s law says extension x is proportional to force F (F = kx), but only up to the limit of proportionality. Beyond that point the proportionality fails.
Q2. A wire is loaded gradually. Point P on its force-extension graph is the limit of proportionality and point E is the elastic limit. Which statement is correct?
- (A) E is reached at a smaller load than P
- (B) Beyond P, the wire no longer returns to its original length when unloaded
- (C) Between P and E, the extension is proportional to the load
- (D) P is reached first; between P and E the wire still behaves elastically, but the extension is not proportional to the load
Show Answer
Answer: D – The elastic limit lies slightly beyond the limit of proportionality. Between the two points Hooke’s law no longer holds, yet the wire still returns to its original length when the load is removed; permanent deformation only begins beyond E.
Q3. During the elastic deformation of a metal wire, the atoms of the metal
- (A) are displaced slightly from their equilibrium positions and return to those positions when the load is removed
- (B) slide past each other into completely new positions
- (C) break away from the lattice altogether
- (D) remain permanently further apart after unloading
Show Answer
Answer: A – In elastic deformation the interatomic bonds are stretched but not broken. The atoms are pulled slightly from their equilibrium separations and spring back when the load is removed, so there is no permanent change of length.
Q4. A metal undergoes plastic deformation because
- (A) its atoms vibrate with a larger amplitude
- (B) every interatomic bond in the metal is broken at once
- (C) planes of atoms slide over one another and stay in their new positions
- (D) its atoms are displaced but always return to their original positions
Show Answer
Answer: C – Under a large enough stress, whole planes of atoms slip over neighbouring planes. The atoms settle into new equilibrium positions, so the change of shape remains after the load is removed.
Q5. A spring extends by 3.0 cm when a force of 6.0 N is applied to it. The spring obeys Hooke’s law. What extension is produced by a force of 10 N?
- (A) 3.6 cm
- (B) 5.0 cm
- (C) 6.0 cm
- (D) 18 cm
Show Answer
Answer: B – k = F / x = 6.0 / 0.030 = 200 N m−1. Then x = F / k = 10 / 200 = 0.050 m = 5.0 cm. (Or simply scale up: 3.0 cm × 10 / 6.0 = 5.0 cm.)
Q6. Just before a ductile wire fractures, necking occurs. Necking means that
- (A) the wire becomes uniformly thinner along its whole length
- (B) the stress in the wire falls to zero
- (C) the wire returns to its original length
- (D) the cross-sectional area decreases sharply at one localised point along the wire
Show Answer
Answer: D – After the ultimate tensile stress is reached, the wire narrows rapidly at one weak point. The stress concentrates at this neck until the wire finally snaps there.
Q7. The elastic limit of a wire is best defined as the
- (A) maximum load that can be applied so that the wire still returns to its original length when the load is removed
- (B) load at which the wire breaks
- (C) load beyond which the extension is no longer proportional to the load
- (D) load at which necking begins
Show Answer
Answer: A – Up to the elastic limit all deformation is recoverable. Option (C) defines the limit of proportionality, which is a different (and slightly smaller) load.
Q8. A wire is loaded beyond its elastic limit and the load is then completely removed. Afterwards, the length of the wire is
- (A) exactly equal to its original length
- (B) shorter than its original length
- (C) greater than its original length, by a permanent extension
- (D) reduced to zero
Show Answer
Answer: C – Loading past the elastic limit causes plastic deformation. On unloading, only the elastic part of the extension is recovered, leaving a permanent extension (a permanent set).
Q9. A rubber band can be stretched to several times its original length and still returns to its original length when released, but its force-extension graph is a curve rather than a straight line. Rubber is therefore best described as
- (A) plastic but not elastic
- (B) a material that obeys Hooke’s law
- (C) a brittle material
- (D) elastic, but not obeying Hooke’s law
Show Answer
Answer: D – Elastic means the material recovers its original size and shape on unloading; it does not require proportionality. Rubber recovers fully (elastic) but its extension is not proportional to the load, so it does not obey Hooke’s law.
Q10. A steel spring and a straight steel wire are both loaded within the limit of proportionality, and both obey Hooke’s law. The main difference in their behaviour is that
- (A) the spring stores no elastic potential energy
- (B) the wire behaves plastically while the spring behaves elastically
- (C) the spring shows a much larger extension for the same load because of its coiled shape
- (D) the extension of the wire is not proportional to the load
Show Answer
Answer: C – In a spring, the applied force twists and bends the coiled wire, giving a large, easily measured extension (a small spring constant). A straight wire of the same steel extends only by a tiny amount for the same load, but both extensions are proportional to the load.
Force-Extension Graphs & Strain Energy
Q11. The gradient of a force-extension graph for a spring is equal to the
- (A) elastic potential energy stored in the spring
- (B) Young modulus of the spring material
- (C) spring constant k
- (D) strain in the spring
Show Answer
Answer: C – From F = kx, the gradient F / x is the spring constant k, in N m−1. The area under the graph, not the gradient, gives the stored energy.
Q12. The force-extension graph of a wire is a straight line through the origin passing through the point (extension 2.0 mm, force 50 N). What is the spring constant of the wire?
- (A) 2.5 × 104 N m−1
- (B) 25 N m−1
- (C) 4.0 × 10−5 N m−1
- (D) 2.5 × 103 N m−1
Show Answer
Answer: A – k = F / x = 50 / (2.0 × 10−3) = 50 / 0.0020 = 2.5 × 104 N m−1. The extension must first be converted from mm to m.
Q13. A force of 12 N stretches a spring by 3.0 cm. What is the elastic potential energy stored in the spring?
- (A) 0.36 J
- (B) 36 J
- (C) 0.72 J
- (D) 0.18 J
Show Answer
Answer: D – E = 0.5 F x = 0.5 × 12 × 0.030 = 0.18 J. The factor 0.5 appears because the force grows from zero as the spring is stretched; 0.36 J (option A) wrongly uses the full force over the whole extension.
Q14. A spring of spring constant 400 N m−1 is stretched by 5.0 cm. What is the elastic potential energy stored?
- (A) 1000 J
- (B) 0.50 J
- (C) 1.0 J
- (D) 20 J
Show Answer
Answer: B – E = 0.5 k x2 = 0.5 × 400 × (0.050)2 = 0.5 × 400 × 0.0025 = 0.50 J.
Q15. On a force-extension graph, the area under the line up to a chosen extension is counted as 48 grid squares. Each square represents 0.010 J. What is the work done in stretching the spring to this extension?
- (A) 0.48 J
- (B) 4.8 J
- (C) 48 J
- (D) 0.048 J
Show Answer
Answer: A – The area under a force-extension graph equals the work done (the stored elastic potential energy): 48 × 0.010 J = 0.48 J.
Q16. A metal wire is loaded in its elastic region with a stress of 2.0 × 108 Pa and a strain of 1.0 × 10−3. What is the elastic strain energy stored per unit volume?
- (A) 2.0 × 105 J m−3
- (B) 2.0 × 1011 J m−3
- (C) 1.0 × 105 J m−3
- (D) 1.0 × 1011 J m−3
Show Answer
Answer: C – Energy per unit volume = 0.5 × stress × strain = 0.5 × (2.0 × 108) × (1.0 × 10−3) = 1.0 × 105 J m−3. This is the area of the triangle under the stress-strain graph.
Q17. For a material loaded within its elastic region, the strain energy per unit volume is given by the area under which graph?
- (A) A force-extension graph
- (B) A load-time graph
- (C) A stress-extension graph
- (D) A stress-strain graph
Show Answer
Answer: D – Stress × strain has units of (N m−2) × (dimensionless) = J m−3, so the area under a stress-strain graph is energy per unit volume. The area under a force-extension graph gives the total energy for that particular specimen.
Q18. A wire is loaded into its plastic region and then unloaded. On the force-extension graph, the unloading line
- (A) follows the loading line exactly back to the origin
- (B) is parallel to the original elastic part of the loading line and meets the extension axis at a permanent extension
- (C) is horizontal, at the maximum force reached
- (D) has a steeper gradient than the initial loading line
Show Answer
Answer: B – During unloading only the elastic strain is recovered, so the unloading line has the same gradient (the same k) as the initial linear section. It reaches zero force at a non-zero extension: the permanent set left by the plastic deformation.
Q19. A spring of spring constant 250 N m−1 is compressed by 8.0 cm. What is the elastic potential energy stored in the spring?
- (A) 0.80 J
- (B) 1.6 J
- (C) 20 J
- (D) 0.16 J
Show Answer
Answer: A – E = 0.5 k x2 = 0.5 × 250 × (0.080)2 = 0.5 × 250 × 0.0064 = 0.80 J. The same formula applies to a compression as to an extension.
Q20. A spring is stretched elastically by a force F and stores energy E. The force is then doubled, staying within the limit of proportionality. What is the new stored energy?
- (A) E
- (B) 2E
- (C) 4E
- (D) 8E
Show Answer
Answer: C – Doubling F doubles the extension x as well. Since E = 0.5 F x (or 0.5 k x2), the energy increases by 2 × 2 = 4 times, giving 4E.
Materials: Brittle, Ductile & Polymeric
Q21. The stress-strain graph of a brittle material, such as glass, is best described as
- (A) a long curve with a large plastic region before fracture
- (B) a curve showing a very large strain even at small stress
- (C) a straight line followed by a long horizontal plastic section
- (D) a straight line that ends suddenly at fracture, with no plastic deformation
Show Answer
Answer: D – A brittle material obeys Hooke’s law almost up to its breaking point and then fractures suddenly at a small strain. There is no plastic region and no warning before failure.
Q22. Copper is described as a ductile material. This means that copper
- (A) fractures after only a very small extension
- (B) can be drawn out into a wire, undergoing a large plastic deformation before it fractures
- (C) returns to its original length after any load, however large
- (D) has the highest possible Young modulus of any metal
Show Answer
Answer: B – Ductile materials extend a great deal plastically before breaking, which is why metals such as copper can be drawn into wires and pressed into sheets.
Q23. A polymeric material, such as rubber, shows which stress-strain behaviour?
- (A) Very large strains produced by small stresses, with a non-linear (curved) graph
- (B) A straight line ending in sudden fracture at a very small strain
- (C) Only a very small strain, even under a very large stress
- (D) A long horizontal plastic plateau at exactly constant stress
Show Answer
Answer: A – Polymers stretch enormously (strains of 1 or more) under modest stresses as their long chain molecules uncoil and align. The graph is curved, and much of the deformation can be elastic.
Q24. The stress-strain graphs of three materials P, Q and R are drawn on the same axes. In the elastic region the graphs are straight lines: P has the steepest gradient and R the least steep. Which material is the stiffest (has the largest Young modulus)?
- (A) R
- (B) Q
- (C) P
- (D) It cannot be decided from the gradients alone
Show Answer
Answer: C – The Young modulus is the gradient of the linear part of a stress-strain graph. The steepest gradient (P) means the largest stress is needed for a given strain, so P is the stiffest.
Q25. The strength (ultimate tensile strength) of a material is read from its stress-strain graph as the
- (A) maximum stress reached on the graph
- (B) gradient of the straight section
- (C) strain at which fracture occurs
- (D) total area under the graph
Show Answer
Answer: A – The ultimate tensile strength is the peak stress the material can withstand. The gradient gives the Young modulus (stiffness), and the total area gives the toughness.
Q26. The toughness of a material is measured by the
- (A) gradient of its stress-strain graph
- (B) maximum stress on its stress-strain graph
- (C) breaking strain only
- (D) total area under its stress-strain graph up to fracture
Show Answer
Answer: D – Toughness is the energy absorbed per unit volume before fracture, which is the whole area under the stress-strain curve. A tough material combines high stress with a large strain to fracture; a brittle material has a tiny area and so is not tough.
Q27. Cast iron is suitable for an engine block but not for a lifting cable, because cast iron is
- (A) ductile and weak in tension
- (B) brittle: it fractures suddenly with little or no plastic deformation
- (C) a polymeric material
- (D) elastic up to extremely large strains
Show Answer
Answer: B – Like glass, cast iron has a stress-strain graph that is nearly a straight line ending in sudden fracture. A cable must stretch visibly (ductile behaviour) before failing, so cast iron would be dangerous.
Q28. Material X has a stress-strain graph that rises to a high stress and then continues through a long plastic region, fracturing at a strain of 0.35. Material Y has a straight-line graph that ends at a strain of 0.002. Which statement is correct?
- (A) X is brittle and Y is ductile
- (B) Y is tougher than X
- (C) X is ductile, and its much larger area under the graph makes it far tougher than Y
- (D) Y undergoes a large plastic deformation before breaking
Show Answer
Answer: C – The long plastic region and large breaking strain make X ductile. Toughness is the area under the graph; X has a high stress sustained over a large strain, so its area (and toughness) is far greater than that of the brittle material Y.
Q29. A ductile material is safer than a brittle material for the cable of a crane because the ductile material
- (A) never deforms under load
- (B) stretches visibly before it fails, giving warning that it is overloaded
- (C) always has a lower density
- (D) cannot corrode
Show Answer
Answer: B – A ductile cable undergoes obvious plastic extension (and necking) before fracture, so the overload can be spotted in time. A brittle cable would snap without any warning.
Q30. Samples of two materials are tested to fracture. Material A gives a straight-line stress-strain graph ending at a strain of 0.002. Material B extends to a strain of about 5 under a small stress and returns to its original length when unloaded. The correct conclusion is:
- (A) A is ductile and B is brittle
- (B) A and B are both brittle
- (C) A is polymeric and B is ductile
- (D) A is brittle and B is polymeric
Show Answer
Answer: D – A fractures at a tiny strain with no plastic region: brittle behaviour. B shows the huge, recoverable strains at low stress that are characteristic of a polymeric material such as rubber.
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