A Level Physics 6.1 Stress and Strain MCQs – 30 Practice Questions with Answers (Cambridge 9702)

Free practice MCQs for Cambridge International AS & A Level Physics (9702), Section 6.1 Stress and Strain (Topic 6: Deformation of Solids). Every question is original, with a fully worked answer – just tap “Show Answer” to check yourself. Unless stated otherwise, take g = 10 N/kg.

Formula recap: stress σ = F / A · strain ε = ΔL / L · Young modulus E = σ / ε = FL / (AΔL). See every equation in context on our A Level Physics Formula Sheet.

Stress

Q1. Tensile stress in a wire is defined as

  • (A) the force multiplied by the cross-sectional area
  • (B) the force per unit cross-sectional area
  • (C) the extension per unit original length
  • (D) the force per unit original length
Show Answer

Answer: B – Stress σ = F / A: the tensile force divided by the cross-sectional area of the wire. Option C is the definition of strain, not stress.

Q2. The SI unit of stress is

  • (A) N m
  • (B) kg m−3
  • (C) Pa (N m−2)
  • (D) J m−3
Show Answer

Answer: C – Stress = force / area, so its unit is N / m2 = N m−2, the pascal (Pa). This is the same unit as pressure, because pressure is also a force per unit area.

Q3. A tensile force of 250 N acts on a wire of cross-sectional area 5.0 × 10−6 m2. The tensile stress in the wire is

  • (A) 5.0 × 107 Pa
  • (B) 1.3 × 103 Pa
  • (C) 2.0 × 10−8 Pa
  • (D) 5.0 × 105 Pa
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Answer: A – σ = F / A = 250 / (5.0 × 10−6) = 5.0 × 107 Pa. Option B comes from multiplying F by A instead of dividing.

Q4. A wire of diameter 2.0 mm carries a tensile force of 100 N. The stress in the wire is approximately

  • (A) 7.9 × 106 Pa
  • (B) 32 Pa
  • (C) 1.6 × 107 Pa
  • (D) 3.2 × 107 Pa
Show Answer

Answer: D – First convert the diameter: 2.0 mm = 2.0 × 10−3 m, so the radius r = 1.0 × 10−3 m. Area A = πr2 = π × (1.0 × 10−3)2 = 3.14 × 10−6 m2. Stress = 100 / (3.14 × 10−6) = 3.2 × 107 Pa. Option A is the classic trap: using the diameter as the radius makes the area 4 times too big, so the stress comes out 4 times too small.

Q5. Which statement about tensile and compressive stress is correct?

  • (A) Tensile stress is caused by forces that squash a material so its length decreases
  • (B) Compressive stress is measured in newtons, tensile stress in pascals
  • (C) Tensile stress arises from pulling forces that tend to increase the length of a material; compressive stress arises from pushing forces that tend to decrease its length
  • (D) A material can experience tensile stress but never compressive stress
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Answer: C – Both are calculated as force per unit area and both are measured in pascals. They differ only in the direction of the applied forces: pulling apart (tension) or pushing together (compression).

Q6. Two wires X and Y are loaded with the same tensile force. Wire Y has half the cross-sectional area of wire X. Compared with the stress in wire X, the stress in wire Y is

  • (A) twice as large
  • (B) the same
  • (C) half as large
  • (D) four times as large
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Answer: A – σ = F / A. With F fixed, halving A doubles the stress. This is why a thinner wire reaches its breaking stress under a much smaller load.

Q7. The ultimate tensile stress of a material is

  • (A) the stress at which the material starts to deform plastically
  • (B) the stress when the extension is exactly 1.0 mm
  • (C) the maximum strain the material can withstand
  • (D) the maximum tensile stress the material can withstand before it breaks
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Answer: D – The ultimate tensile stress (breaking stress) is the peak stress on the stress-strain graph – the greatest stress the material survives before fracture.

Q8. A metal rod of diameter 4.0 mm breaks when a tensile force of 6.3 kN is applied. The breaking stress of the metal is approximately

  • (A) 1.3 × 108 Pa
  • (B) 5.0 × 108 Pa
  • (C) 5.0 × 105 Pa
  • (D) 2.0 × 109 Pa
Show Answer

Answer: B – Radius r = 2.0 × 10−3 m, so A = πr2 = π × (2.0 × 10−3)2 = 1.26 × 10−5 m2. Breaking stress = 6300 / (1.26 × 10−5) = 5.0 × 108 Pa. Note the force must also be converted: 6.3 kN = 6300 N.

Q9. A cable of cross-sectional area 2.0 × 10−5 m2 supports a load of mass 200 kg. The tensile stress in the cable is

  • (A) 1.0 × 105 Pa
  • (B) 1.0 × 107 Pa
  • (C) 1.0 × 108 Pa
  • (D) 4.0 × 108 Pa
Show Answer

Answer: C – The force is the weight: F = mg = 200 × 10 = 2000 N. Stress = 2000 / (2.0 × 10−5) = 1.0 × 108 Pa. Forgetting to convert mass to weight (using 200 instead of 2000) gives option B.

Q10. Which wire experiences the greatest tensile stress?

  • (A) force 400 N, area 2.0 × 10−6 m2
  • (B) force 800 N, area 8.0 × 10−6 m2
  • (C) force 100 N, area 1.0 × 10−6 m2
  • (D) force 600 N, area 4.0 × 10−6 m2
Show Answer

Answer: A – Work out each stress: A: 400 / 2.0 × 10−6 = 2.0 × 108 Pa; B: 800 / 8.0 × 10−6 = 1.0 × 108 Pa; C: 100 / 1.0 × 10−6 = 1.0 × 108 Pa; D: 600 / 4.0 × 10−6 = 1.5 × 108 Pa. The biggest force (option B) does not give the biggest stress – the area matters just as much.

Strain

Q11. Tensile strain is defined as

  • (A) the extension multiplied by the original length
  • (B) the extension divided by the original length
  • (C) the force divided by the cross-sectional area
  • (D) the original length divided by the extension
Show Answer

Answer: B – Strain ε = ΔL / L, the extension (change in length) as a fraction of the original length. Option C is stress.

Q12. Strain has no unit because

  • (A) extensions are always measured in metres
  • (B) it is too small a quantity to need a unit
  • (C) the units of force and area cancel out
  • (D) it is a ratio of two lengths, so the units cancel
Show Answer

Answer: D – Strain = extension / original length = (length) / (length). The units are identical top and bottom, so they cancel, leaving a pure number.

Q13. A wire of original length 1.5 m extends by 3.0 mm when loaded. The strain in the wire is

  • (A) 2.0 × 10−3
  • (B) 2.0
  • (C) 4.5 × 10−3
  • (D) 5.0 × 10−4
Show Answer

Answer: A – Convert the extension: 3.0 mm = 3.0 × 10−3 m. Strain = (3.0 × 10−3) / 1.5 = 2.0 × 10−3. Option B is the trap: dividing 3.0 by 1.5 without converting mm to m.

Q14. A specimen has a tensile strain of 4.0 × 10−3. Expressed as a percentage, this strain is

  • (A) 4.0%
  • (B) 0.0040%
  • (C) 0.40%
  • (D) 40%
Show Answer

Answer: C – Percentage strain = strain × 100 = 4.0 × 10−3 × 100 = 0.40%. The specimen has stretched by 0.40% of its original length.

Q15. A wire of original length 2.5 m extends by 1.25 mm under load. The strain is

  • (A) 5.0 × 10−3
  • (B) 5.0 × 10−4
  • (C) 2.0 × 10−3
  • (D) 3.1 × 10−3
Show Answer

Answer: B – Strain = (1.25 × 10−3) / 2.5 = 5.0 × 10−4. Again, the extension must be in metres before dividing.

Q16. A cable of length 4.0 m is subjected to a strain of 2.0 × 10−3. Its extension is

  • (A) 0.80 mm
  • (B) 2.0 mm
  • (C) 0.50 mm
  • (D) 8.0 mm
Show Answer

Answer: D – Extension = strain × original length = 2.0 × 10−3 × 4.0 = 8.0 × 10−3 m = 8.0 mm.

Q17. Two wires of the same material are stretched by the same amount (same extension). Wire P has an original length of 1.0 m and wire Q an original length of 3.0 m. Which statement is correct?

  • (A) Wire P has the larger strain
  • (B) Both wires have the same strain
  • (C) Wire Q has the larger strain
  • (D) The strains are equal to the extensions
Show Answer

Answer: A – Strain = extension / original length. The extension is the same for both, so the shorter wire (P) has the larger strain – its length has changed by a bigger fraction.

Q18. A material with Young modulus 2.0 × 1011 Pa is under a tensile stress of 1.0 × 108 Pa (within its limit of proportionality). The strain is

  • (A) 2.0 × 103
  • (B) 2.0 × 10−4
  • (C) 5.0 × 10−4
  • (D) 5.0 × 10−3
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Answer: C – Since E = σ / ε, strain ε = σ / E = (1.0 × 108) / (2.0 × 1011) = 5.0 × 10−4.

Q19. A concrete pillar of height 3.0 m shortens by 0.60 mm under a compressive load. The compressive strain in the pillar is

  • (A) 5.0 × 10−4
  • (B) 2.0 × 10−4
  • (C) 0.20
  • (D) 1.8 × 10−3
Show Answer

Answer: B – Strain is calculated the same way in compression, using the change in length: (0.60 × 10−3) / 3.0 = 2.0 × 10−4.

Q20. Which of the following could be the strain in a stretched wire?

  • (A) 1.5 × 10−3 m
  • (B) 1.5 × 10−3 Pa
  • (C) 1.5 × 10−3 N m−2
  • (D) 1.5 × 10−3
Show Answer

Answer: D – Strain is a dimensionless ratio, so it is written as a pure number with no unit at all. Options B and C have the unit of stress.


The Young Modulus

Q21. The Young modulus of a material is defined as

  • (A) stress × strain
  • (B) force × extension
  • (C) stress / strain, within the limit of proportionality
  • (D) the maximum stress before the material breaks
Show Answer

Answer: C – E = σ / ε. It is constant for a material only while stress is proportional to strain (Hooke’s law region), which is why the definition includes the limit of proportionality.

Q22. The SI unit of the Young modulus is

  • (A) Pa
  • (B) N
  • (C) no unit
  • (D) J
Show Answer

Answer: A – E = stress / strain. Strain is dimensionless, so E has the same unit as stress: the pascal (Pa). Values are huge, so they are often quoted in GPa (109 Pa).

Q23. In a Young modulus experiment, a wire of length 2.0 m and diameter 0.40 mm extends by 2.0 mm when a force of 25 N is applied. The Young modulus of the wire material is approximately

  • (A) 2.0 × 109 Pa
  • (B) 5.0 × 1010 Pa
  • (C) 8.0 × 1010 Pa
  • (D) 2.0 × 1011 Pa
Show Answer

Answer: D – Radius r = 0.20 mm = 2.0 × 10−4 m, so A = πr2 = π × (2.0 × 10−4)2 = 1.26 × 10−7 m2. Stress = 25 / (1.26 × 10−7) = 2.0 × 108 Pa. Strain = (2.0 × 10−3) / 2.0 = 1.0 × 10−3. E = (2.0 × 108) / (1.0 × 10−3) = 2.0 × 1011 Pa – the value for steel. Using the diameter 0.40 mm as the radius is the classic error here: it makes A four times too large and E four times too small.

Q24. The straight-line (elastic) region of a stress-strain graph passes through the point where the stress is 1.5 × 108 Pa and the strain is 7.5 × 10−4. The Young modulus of the material is

  • (A) 1.1 × 105 Pa
  • (B) 2.0 × 1011 Pa
  • (C) 5.0 × 10−12 Pa
  • (D) 2.0 × 1010 Pa
Show Answer

Answer: B – The Young modulus is the gradient of the stress-strain graph: E = σ / ε = (1.5 × 108) / (7.5 × 10−4) = 2.0 × 1011 Pa.

Q25. A material with a very large Young modulus is best described as

  • (A) stiff – it needs a very large stress to produce a small strain
  • (B) weak – it breaks under a very small stress
  • (C) stretchy – it produces a large strain for a small stress
  • (D) ductile – it can be drawn into long wires
Show Answer

Answer: A – E = σ / ε, so a large E means a large stress is needed per unit strain. Steel (E ≈ 2.0 × 1011 Pa) is stiff; rubber (E ≈ 107 Pa) is not. Note that stiffness (E) is not the same as strength (breaking stress).

Q26. A steel wire (E = 2.0 × 1011 Pa) of length 3.0 m and cross-sectional area 1.0 × 10−6 m2 is stretched by a force of 200 N. The extension of the wire is

  • (A) 0.30 mm
  • (B) 30 mm
  • (C) 3.0 mm
  • (D) 0.030 mm
Show Answer

Answer: C – Stress = 200 / (1.0 × 10−6) = 2.0 × 108 Pa. Strain = σ / E = (2.0 × 108) / (2.0 × 1011) = 1.0 × 10−3. Extension = strain × length = 1.0 × 10−3 × 3.0 = 3.0 × 10−3 m = 3.0 mm. (Or in one step: ΔL = FL / (AE).)

Q27. In Searle’s apparatus for measuring the Young modulus, two wires are hung side by side from the same support: the test wire, which carries the loads, and an identical control (reference) wire. The purpose of the control wire is to

  • (A) double the extension so it is easier to measure
  • (B) compensate for any extension caused by temperature changes (and sagging of the support), so only the load’s effect is measured
  • (C) carry half the load so the test wire does not break
  • (D) keep the test wire vertical
Show Answer

Answer: B – The vernier scale measures the movement of the test wire relative to the control wire. If the room warms up, both wires expand equally, so the relative reading is unaffected – temperature effects cancel out.

Q28. A wire of length 1.0 m and cross-sectional area 2.0 × 10−6 m2 extends by 0.50 mm under a force of 40 N. The Young modulus of the material is

  • (A) 4.0 × 107 Pa
  • (B) 1.0 × 1010 Pa
  • (C) 2.5 × 1010 Pa
  • (D) 4.0 × 1010 Pa
Show Answer

Answer: D – E = FL / (AΔL) = (40 × 1.0) / (2.0 × 10−6 × 0.50 × 10−3) = 40 / (1.0 × 10−9) = 4.0 × 1010 Pa. Remember to convert the extension: 0.50 mm = 0.50 × 10−3 m.

Q29. Two materials X and Y are tested in tension. For the same stress, the strain in X is half the strain in Y. It follows that

  • (A) the Young modulus of X is half that of Y
  • (B) the Young modulus of X equals that of Y
  • (C) the Young modulus of X is twice that of Y
  • (D) material X must break at a lower stress than Y
Show Answer

Answer: C – E = σ / ε. With the same σ, a strain half as big means E is twice as big: EX = 2EY. X is the stiffer material. (Nothing can be concluded about breaking stress from this information alone.)

Q30. The Young modulus of a material depends on

  • (A) the material only – not on the length, thickness or shape of the specimen
  • (B) the length of the specimen only
  • (C) the cross-sectional area of the specimen only
  • (D) both the length and the cross-sectional area of the specimen
Show Answer

Answer: A – E is a property of the material itself. A longer wire extends more and a thicker wire extends less, but when F, A, ΔL and L are combined as FL / (AΔL), the same value of E is always obtained for the same material (at the same temperature).

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