Young modulus questions in A Level Physics (9702) are some of the most predictable on the paper. They ask you to define stress, strain or the Young modulus, calculate it from a force, an extension and a wire's dimensions, and link it to the spring constant and the energy stored. The physics is short. The marks are lost in unit conversions, in mixing up diameter and radius, and in definitions that are nearly right. This post goes through what recent mark schemes credit, works through one question in full, and gives you three real questions to try.
The definitions examiners credit
These are the wordings accepted in recent 9702 schemes:
| Term | Wording in the mark scheme |
|---|---|
| Strain | extension / original length |
| Young modulus | ratio of stress to strain, or stress per unit strain |
| Hooke's law | extension is proportional to (applied) force |
| Stress | force / area, from the working line \sigma = F/A |
| Spring constant | k = F/x, or the gradient of a force-extension graph |
Strain has no unit, because it is a length divided by a length. Stress and the Young modulus are both in pascals.
Putting the definitions together gives the equation you use most:
$$E = \frac{\sigma}{\varepsilon} = \frac{F/A}{x/L} = \frac{FL}{Ax}$$What the mark schemes reward
The equation, then the substitution. 9702 calculation marks are usually C marks for working and an A mark for the answer. In Paper 21, May/June 2024, the Young modulus calculation gave C1 for "Young modulus = stress / strain", C1 for the substitution, and A1 for the answer. Write the equation before the numbers.
Linking E and k. Rearranging E = FL/(Ax) with k = F/x gives
$$k = \frac{EA}{L}$$Paper 21, Oct/Nov 2025 asked you to show this, with C1 for "k = F/x" and A1 for the rearrangement. The same idea explains why a thicker sample of the same material is stiffer: Paper 23, May/June 2024 credited "the spring constant (of the second sample) will be greater", after a first mark for the reason, such as a greater force being needed for the same extension.
Energy from the graph. The area under a force-extension graph is the work done in stretching. Paper 23, Oct/Nov 2023 gave two marks for explaining this: the area represents the work done, and that work done equals the elastic potential energy. Within the limit of proportionality the graph is a straight line, so
$$E_P = \tfrac{1}{2}Fx = \tfrac{1}{2}kx^2$$Reading a force-extension graph
Many questions give you a graph and expect you to take values from it. Paper 23, May/June 2024, Question 3 is a good example of what is asked:
- Limit of proportionality. The point where the straight section ends. On that graph the scheme placed it at (60, 5.4), an extension of 60 mm and a force of 5.4 N.
- Elastic limit. The point beyond which the sample no longer returns to its original length. The question stated that the sample behaved elastically up to 80 mm, and the scheme placed it at (80, 5.9).
- Spring constant. The gradient of the straight section. The scheme's working was 5.4 / 0.060, giving 90 N m⁻¹. Note that 60 mm became 0.060 m.
- Young modulus. With a cross-sectional area of 0.40 mm², which is 4.0 \times 10^{-7} m², and a length of 3.2 m, the scheme's working was (5.4 \times 3.2) / (4.0 \times 10^{-7} \times 0.06), giving 7.2 \times 10^8 Pa.
- Work done up to breaking. The area under the whole line, including the curved part, so it is an estimate. The scheme accepted (1.0 \pm 0.2) J.
The limit of proportionality and the elastic limit are different points, and a question can ask for both on the same line.
Worked walk-through: Paper 21, May/June 2024, Question 4
This question is worth 8 marks and covers the core of the topic.
(a) Define strain, 1 mark. Extension divided by original length. "Change in length over length" is risky. Say "original".
(b)(i) Calculate the Young modulus, 3 marks. A copper wire of length 4.0 m and cross-sectional area 4.5 \times 10^{-7} m² extends by 1.4 mm under a tensile force of 18 N, up to its limit of proportionality.
Start with the equation (C1): Young modulus = stress / strain.
Work out each part, converting the extension to metres first:
$$\sigma = \frac{18}{4.5 \times 10^{-7}} = 4.0 \times 10^{7} \text{ Pa}$$$$\varepsilon = \frac{1.4 \times 10^{-3}}{4.0} = 3.5 \times 10^{-4}$$That substitution is the second C1. Then divide:
$$E = \frac{4.0 \times 10^{7}}{3.5 \times 10^{-4}} = 1.1 \times 10^{11} \text{ Pa}$$That is the A1. If you forget to convert 1.4 mm to metres, your answer is a thousand times too large. Copper is a metal, so an answer around 10^{11} Pa is sensible, and one around 10^{14} Pa should make you check the units.
(b)(ii) Draw the stress-strain line, 2 marks. One mark is for a straight line through the origin. The other is for ending it at the point (3.5, 4.0) on the grid's scales, which is the strain and stress you have just calculated. Up to the limit of proportionality stress is proportional to strain, so the line must be straight.
(c) Compare a thicker wire, 2 marks. A second copper wire has the same length but a larger diameter, and carries the same 18 N. Stress is less in the second wire, because the same force acts on a larger area. Strain is also less, because the material and its Young modulus are the same, so less stress means less strain. The table is worth two marks, and the scheme's answer is "less in second wire" in both rows.
Try the question yourself, including the graph, and see which marks your answer earns.
Common ways marks are lost
- Not converting units. Millimetres to metres, and square millimetres to square metres. Note that 1 mm² is 10^{-6} m², not 10^{-3} m².
- Using the diameter as the radius. Paper 21, Oct/Nov 2025 gave a diameter of 1.6 mm, and the scheme's working used a radius of 0.80 \times 10^{-3} m in \pi r^2.
- "Change in length over length". Strain uses the original length.
- Saying a thicker wire has a bigger Young modulus. The Young modulus belongs to the material. The spring constant changes with the dimensions.
- Forgetting the half in the energy. The area under a straight line from the origin is a triangle: \frac{1}{2}Fx, not Fx.
- Leaving out the equation. If the answer is wrong, the C marks for working are all you can collect.
Two more real questions to try
Paper 23, May/June 2026, Question 2 is worth 12 marks. A polythene thread hangs from one end. You use its density to show its length, show the stress at the top due to its own weight, and suggest why the stress varies linearly along it. Then the thread is laid flat with a uniform stress, and you calculate its extension and the elastic potential energy, before deciding whether the hanging thread stretched more or less. That last part is a reasoning mark: the scheme says less, because the stress in the hanging thread is less than the top value everywhere except at the top. For the "suggest an explanation" part, the method in how to answer explain questions in 9702 applies.
Paper 21, Oct/Nov 2025, Question 3 joins Young modulus to resistivity. You define the Young modulus, show that k = EA/L, sketch how resistance and spring constant vary with force, then use a copper wire's data to find its length and its spring constant, given as 6.9 \times 10^4 N m⁻¹ in the scheme.
You do not need an account to try these: as a guest, each question part can be marked twice for free. Write by hand or type, and the C and A marks are shown against your working.
Practise more stress and strain questions
The free 9702 stress and strain topical page lists 39 questions from 30 papers, from 2023 to 2026, each linked to its paper. Most are multiple choice, and 15 are structured. The neighbouring elastic and plastic behaviour page covers the energy and elastic limit questions. For full papers, open the A Level Physics past papers and choose a Paper 2.
Frequently asked questions
- What is the definition of the Young modulus for 9702?
- Recent 9702 mark schemes accept the ratio of stress to strain, or stress per unit strain. Stress is force per unit cross-sectional area, and strain is extension divided by original length.
- Does the Young modulus change if the wire is thicker?
- No. The Young modulus is a property of the material. A thicker wire of the same material has the same Young modulus, but under the same force it has less stress and less strain, and its spring constant is greater.
- How do I get the elastic potential energy from a force-extension graph?
- Find the area under the graph, which is the work done in stretching. Within the limit of proportionality the graph is a straight line through the origin, so the area is a triangle and the energy is one half times force times extension.
- What units should I use in Young modulus calculations?
- Use SI units throughout: force in newtons, area in square metres, extension and length in metres. Convert millimetres to metres and square millimetres to square metres before you substitute, and use the radius, not the diameter, when you calculate a cross-sectional area. The answer is in pascals.