AQA A-Level Physics Paper 1, November 2020: Question 26

1 mark · Medium difficulty · Multiple Choice

Identify what the gradient of an extension against weight graph represents for a vertical wire of initial length L and cross-sectional area A.

Practise this question

Question

Multiple choice question 26. An experiment determines the Young modulus E of steel using a wire of length L and cross-sectional area A with weights suspended from it, showing a linear graph of extension against weight. Below are four options for what the gradient represents: A equals E, B equals 1 over E, C equals E A over L, and D equals L over E A.
Question text

26 An experiment is carried out to determine the Young modulus E of steel using a vertical

wire of initial length L and cross-sectional area A. Various weights are suspended from

the wire. A graph of extension against weight is plotted.

What does the gradient of the graph represent?

[1 mark]

A E

B

E

EA

C

L

L

D

EA

Mark scheme

Show the mark scheme Mark scheme indicating the correct answer for question 26 is option D.

26 D

How to answer it

Young Modulus Graph Analysis

What this question tests

This question assesses your ability to manipulate the definition of the Young modulus, link physical variables to graphical representations, and correctly identify gradients from plotted experimental data.

Question 26 • Multiple Choice [1 Mark]

Determining the Gradient of an Extension-Weight Graph

✅ Correct Answer: D

The correct option is D ( L / EA ).

💡 Key Knowledge

  • The formula for Young modulus is E = (F × L) / (A × ΔL) .
  • Tensile force F is equivalent to the applied weight W .
  • Extension ΔL is plotted on the y-axis, and weight W is on the x-axis.

🧠 Exam Technique

Always start by writing out the defining equation containing your variables. Rearrange it into the linear format y = mx to explicitly isolate the gradient ( m ).

❌ Common Errors

Many students confuse Young modulus ( E ) with the gradient of this graph or invert the fraction, choosing option C ( EA / L ) instead, which represents the gradient of a force-extension graph.

📐 Derivation Step-by-Step

  1. Start with the definition: E = (F × L) / (A × ΔL)
  2. Substitute force F for weight W : E = (W × L) / (A × ΔL)
  3. Rearrange to isolate extension ( ΔL ) on the y-axis: ΔL = (L / EA) × W
  4. Compare with the straight-line equation y = mx :
    • y = extension ( ΔL )
    • x = weight ( W )
    • Gradient m = L / EA
Examiner Note: Top-level students rapidly derive this relationship in their heads or on rough paper rather than guessing, ensuring they never mix up L/EA with EA/L .

Topics

Physics · Required Practicals · 3.4 Mechanics and materials · AS practicals (1–6)

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.