AQA AS Level Physics Paper 2, June 2023: Question 27

1 mark · Medium difficulty · Multiple Choice

Determine the Young modulus of a material from its stress-strain graph.

Practise this question

Question

Multiple choice question 27 shows a stress-strain graph for a material with stress on the y-axis (ranging from 0 to 3.8 GPa) and strain on the x-axis (ranging from 0 to 0.030). The curve is initially linear before starting to bend towards a maximum stress of 3.6 GPa at a strain of 0.030. Below the graph, the question asks for the Young modulus of the material, with four multiple-choice options A, B, C, and D giving values in Pa.
Question text

27 The graph shows the variation of stress with strain for a material.

What is the Young modulus of the material?

[1 mark]

A 1.2 × 105 Pa

B 1.5 × 105 Pa

C 1.2 × 1011 Pa

D 1.5 × 1011 Pa

Mark scheme

Show the mark scheme The mark scheme table shows question 27 with the correct answer D and the corresponding value of 1.5 x 10^11 Pa.

27 D 1.5 × 10 Pa

How to answer it

Determining Young Modulus from a Stress-Strain Graph

What this question tests

This question assesses your ability to interpret graphical data involving material deformation. Specifically, it tests your understanding that the Young modulus is represented by the gradient of the linear region of a stress-strain graph, as well as your competence in handling standard form prefixes (like Giga, G ).

Question 27 Analysis

Exam Part: Multiple Choice [1 mark]

✅ Correct Answer

D ( 1.5 × 10¹¹ Pa )

Mark Awarded: 1 / 1 for selecting option D.

💡 Key Knowledge

  • Definition: Young modulus E is defined as tensile stress divided by tensile strain ( E = σ / ε ).
  • Graphical Link: Since stress is on the y-axis and strain is on the x-axis, the gradient ( Δy / Δx ) of the straight-line portion equals the Young modulus.
  • Prefix Conversion: 1 GPa = 1 × 10⁹ Pa . Failing to convert this prefix is the single most common reason students pick the wrong power of ten.

🧠 Exam Technique

  • Choose a large triangle along the straight, linear section of the graph to minimise reading uncertainties.
  • Avoid using the curved region at the top where Hooke's Law no longer strictly applies.
  • Always check the axes labels carefully for multiplier prefixes like / GPa .

❌ Common Errors

  • Power of Ten Traps: Forgetting the 10⁹ multiplier for GPa leads to an answer scaled incorrectly by a factor of a million (e.g., choosing option B).
  • Reading the Wrong Region: Attempting to calculate the gradient after the curve flattens out (past strain 0.015 ), which no longer gives the Young modulus.

📐 Step-by-Step Calculation

  1. Identify linear coordinates: At strain ( ε ) = 0.000 , stress ( σ ) = 0.0 GPa . At strain ( ε ) = 0.020 , the straight line reaches a stress ( σ ) of 3.0 GPa . (Note: You can use any clear point on the straight line segment, e.g., strain = 0.010, stress = 1.5 GPa).
  2. Apply the gradient formula:
    E = Δσ / Δε = (3.0 × 10⁹ Pa - 0) / (0.020 - 0)
  3. Compute the value:
    E = (3.0 × 10⁹) / 0.020 = 150,000,000,000 Pa
  4. Convert to standard form:
    1.5 × 10¹¹ Pa , which matches option D.

Topics

Physics · Practical skills · 3.4 Mechanics and materials · Data analysis

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.