AQA AS Level Physics Paper 2, November 2021: Question 32

1 mark · Hard difficulty · Multiple Choice

Determine the Young modulus of a wire given its density, mass, initial length, tensile force, and extension.

Practise this question

Question

A multiple choice question asking for the Young modulus of a wire made from a material of density rho, mass m, initial length L, tensile force F, and extension delta L. Four options A, B, C, and D present different algebraic combinations of these variables.
Question text

32 A wire is made from a material of density ρ.

The wire has a mass m and an initial length L.

When the tensile force in the wire is F the extension of the wire is ΔL.

What is the Young modulus of the material?

[1 mark]

FρL2

A

m∆L

FL2

B

mρ∆L

Fρ

C

m∆L

FmL2

D

ρ∆L

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer is option A, which is the expression F rho L squared over m delta L.

FρL2

32 A

m L

How to answer it

Question 32: Young Modulus Algebraic Derivation

What this question tests

This question tests your ability to combine core definitions of material properties (Young modulus and density) and substitute variables algebraically to eliminate unknown dimensions like cross-sectional area (A) or volume (V).

Question Part 32

Determining the Correct Expression

✅ Correct Answer: A

E = (F * ρ * L²) / (m * ΔL)

Worth 1 mark. Requires fluent combination of multiple equations without making algebraic transposition errors.

💡 Key Knowledge Required

  • Young Modulus: E = (F * L) / (A * ΔL)
  • Density: ρ = m / V
  • Volume of a wire: V = A * L (treating the wire as a cylinder)

🧠 Exam Technique

Since the cross-sectional area A is not given in the question, your first priority in algebraic multiple-choice questions is to substitute out any unlisted variables using bridging equations.

❌ Common Errors

Students often forget that the volume of a wire is A * L . Mistakenly substituting V = m / ρ directly into the Young modulus equation without accounting for length dimensions leads straight to incorrect distractor options like B or C.

📐 Step-by-Step Derivation

  1. Start with the standard definition of the Young modulus:
    E = (F * L) / (A * ΔL)
  2. Express the volume of the wire using its cross-sectional area and initial length:
    V = A * L
  3. Relate volume to mass and density:
    ρ = m / V  ⇒  V = m / ρ
  4. Equate the two expressions for volume:
    A * L = m / ρ
  5. Rearrange to make cross-sectional area A the subject:
    A = m / (ρ * L)
  6. Substitute this expression for A back into the Young modulus equation:
    E = (F * L) / [ (m / (ρ * L)) * ΔL ]
  7. Simplify the compound fraction by bringing (ρ * L) up to the numerator:
    E = (F * L * ρ * L) / (m * ΔL) = (F * ρ * L²) / (m * ΔL)

Topics

Physics · 3.4 Mechanics and materials

Question and mark scheme from the AQA AS Level Physics examination, Paper 2, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.