AQA AS Level Physics Paper 2, June 2025: Question 21

Young Modulus of Joined Wires in Series

AQA AS Physics • Mechanics & Materials • Multiple Choice

What this question tests

This question tests your conceptual understanding of the definition of the Young modulus and whether you can distinguish between intensive material properties and extensive geometrical dimensions.

Recognising that the ratio of tensile stress to tensile strain defines the Young modulus ( E = σ / ε ).
Understanding that the Young modulus is a property of the material only, completely independent of wire dimensions (length, diameter, cross-sectional area).
Filtering out superfluous "distractor" numerical data in multiple-choice questions.

Question 21 Walkthrough

Analysis of stress-to-strain ratio for connected wires

✅ Correct Answer

B ( X_P )

Since stress / strain is the definition of the Young modulus ( E ), and both wires are made of the same material, their Young moduli are identical:
X_Q = X_P .

💡 Key Knowledge

Tensile stress ( σ ): Force per unit area, σ = F / A (units: Pa or N m⁻²).
Tensile strain ( ε ): Extension per unit length, ε = ΔL / L (dimensionless).
Young Modulus ( E ): E = σ / ε . It depends strictly on the material and its temperature, not on its length or cross-sectional area.

📐 Step-by-Step Breakdown

Identify the defined quantities:
The question states:
X_P = stress in P / strain of P
X_Q = stress in Q / strain of Q
Recall the physical definition:
By definition, stress / strain = E (the Young modulus). Therefore, X_P = E_P and X_Q = E_Q .
Apply the material condition:
The stem states explicitly: "P and Q are wires made from the same material."
Conclusion:
Because they share identical composition, E_Q = E_P . Hence, X_Q = X_P .

❌ Common Traps & Mistakes

Falling for the "red herrings": The examiners gave lengths ( L_Q = 2L_P ) and diameters ( d_Q = 2d_P ) purely to bait students into spending time computing ratios.
Confusing Young Modulus with Stiffness: Stiffness ( k = F / ΔL ) depends on dimensions ( k = EA / L ), whereas the Young modulus ( E ) does not.
Calculating diameter-to-area changes: Many candidates calculated that area increases by 2² = 4 , then picked A ( X_P / 4 ) or D ( 4X_P ), mistakenly thinking they were asked for stress or extension.

🧠 Exam Technique: Spotting Intensive Properties

When an AQA multiple-choice question provides several geometric values (length, radius, diameter) alongside the phrase "made from the same material", stop and look at the exact quantity being requested:

If the question asks for Young Modulus or Resistivity: these are material constants. The answer is simply equal (ratio = 1).
If the question asks for extension ( ΔL ), strain ( ε ), stress ( σ ), or resistance ( R ): then and only then do you need to calculate using the geometric ratios!

Mark Scheme Breakdown (1 Mark Total):
• [1 mark] selects option B (AO1 — recall of Young modulus as an intensive material property).