AQA A-Level Physics Paper 1, June 2025: Question 26

1 mark · Easy difficulty · Multiple Choice

Determine the resistivity of a metal rod that has the same dimensions as two rods connected in series and the same total resistance.

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Question

Diagram showing two cylindrical metal rods W1 and W2 connected in series end-to-end, each having length L, diameter d, and resistivities rho 1 and rho 2 respectively. Below them is a single rod X of length L and diameter d. The question asks for the resistivity of X if its resistance equals the combined resistance of W1 and W2. Four multiple-choice options are provided: A (rho 1 + rho 2), B ((rho 1 + rho 2) / 2), C ((rho 1 + rho 2) / (rho 1 * rho 2)), and D ((rho 1 * rho 2) / (rho 1 + rho 2)).
Question text

26 Two thin metal rods W1 and W2, each of length L and diameter d, are connected in series.

The resistivity of W1 is ρ1 and the resistivity of W2 is ρ2.

A single metal rod X, also of length L and diameter d, has the same resistance as the

series combination of W1 and W2.

What is the resistivity of X?

[1 mark]

A 1 + 2

1 + 2

B

1 + 2

C

D

1 + 2

Mark scheme

Show the mark scheme Mark scheme entry for question 26 indicating that option A (rho 1 + rho 2) is the correct answer, assessed under assessment objective AO2.

26 A ρ1 + ρ2 AO2

How to answer it

Equivalent Resistivity of Series Conductors

📋 What this question tests

This question assesses your ability to apply the resistivity equation ( R = ρL / A ) alongside series resistance rules ( R_total = R₁ + R₂ ). You must correctly equate physical dimensions and manipulate algebraic expressions to find an effective material property.

Question 26

Multiple Choice Analysis

Determining the Resistivity of Rod X

✅ Correct Answer

A: ρ₁ + ρ₂

Awarded 1 Mark (AO2)

💡 Key Knowledge

  • Resistivity formula: R = ρL / A
  • Series resistance: When two components are in series, their resistances add directly: R_total = R₁ + R₂
  • Cross-sectional area: Since diameter d is identical for all three rods, the cross-sectional area A = π(d/2)² is identical for all rods.

📐 Step-by-Step Derivation

  1. Determine cross-sectional area:
    All three rods (W₁, W₂, and X) have the same diameter d , so they all have the same cross-sectional area A .
  2. Calculate resistance of rods W₁ and W₂:
    For rod W₁: R₁ = ρ₁L / A
    For rod W₂: R₂ = ρ₂L / A
  3. Find total resistance of the series combination:
    R_total = R₁ + R₂ = (ρ₁L / A) + (ρ₂L / A) = (ρ₁ + ρ₂)(L / A)
  4. Express resistance of rod X:
    Rod X has length L , area A , and resistivity ρ_X :
    R_X = ρ_X L / A
  5. Equate R_X to R_total and solve for ρ_X:
    ρ_X L / A = (ρ₁ + ρ₂)(L / A)
    Cancelling L / A from both sides gives:
    ρ_X = ρ₁ + ρ₂

🧠 Exam Technique

  • Inspect dimensions first: Notice that rod X has length L , NOT 2L . It replaces the total resistance of the two rods in half the total physical length!
  • Dimensional sanity check: Resistivity has units of Ω m . Options C and D give incorrect units ( 1 / (Ω m) and dimensionless, respectively) if treated as raw values, immediately eliminating them.

❌ Common Traps & Distractor Analysis

  • Option B [(ρ₁ + ρ₂) / 2]: The most common wrong answer! Students intuitively think of the "average" resistivity because a single rod of the same material of length 2L would need an average resistivity of (ρ₁ + ρ₂)/2 . However, rod X is only of length L , so its resistivity must be twice that average.
  • Option C & D: Confusing the series resistance formula with the parallel formula ( 1/R_total = 1/R₁ + 1/R₂ or the "product over sum" shortcut).

Topics

Physics · 3.5 Electricity

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.