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AS Daily A Level Physics question

2026-04-07 OCR A Mechanics & Materials (M3) Module 3.4 Materials — Young modulus; force–extension practical Module 3.3 Work, energy and power — elastic strain energy Module 2.1 Practical skills — selection of apparatus length to improve sensitivity

A student is measuring Young modulus for a metal wire. With a 1.50 m test wire (constant diameter), adding a 10 N load produces an extra extension of 1.2 mm and the student’s lab book records the elastic energy stored as 0.0060 J. They repeat with the same material and diameter but a 3.00 m test wire, using the same 10 N load. Which outcome is most consistent with correct reasoning?

  1. A ΔL = 1.2 mm; energy = 0.0060 J
  2. B ΔL = 1.2 mm; energy = 0.012 J
  3. C ΔL = 2.4 mm; energy = 0.0060 J
  4. D ΔL = 2.4 mm; energy = 0.012 J (correct)

Answer

The correct answer is D.

Correct: D — ΔL = 2.4 mm; energy = 0.012 J. A doubling the wire length doubles the extension for the same force (ΔL ∝ L), and with the same force the energy stored is proportional to the extension, so both quantities double. A assumes extension and energy are unaffected by wire length, ignoring ΔL ∝ L. B keeps the extension the same but doubles the energy, which is inconsistent because for a fixed force the energy change requires a change in extension. C doubles the extension but keeps the energy the same, neglecting that for fixed force the energy scales with ΔL. D correctly applies ΔL ∝ L (1.2 mm → 2.4 mm) and then doubles the stored energy (0.0060 J → 0.012 J).

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