A2 Daily A Level Physics question
In a required practical to determine Young’s modulus, a student plots extension x against hanging mass m for a long metal wire. For Wire 1, length is L and diameter is d. For Wire 2, made of the same material, length is 2L and diameter is d/2. The same mass steps and instruments are used, and all data are within the elastic limit. Ignoring any zero-offset, which statement must be true for the x–m graph of Wire 2 compared with Wire 1, and why?
Answer
The correct answer is A.
Correct: A — Its gradient is 8 times larger; with the same extension resolution, the percentage uncertainty in the gradient is smaller. In the elastic region, x ∝ (L/A)F and F = mg, so the x–m gradient ∝ (L/A)g; with L → 2L and A → A/4 (since d → d/2), L/A increases by 2 ÷ (1/4) = 8, making the gradient 8×; a larger signal per mass step reduces percentage uncertainty for the same absolute x resolution. B is wrong because it considers only doubling L and ignores the area change; halving diameter quarters A, giving an 8×, not 2×, change. C is wrong in both direction and factor: a thinner wire (smaller A) extends more per unit mass, not less, and the factor is 8× larger, not 4× smaller. D is wrong because, although the material (E) is the same, the geometry (L and A) also sets the slope, so changing length and diameter does affect the gradient.