AS Daily A Level Physics question
A metal wire, P, has length L and diameter d. A second wire, Q, is made from the same material but has length 2L and diameter 2d. The same tensile force is applied to both wires, and both are observed to obey Hooke's law. What is the ratio of the elastic potential energy stored in wire Q to the energy stored in wire P?
Answer
The correct answer is B.
Correct: B — 0.50. Stored energy E is proportional to extension x (since the force F is constant, E = ½Fx). From the definition of the Young modulus, extension x is proportional to L/A, where L is length and A is cross-sectional area. Since area A is proportional to diameter squared (d²), the extension x is proportional to L/d². For wire Q, L is doubled and d is doubled, so the factor affecting extension is 2L / (2d)² = 2L / 4d² = 0.5 * (L/d²). Therefore, the extension and the stored energy of wire Q are half that of wire P. A is incorrect; this value of 0.25 (or 1/4) arises if energy is assumed to be inversely proportional to the area alone, ignoring the change in length. C is incorrect; a ratio of 1.0 could arise from cancelling errors, for instance by incorrectly assuming extension is proportional to L/d instead of L/d². D is incorrect; this value of 2.0 arises if only the change in length is considered, ignoring the quadrupled cross-sectional area.