Two wires A and B are made of the same material. Their lengths satisfy $L_A/L_B$ = 1/3 and their diameters satisfy $d_A/d_B$ = 2. If the same tensile force is applied to both wires, the ratio of elongations Δ $L_A:ΔL_B$ is:
Options

Question

Two wires A and B are made of the same material. Their lengths satisfy $L_A/L_B$ = 1/3 and their diameters satisfy $d_A/d_B$ = 2. If the same tensile force is applied to both wires, the ratio of elongations Δ $L_A:ΔL_B$ is:

Options

G1 Physics · Accepted Answer

For a wire, ΔL = FL/(AY). The wires have the same material, so Y is the same, and the same force F is applied. Therefore, Δ L A / Δ L B L_A/ΔL_B L A ​ /Δ L B ​ = ( L A / L B ) ( A B / A A ) . L_A/L_B)(A_B/A_A). L A ​ / L B ​ ) ( A B ​ / A A ​ ) . Since area A ∝ d², A B / A A A_B/A_A A B ​ / A A ​ = ( d B / d A ) 2 d_B/d_A)² d B ​ / d A ​ ) 2 = (1/2)² = 1/4. Thus, Δ L A / Δ L B L_A/ΔL_B L A ​ /Δ L B ​ = (1/3)(1/4) = 1/12. Hence the ratio is 1:12. Correct option: B.