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Two wires A and B are made of the same material. The ratio of their lengths is L_A: L_B = 1: 2 and the ratio of their diameters is d_A: d_B = 2: 1. When the same force F is applied to each wire, find the ratio of the increase in length of wire A to that of wire B (delta_L_A: delta_L_B).
- 2: 1
- 1: 4
- 1: 8
- 8: 1
Correct answer: 1: 8
Solution
delta_L = FL/(YA) = FL/(Y * pi * d²/4) = 4FL/(Y*pi*d²). So delta_L is proportional to L/d². delta_L_A/delta_L_B = (L_A/d_A²)/(L_B/d_B²) = (L_A*d_B²)/(L_B*d_A²) = (1/2)*(1/4) = 1/8. So ratio = 1:8.
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