Two separate wires A and B are stretched by 2 mm and 4 mm respectively, when they are subjected to a force of 2 N. Assume that both the wires are made up of same material and the radius of wire B is 4 times that of the radius of wire A. The length of the wires A and B are in the ratio of a : b. Then a/b can be expressed as 1/x where x is _________.

Solution

Physics sir · 4 steps

Same material and same force, so length goes as extension times area. Do not forget to square the radius.

1. Start from Young's modulus. For a wire, Y = (F/A)/(Δ L/L). Rearranging gives the length in terms of the things we know.

   L = (Y A Δ L)/F = (Y π r² Δ L)/F

2. Keep what changes. Both wires have the same Y (same material) and the same F = 2 N. So L ∝ r² Δ L.
3. Take the ratio. With r_B = 4r_A, Δ L_A = 2 mm and Δ L_B = 4 mm: L_A/L_B = (r_A² × 2)/((4r_A)² × 4) = 2/64 = 1/32.
4. So. a/b = 1/32, which gives x = 32.

The common slip: 8. Using the radius instead of the area gives 2/(4 × 4) = 1/8, so x = 8. Area depends on r², so the factor is 16, not 4.

The idea to keep. When the material and the load are the same for both wires, L ∝ A Δ L. Area goes as radius squared, so a factor of 4 in radius means a factor of 16 in area.

Answer from the NTA final answer key. Solution by PrepWiser.