Two wires A and B made of different materials of lengths 6.0 cm and 5.4 cm , respectively and area of cross sections 3.0× 10⁻⁵ m² and 4.5× 10⁻⁵ m², respectively are stretched by the same magnitude under a given load. The ratio of the Young's modulus of A to that of B is x:3. The value of x is ____

Solution

Physics sir · 4 steps

Same load, same stretch, so Young's modulus goes as length over area. Just compare L/A for the two wires.

1. Write Young's modulus. Stress is F/A and strain is Δ L/L, so Young's modulus is their ratio.

   Y = (F/A)/(Δ L/L) = (F L)/(A Δ L)

2. Cancel common terms. The load F and the extension Δ L are the same for both wires. They cancel in the ratio, so Y_A/Y_B = (L_A/A_A)/(L_B/A_B).
3. Put in numbers. L_A/A_A = 6.0/3.0 = 2 and L_B/A_B = 5.4/4.5 = 1.2, using the same units for both. So Y_A/Y_B = 2/1.2 = 5/3.
4. So. The ratio is 5:3, so x = 5, which is option (d).

The idea to keep. When the load and the extension are equal, Y ∝ L/A. A longer or thinner wire must be stiffer to stretch by the same amount.

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