Young's moduli of the material of wires A and B are in the ratio of 1: 4, while its area of cross sections are in the ratio of 1: 3. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires A and B will be in the ratio of
[Assume length of wires A and B are same]

Solution

Physics sir · 4 steps

Same load and same length, so elongation goes as 1/(YA): just flip the product of the two ratios.

1. Elongation formula. Young's modulus is stress over strain, Y = (F/A)/(Δ L/L). Rearranging gives the elongation of a loaded wire.

   Δ L = FL/AY

2. Take the ratio. F and L are the same for both wires, so they cancel. That leaves (Δ L_A)/(Δ L_B) = (A_B Y_B)/(A_A Y_A).
3. Put in numbers. With Y_A : Y_B = 1 : 4 and A_A : A_B = 1 : 3, we get (Δ L_A)/(Δ L_B) = (3 × 4)/(1 × 1) = 12.
4. So. The elongations of A and B are in the ratio 12 : 1, option (c). Wire A is both thinner and less stiff, so it stretches more.

The idea to keep. For the same load and length, Δ L ∝ 1/AY. A thinner, softer wire stretches more, so check that your ratio favours the weaker wire.

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