Two wires A and B are made of same material having ratio of lengths L_A/L_B=1/3 and their diameters ratio d_A/d_B=2. If both the wires are stretched using same force, what would be the ratio of their respective elongations?

Solution

Physics sir · 3 steps

Same material and same force, so elongation depends only on length divided by cross-sectional area.

1. Elongation formula. Young's modulus is Y = (F/A)/(Δ L/L), so Δ L = FL/AY. Here F and Y are the same for both wires, so Δ L ∝ L/A.
2. Area from diameter. Area is A = (π d²)/4, so A ∝ d². With d_A/d_B = 2, the area ratio is A_A/A_B = 2² = 4.

   (Δ L_A)/(Δ L_B) = L_A/L_B × A_B/A_A = 1/3 × 1/4 = 1/12

3. So. The elongations are in the ratio 1 : 12, which is option (b).

The idea to keep. For the same material and force, Δ L ∝ L/d². Always square the diameter ratio, because area grows as the square of the diameter.

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