A wire of length ' L ' and radius ' r ' is clamped rigidly at one end. When the other end of the wire is pulled by a force f, its length increases by ' l '. Another wire of same material of length ' 2 L ' and radius ' 2 r ' is pulled by a force ' 2 f '. Then the increase in its length will be :

Solution

Physics sir · 4 steps

Same material means same Y, so write ΔL = FL/(πr²Y) and just scale F, L and r² by their factors.

1. Extension formula. Young's modulus is Y = (F/A)/(Δ L/L), so the extension is Δ L = FL/AY with A = π r². Both wires are of the same material, so Y is the same for both.

   Δ L = FL/(π r² Y)

2. Scale each factor. The force doubles (factor 2) and the length doubles (factor 2), so the top grows by 2 × 2 = 4. The radius doubles, so r² grows by 2² = 4.
3. Take the ratio. (Δ L₂)/l = (2f × 2L)/(2r)² × r²/fL = 4/4 = 1.
4. So. The second wire stretches by l, the same as the first, which is option (c).

The idea to keep. For stretching wires, extension goes as FL/r². Same material keeps Y fixed, and do not forget the radius enters squared through the area.

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