A wire of length L and radius r is clamped at one end. If its other end is pulled by a force F, its length increases by l. If the radius of the wire and the applied force both are reduced to half of their original values keeping original length constant, the increase in length will become:

Solution

Physics sir · 3 steps

Write the extension as FL/(π r² Y), then watch what halving F and halving r each do to it.

1. Extension formula. Young's modulus is stress over strain, Y = (F/π r²)/(l/L). Rearranging gives the extension directly.

   l = FL/(π r² Y)

2. Apply the changes. L and Y stay the same, since the material is unchanged. Halving F halves l. Halving r makes r² one quarter, which multiplies l by 4.
3. So. The new extension is l' = (F/2)L/(π (r/2)² Y) = (1/2)/(1/4) × l = 2l. The extension becomes 2 times, option (a).

The idea to keep. Extension goes as F/r² for the same wire material and length. The radius enters squared through the area, so always square its change before combining it with the force.

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