Increase in length of a wire is 1 mm when suspended by a weight. If the same weight is suspended on a wire of double its length and double its radius, the increase in length will be

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2 mm

0.5 mm

4 mm

0.25 mm

answer is B.

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Detailed Solution

$\begin{array}{l} l = \frac{FL}{AY} \Rightarrow l \propto \frac{L}{r^{2}} (F and Y are same) \\ ∴ \frac{l_{2}}{l_{1}} = \frac{L_{2}}{L_{1}} {(\frac{r_{1}}{r_{2}})}^{2} = 2 \times {(\frac{1}{2})}^{2} = \frac{1}{2} \Rightarrow l_{2} = \frac{l_{1}}{2} = \frac{l}{2} = 0 .5 mm \end{array}$

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