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A wire is stretched by 0.01m by a certain force 'F' another wire of same material whose diameter and lengths are double to original wire is stretched by double the force then its elongation will be

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0.005 m

0.01 m

0.02 m

0.04 m

answer is B.

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Detailed Solution

$e = \frac{F L}{π r^{2} y}$ , $e \propto \frac{F L}{r^{2}}$

$∴ \frac{e_{1}}{e_{2}} = \frac{F_{1}}{F_{2}} \times \frac{L_{1}}{L_{2}} \times {(\frac{r_{2}}{r_{1}})}^{2} = (\frac{F}{2 F}) (\frac{L}{2 L}) \times {(\frac{2 r}{r})}^{2} = \frac{1}{2} \times \frac{1}{2} \times 2^{2} = 1 ∴ e_{1} = e_{2} = 0.01 m$

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