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Two wires of the same length and radius are stretched by the same load and the difference of their elongations produced is found to be 0.25 cm . If their modulus of elasticity are in the ratio 5:3. Then the individual elongations are

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0.375cm, 0.625 cm

0.625 cm, 0.650 cm

0.325 cm, 0.350cm

0.675 cm, 0.325cm

answer is A.

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Detailed Solution

$e_{2} - e_{1} = 0.25 c m (∵ Y_{1} > Y_{2})$

$Y = \frac{F l}{A e}$

$\Rightarrow e \propto \frac{1}{y} (∵ Y, F, l a r e s a m e)$

$\Rightarrow \frac{e_{1}}{e_{2}} = \frac{y_{2}}{y_{1}} = \frac{3}{5}$

$e_{1} = \frac{3}{5} e_{2} \Rightarrow e_{1} = \frac{3}{5} (e_{1} + 0.25)$

$\Rightarrow 5 e_{1} = 3 e_{1} + 0.75 \Rightarrow 2 e_{1} = 0.75$

$\Rightarrow e_{1} = 0.375 c m \Rightarrow e_{2} = e_{1} + 0.25 = 0.375 + 0.25 = 0.625 c m$

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