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Two wires of the same length and radius are stretched by the same load the difference of their elongations produced is formed to be 0.25cm, if their modules of elasticity are in the ratio 5:3 then the individual Elongation are

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0.375cm, 0.625 cm

0.625cm, 0.650 cm

0.325 cm, 0.350cm

0.675 cm, 0.325 cm

answer is A.

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Detailed Solution

$L_{1} = L_{2} = L$

$r_{1} = r_{2} = r$

$F_{1} = F_{2} = F$

according to question e₁-e₂=0.25cm

$e = F L / A Y$

$e \propto \frac{1}{Y}$

$\Rightarrow \frac{e_{1}}{e_{2}} = \frac{Y_{2}}{Y_{1}} \Rightarrow \frac{0.25 + e_{2}}{e_{2}} = \frac{3}{5} \Rightarrow e_{2} = 0.625$

by solving e₁=0.375,

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