Two rods one of aluminium of length $l_{1}$ having coefficient of linear expansion $α_{a}$ , and other steel of length $l_{2}$ having coefficient of linear expansion $α_{s}$ are joined end to end. The expansion in both the rods is same on variation of temperature. Then the value of $\frac{l_{1}}{l_{1} + l_{2}}$ is

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$\frac{α_{a} + α_{s}}{α_{a}}$

None of these

$\frac{α_{s}}{α_{a} + α_{s}}$

$\frac{α_{a}}{α_{a} - α_{s}}$

answer is A.

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Detailed Solution

$∆ l_{1} = ∆ l_{2}$

$l_{1} α_{1} ∆ T = l_{2} α_{2} ∆ T$

$l_{1} α_{1} = l_{2} α_{2}$

where

$l_{1} \to$ Length of aluminium rod

$α_{a} \to$ Coefficient of linear expansion of aluminium rod

$l_{2} \to$ Length of steel rod

$α_{s} \to$ Coefficient of linear expansion of steel rod

$\frac{l_{1}}{l_{1} + l_{2}} = \frac{l_{1} / l_{2}}{l_{1} / l_{2} + 1} = \frac{α_{2} / α_{1}}{α_{2} / α_{1} + 1} = \frac{α_{2}}{α_{1} + α_{2}}$

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