# $\;l_{1}\;$ and $\;l_{2}\;$ are the lengths of two rods at $\;0^{0}C\;$ and have coefficients of linear expansion $\;\alpha_{1}\;$ and $\;\alpha_{2}\;$ respectively . If the difference between the lengths X is the same at all temperatures then $\;l_{1}\;$ is

$(a)\;\large\frac{\alpha_{2}\;x}{\alpha_{1} -\alpha_{2}}\qquad(b)\;\large\frac{\alpha_{1}\;x}{\alpha_{1} -\alpha_{2}}\qquad(c)\;\large\frac{\alpha_{2}}{\alpha_{1}}\;x\qquad(d)\;\large\frac{\alpha_{1}}{\alpha_{2}}$

Answer : $\;\large\frac{\alpha_{2}\;x}{\alpha_{1} -\alpha_{2}}$
Explanation :
$l_{1}^{'} = l_{1} \;(1 + \alpha_{1} \bigtriangleup t)$
$l_{2}^{'} =l_{2}\; (1 + \alpha_{1} \bigtriangleup t)$
$l_{1}^{'} - l_{2}^{'} = (l_{1} + l_{2}) + (l_{1} \alpha_{1} -l_{2} \alpha_{2}) \; \bigtriangleup t$
$x =x + (l_{1} \alpha_{1} + l_{2} \alpha_{2}) \; \bigtriangleup t$
$l_{1} \alpha_{1} = l_{2} \alpha_{2}$
Therefore , $\rho_{1} = \large\frac{\alpha_{2}}{\alpha_{1}} l_{2} = \large\frac{\alpha_{2}}{\alpha_{1}} (l_{1} -x)$
$\rho_{1} = \large\frac{\alpha_{2} x}{\alpha_{1} - \alpha_{2}} x$