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Two different wires having lengths $L _{1}$ and $L _{2}$ and respective temperature coefficient of linear expansion $\alpha_{1}$ and $\alpha_{2},$ are joined end-to-end. Then the effective temperature coefficient of linear expansion is
$4 \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}} \frac{ L _{2} L _{1}}{\left( L _{2}+ L _{1}\right)^{2}}$
$2 \sqrt{\alpha_{1} \alpha_{2}}$
$\frac{\alpha_{1}+\alpha_{2}}{2}$
$\frac{\alpha_{1} L_{1}+\alpha_{2} L_{2}}{L_{1}+L_{2}}$
Solution
At $T^{\circ} C \quad L = L _{1}+ L _{2}$
At $T +\Delta T \quad L _{ eq }= L _{1}+ L _{2}$
where $L _{1}= L _{1}\left(1+\alpha_{1} \Delta T \right)$
$L _{2}^{\prime}= L _{2}\left(1+\alpha_{2} \Delta T \right)$
$L _{ eq }^{\prime}=\left( L _{1}+ L _{2}\right)\left(1+\alpha_{ avg } \Delta T \right)$
$\Rightarrow\left(L_{1}+L_{2}\right)\left(1+\alpha_{avg } \Delta T \right)= L _{1}+ L _{2}+ L _{1} \alpha_{1} \Delta T + L _{2} \alpha_{2} \Delta T$
$\Rightarrow\left( L _{1}+ L _{2}\right) \alpha_{ avg }= L _{1} \alpha_{1}+ L _{2} \alpha_{2}$
$\Rightarrow \alpha_{ avg }=\frac{ L _{1} \alpha_{1}+ L _{2} \alpha_{2}}{ L _{1}+ L _{2}}$
