Two different wires having lengths $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$, 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 :
Correct Option: , 4
where $\mathrm{L}_{1}^{\prime}=\mathrm{L}_{1}\left(1+\alpha_{1} \Delta \mathrm{T}\right)$
$\mathrm{L}_{2}^{\prime}=\mathrm{L}_{2}\left(1+\alpha_{2} \Delta \mathrm{T}\right)$
$\mathrm{L}_{\mathrm{eq}}^{\prime}=\left(\mathrm{L}_{1}+\mathrm{L}_{2}\right)\left(1+\alpha_{\mathrm{avg}} \Delta \mathrm{T}\right)$
$\Rightarrow\left(\mathrm{L}_{1}+\mathrm{L}_{2}\right)\left(1+\alpha_{\operatorname{avg}} \Delta \mathrm{T}\right)=\mathrm{L}_{1}+\mathrm{L}_{2}+\mathrm{L}_{1} \alpha_{1} \Delta \mathrm{T}+\mathrm{L}_{2} \alpha_{2} \Delta \mathrm{T}$
$\Rightarrow\left(\mathrm{L}_{1}+\mathrm{L}_{2}\right) \alpha_{\mathrm{avg}}=\mathrm{L}_{1} \alpha_{1}+\mathrm{L}_{2} \alpha_{2}$
$\Rightarrow \alpha_{\mathrm{avg}}=\frac{\mathrm{L}_{1} \alpha_{1}+\mathrm{L}_{2} \alpha_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$
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