Two different wires having lengths L _{1} and | vedclass

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Question

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 \rig

Vedclass · Accepted Answer

$\frac{\alpha_{1} L_{1}+\alpha_{2} L_{2}}{L_{1}+L_{2}}$

Vedclass · Answer

At $T^{\circ} C \quad L = L _{1}+ L _{2}$

At $T +\Delta T \quad L _{ eq }= L _{1}+ L _{2}$

where $L _{1}= L _{1}\left(1+\alpha_{1} \Delta T \right)$

$L _{2}^{\prime}= L _{2}\left(1+\alpha_{2} \Delta T \right)$

$L _{ eq }^{\prime}=\left( L _{1}+ L _{2}\right)\left(1+\alpha_{ avg } \Delta T \right)$

$\Rightarrow\left(L_{1}+L_{2}\right)\left(1+\alpha_{avg } \Delta T \right)= L _{1}+ L _{2}+ L _{1} \alpha_{1} \Delta T + L _{2} \alpha_{2} \Delta T$

$\Rightarrow\left( L _{1}+ L _{2}\right) \alpha_{ avg }= L _{1} \alpha_{1}+ L _{2} \alpha_{2}$

$\Rightarrow \alpha_{ avg }=\frac{ L _{1} \alpha_{1}+ L _{2} \alpha_{2}}{ L _{1}+ L _{2}}$

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

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