Question

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 :

• A. $4 \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}} \frac{ L _{2} L _{1}}{\left( L _{2}+ L _{1}\right)^{2}}$
• B. $2 \sqrt{\alpha_{1} \alpha_{2}}$
• C. $\frac{\alpha_{1}+\alpha_{2}}{2}$
• D. $\frac{\alpha_{1} L_{1}+\alpha_{2} L_{2}}{L_{1}+L_{2}}$

Answer 1

Answer: (D)

where $ L_{1}'=L_{1}\left(1+\alpha_{1} \Delta T\right) $
$L_{2}'=L_{2}\left(1+\alpha_{2} \Delta T\right) $
$ L_{e q}'=\left(L_{1}+L_{2}\right)\left(1+\alpha_{\text {avg }} \Delta T\right) $
$ \Rightarrow \left(L_{1}+L_{2}\right)\left(1+\alpha_{\text {avg }} \Delta T\right)=L_{1}+L_{2}+L_{1} \alpha_{1} \Delta T+L_{2} a_{2} \Delta T $
$ \Rightarrow \left(L_{1}+L_{2}\right) \alpha_{\text {avg }}=L_{1} \alpha_{1}+L_{2} \alpha_{2} $
$ \Rightarrow \alpha_{\text {avg }}=\frac{L_{1} \alpha_{1}+L_{2} \alpha_{2}}{L_{1}+L_{2}} $