 # Two different wires having lengths 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 :

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

2. $2 \sqrt{\alpha_{1} \alpha_{2}}$

3. $\frac{\alpha_{1}+\alpha_{2}}{2}$

4. $4 \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}} \frac{L_{2} L_{1}}{\left(L_{2}+L_{1}\right)^{2}}$

Correct Option: 1

Solution:

(1) Let $L_{1}^{\prime}$ and $L_{2}^{\prime}$ be the lengths of the wire when temperature is changed by $\Delta T^{\circ} \mathrm{C}$.

At $T^{\circ} \mathrm{C}$,

$L_{e q}=L_{1}+L_{2}$

At $T+\Delta^{\circ} \mathrm{C}$

$L_{e q}^{\prime}=L_{1}^{\prime}+L_{2}^{\prime}$

$\therefore L_{e q}\left(1+\alpha_{e q} \Delta T\right)=L_{1}\left(1+\alpha_{1} \Delta T\right)+L_{2}\left(1+\alpha_{2} \Delta T\right)$

$\left[\because L^{\prime}=L(1+\alpha \Delta T)\right]$

$\Rightarrow\left(L_{1}+L_{2}\right)\left(1+\alpha_{e q} \Delta T\right)=L_{1}+L_{2}+L_{1} \alpha_{1} \Delta T+L_{2} \alpha_{2} \Delta T$

$\Rightarrow \alpha_{\mathrm{eq}}=\frac{\alpha_{1} L_{1}+\alpha_{2} L_{2}}{L_{1}+L_{2}}$