Two different wires having lengths L_1 and L_2, and respective
Question:

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 :

  1. $4 \frac{\alpha_{1} \alpha_{2}}{\alpha_{1}+\alpha_{2}} \frac{\mathrm{L}_{2} \mathrm{~L}_{1}}{\left(\mathrm{~L}_{2}+\mathrm{L}_{1}\right)^{2}}$

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

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

  4. $\frac{\alpha_{1} \mathrm{~L}_{1}+\alpha_{2} \mathrm{~L}_{2}}{\mathrm{~L}_{1}+\mathrm{L}_{2}}$


Correct Option: , 4

Solution:

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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