Two different wires having lengths L_{1} and | vedclass

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(D) At temperature $T$,the total length is $L = L_{1} + L_{2}$.When the temperature increases to $T + \Delta T$,the new lengths are:$L_{1}' = L_{1}(1

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) At temperature $T$,the total length is $L = L_{1} + L_{2}$.When the temperature increases to $T + \Delta T$,the new lengths are:$L_{1}' = L_{1}(1 + \alpha_{1} \Delta T)$$L_{2}' = L_{2}(1 + \alpha_{2} \Delta T)$The total new length $L_{eq}'$ is the sum of the individual new lengths:$L_{eq}' = L_{1}' + L_{2}' = L_{1}(1 + \alpha_{1} \Delta T) + L_{2}(1 + \alpha_{2} \Delta T)$$L_{eq}' = L_{1} + L_{1} \alpha_{1} \Delta T + L_{2} + L_{2} \alpha_{2} \Delta T$$L_{eq}' = (L_{1} + L_{2}) + (L_{1} \alpha_{1} + L_{2} \alpha_{2}) \Delta T$For the equivalent system with effective coefficient $\alpha_{avg}$,the new length is:$L_{eq}' = (L_{1} + L_{2})(1 + \alpha_{avg} \Delta T) = (L_{1} + L_{2}) + (L_{1} + L_{2}) \alpha_{avg} \Delta T$Equating the two expressions for $L_{eq}'$:$(L_{1} + L_{2}) \alpha_{avg} \Delta T = (L_{1} \alpha_{1} + L_{2} \alpha_{2}) \Delta T$$\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 coefficients 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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