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Two different wires having lengths $L_{1}$ and $L_{2}$ , and respective temperature coefficient of linear expansion $α_{1}$ and $α_{2}$ are joined end-to-end. Then the effective temperature coefficient of linear expansion is:

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

$2 \sqrt{α_{1} α_{2}}$

$\frac{α_{1} + α_{2}}{2}$

$4 \frac{α_{1} α_{2}}{α_{1} + α_{2}} \frac{L_{2} L_{1}}{{(L_{2} + L_{1})}^{2}}$

answer is A.

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Detailed Solution

$T^{\circ} C L = L_{1} + L_{2}$

$At T + ΔT L_{eq}^{'} = L_{1}^{'} + L_{2}^{'}$ ; $L_{1}^{'} = L_{1} (1 + α_{1} ΔT)$ , $L_{2}^{'} = L_{2} (1 + α_{2} ΔT)$

$L_{eq}^{'} = (L_{1} + L_{2}) (1 + α_{avg} ΔT)$

$\Rightarrow (L_{1} + L_{2}) (1 + α_{avg} ΔT) = L_{1} + L_{2} + L_{1} α_{1} ΔT + L_{2} a_{2} ΔT$

$\Rightarrow (L_{1} + L_{2}) α_{avg} = L_{1} α_{1} + L_{2} α_{2}$

$\Rightarrow α_{avg} = \frac{L_{1} α_{1} + L_{2} α_{2}}{L_{1} + L_{2}}$

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