Two different wires having lengths L₁ and L₂, 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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$\frac{α_{1} {+ α}_{2}}{2}$

$2 \sqrt{α_{1} α_{2}}$

$\frac{α_{1} L_{1} {+ α}_{2} L_{2}}{L_{1} {+ L}_{2}}$

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

answer is A.

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Detailed Solution

Let $L_{1}^{1}$ and $L_{2}^{1}$ be the lengths of the wire when temperature is changed by ${ΔT}^{0} C$ .

At T⁰C, L_eq = L₁ + L₂

At ${T + ΔT}^{0} {C, L}_{eq}^{1} {= L}_{1}^{1} {+ L}_{2}^{1}$

$L_{eq}^{} ({1 + α}_{eq} ΔT) {= L}_{1}_{} ({1+α}_{1} ΔT) {+ L}_{2} ({1 + α}_{2} ΔT)$

$[∴ L^{1} = L (1 + α ΔT)]$

$(L_{1} {+ L}_{2}) ({1+ α}_{eq} ΔT) {= L}_{1} {+ L}_{2} {+ L}_{1} α_{1} {ΔT + L}_{2} α_{2} ΔT$

$α_{eq} = \frac{α_{1} L_{1} {+ α}_{2} L_{2}}{L_{1} {+ L}_{2}}$

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