Two uniform metal rods of length L1 and L2 and their linear coefficients of expansion α1and α2 respectively, are connected to form a single rod of length (L1+L2). When the temperature of the combined rod is raised by "t°C", the length of each rod increased by the same amount then α2α1+α2is

L1α1=L2α2;L1L2=α2α1⇒L2L1=α1α2L2L1+1=α1α2+1⇒L2+L1L1=α1+α2α2α2α1+α2=L1L1+L2

Two uniform metal rods of length L1 and L2 and their linear coefficients of expansion α1and α2 respectively, are connected to form a single rod of length (L1+L2). When the temperature of the combined rod is raised by "t°C", the length of each rod increased by the same amount then α2α1+α2is

L1α1=L2α2;L1L2=α2α1⇒L2L1=α1α2L2L1+1=α1α2+1⇒L2+L1L1=α1+α2α2α2α1+α2=L1L1+L2

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Two uniform metal rods of length L₁ and L₂and their linear coefficients of expansion $α_{1}$ and $α_{2}$ respectively, are connected to form a single rod of length (L₁+L₂). When the temperature of the combined rod is raised by "t°C", the length of each rod increased by the same amount then $[\frac{α_{2}}{α_{1} + α_{2}}]$ is

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

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

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} L_{1} α_{1} = L_{2} α_{2}; \frac{L_{1}}{L_{2}} = \frac{α_{2}}{α_{1}} \Rightarrow \frac{L_{2}}{L_{1}} = \frac{α_{1}}{α_{2}} \\ \frac{L_{2}}{L_{1}} + 1 = \frac{α_{1}}{α_{2}} + 1 \Rightarrow \frac{L_{2} + L_{1}}{L_{1}} = \frac{α_{1} + α_{2}}{α_{2}} \\ \frac{α_{2}}{α_{1} + α_{2}} = \frac{L_{1}}{L_{1} + L_{2}} \end{array}$