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Given $α_{brass} = 18 \times 10^{- 6} {(^{0} C)}^{- 1}, α_{iron} = 12 \times 10^{- 6} {(^{0} C)}^{- 1}$ then what length of brass and iron should be taken so that the difference between them is always 0.1 m for any variation in temperature .

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0.3m,0.4m

0.4m,0.3m

0.2m,0.3m

0.3m,0.2m

answer is C.

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Detailed Solution

$\begin{array}{l} L_{i} - L_{b} = 0.1 m \to 1 \\ △ L_{b} = {ΔL}_{i} \\ L_{b} α_{b} = L_{i} α_{i} \\ \frac{L_{b}}{L_{i}} = \frac{α_{i}}{α_{b}} = \frac{12}{18} = \frac{2}{3} \to 2 \end{array}$
From 1&2 $L_{i} = 0.3 m & L_{b} = 0.2 m$

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