Coefficient of linear expansion of brass and steel rods are $α_{1}$ and $α_{2}$ Lengths of brass and steel rods are $l_{1}$ and $l_{2}$ respectively. If $(l_{2} - l_{1})$ is maintained same at all temperatures, which one of the following relations holds good?

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$α_{1}^{2} l_{2} = α_{2}^{2} l_{1}$

$α_{1} l_{1} = α_{2} l_{2}$

$α_{1} l_{2} = α_{2} l_{1}$

$α_{1} l_{2}^{2} = α_{2} l_{1}^{2}$

answer is B.

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Detailed Solution

$Linear expansion of brass = α_{1}$

$Linear expansion of steel = α_{2}$

$Length of brass rod = l_{1}$

$Length of steel rod = l_{2}$

$O n i n c r e a \sin g t h e t e m p e r a t u r e o f t h e r o d s b y ∆ T, n e w l e n g t h s w o u l d b e$

$l_{1}^{'} = l_{1} (1 + α_{1} Δ T) . . . . . . . . (1)$

$l_{2}^{'} = l_{2} (1 + α_{2} Δ T) . . . . . . . . . (2)$

$Subtracting eqn. (i) from eqn. (ii), we get$

$l_{2}^{'} - l_{1}^{'} = (l_{2} - l_{1}) + (l_{2} α_{2} - l_{1} α_{1}) Δ T$

According to the question

$\begin{matrix} l_{2}^{'} - l_{1}^{'} = l_{2} - l_{1} \\ ∴ l_{2} α_{2} - l_{1} α_{1} = 0 or l_{1} α_{1} = l_{2} α_{2} \end{matrix}$

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