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Coefficient of linear expansion of brass and steel rods are $\alpha_1$ and $\alpha_2$. 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?
A
$\alpha_1l_2^2 = \alpha_2l_1^2$
B
$\alpha_1^2l_2 = \alpha_2^2l_1$
C
$\alpha_1l_1 = \alpha_2l_2$
D
$\alpha_1l_2 = \alpha_2l_1$
Explanation
Both rods must expand by the same amount.
Detailed Solution
Let the temperature increase by $\Delta T$ and the new lengths be $l'_1$ and $l'_2$.
$l'_1 = l_1(1 + \alpha_1\Delta T)$ ...(i), $l'_2 = l_2(1 + \alpha_2\Delta T)$ ...(ii)
Given $l'_1 - l'_2 = l_1 - l_2$
$\Rightarrow l_1(1 + \alpha_1\Delta T) - l_2(1 + \alpha_2\Delta T) = l_1 - l_2$
$\Rightarrow l_1\alpha_1\Delta T - l_2\alpha_2\Delta T = 0$
$\Rightarrow l_1\alpha_1 = l_2\alpha_2$
$l'_1 = l_1(1 + \alpha_1\Delta T)$ ...(i), $l'_2 = l_2(1 + \alpha_2\Delta T)$ ...(ii)
Given $l'_1 - l'_2 = l_1 - l_2$
$\Rightarrow l_1(1 + \alpha_1\Delta T) - l_2(1 + \alpha_2\Delta T) = l_1 - l_2$
$\Rightarrow l_1\alpha_1\Delta T - l_2\alpha_2\Delta T = 0$
$\Rightarrow l_1\alpha_1 = l_2\alpha_2$