The coefficients of linear expansion of brass | vedclass

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(D) The correct option is $D$.Concept: The length of a metallic rod at a temperature $T$ is given by $l = l_0(1 + \alpha \Delta T)$,where $l_0$ is the

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$l_1 \alpha_1 = l_2 \alpha_2$

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(D) The correct option is $D$.Concept: The length of a metallic rod at a temperature $T$ is given by $l = l_0(1 + \alpha \Delta T)$,where $l_0$ is the initial length,$\alpha$ is the coefficient of linear expansion,and $\Delta T$ is the change in temperature.Let the lengths of the brass and steel rods at temperature $T$ be $l_b$ and $l_s$ respectively.$l_b = l_1(1 + \alpha_1 \Delta T)$$l_s = l_2(1 + \alpha_2 \Delta T)$The difference in lengths is given by $l_s - l_b = l_2(1 + \alpha_2 \Delta T) - l_1(1 + \alpha_1 \Delta T)$.$l_s - l_b = (l_2 - l_1) + (l_2 \alpha_2 - l_1 \alpha_1) \Delta T$.Since the difference $(l_2 - l_1)$ is maintained constant at all temperatures,the term involving $\Delta T$ must be zero.Therefore,$l_2 \alpha_2 - l_1 \alpha_1 = 0$,which implies $l_1 \alpha_1 = l_2 \alpha_2$.

The coefficients of linear expansion of brass and steel rods are $\alpha_1$ and $\alpha_2$ respectively. The lengths of the brass and steel rods are $l_1$ and $l_2$ respectively. If $(l_2 - l_1)$ is maintained the same at all temperatures,which one of the following relations is correct?

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