Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The 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 $(L_2-L_1)$ remains the same at all temperatures, which one of the following relations holds good?

$\alpha_1L_2^2=\alpha_2L_1^2$

$\alpha_1^2L_2=\alpha_2^2L_1$

$\alpha_1L_1=\alpha_2L_2$

$\alpha_1L_2=\alpha_2L_1$

Step-by-Step Solution

Let the change in temperature be $\Delta T$ . The new lengths of the brass and steel rods will be: $L_1' = L_1(1 + \alpha_1\Delta T)$ $L_2' = L_2(1 + \alpha_2\Delta T)$ Given that the difference in lengths $(L_2 - L_1)$ remains the same at all temperatures: $L_2' - L_1' = L_2 - L_1$ $L_2(1 + \alpha_2\Delta T) - L_1(1 + \alpha_1\Delta T) = L_2 - L_1$ $L_2 + L_2\alpha_2\Delta T - L_1 - L_1\alpha_1\Delta T = L_2 - L_1$ $L_2\alpha_2\Delta T - L_1\alpha_1\Delta T = 0$ $\implies L_2\alpha_2 = L_1\alpha_1$ $\implies \alpha_1L_1 = \alpha_2L_2$

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