Two rods A and B of identical dimensions are | vedclass

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(A) Let the initial length of both rods be $L_0$ and the initial temperature be $T_i = 30\,^{\circ}C$.The final length of a rod is given by $L = L_0(1

Vedclass · Accepted Answer

$230$

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(A) Let the initial length of both rods be $L_0$ and the initial temperature be $T_i = 30\,^{\circ}C$.The final length of a rod is given by $L = L_0(1 + \alpha \Delta T)$,where $\alpha$ is the coefficient of linear expansion and $\Delta T$ is the change in temperature.Since the final lengths are equal,we have:$L_0(1 + \alpha_A \Delta T_A) = L_0(1 + \alpha_B \Delta T_B)$This simplifies to:$\alpha_A \Delta T_A = \alpha_B \Delta T_B$Given $\frac{\alpha_A}{\alpha_B} = \frac{4}{3}$,$\Delta T_A = 180 - 30 = 150\,^{\circ}C$,and $\Delta T_B = T - 30$.Substituting the values:$\frac{4}{3} = \frac{T - 30}{150}$$T - 30 = \frac{4}{3} 	imes 150$$T - 30 = 4 	imes 50 = 200$$T = 200 + 30 = 230\,^{\circ}C$.

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