The given system consists of two springs. If the temperature of the rod is increased by ΔT, the compression in the left spring is

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$\frac{3}{2}{\text{L}}\alpha \Delta {\text{T}}$

$\frac{3}{4}{\text{L}}\alpha \Delta {\text{T}}$

$\frac{5}{3}{\text{L}}\alpha \Delta {\text{T}}$

$\frac{5}{2}{\text{L}}\alpha \Delta {\text{T}}$

answer is B.

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Detailed Solution

At elevated temperature, the rod is in equilibrium.

$\therefore k{x_L} = 3k.{x_R}$

$\Rightarrow {x_L} = 3{x_R}$ ... (1)

$e = {X_L} + {X_R}\,\,and\,\,e = L\alpha \Delta T$ ... (2)

$\therefore L\alpha \Delta T = {x_L} + \frac{{{x_L}}}{3} = \frac{{4{x_L}}}{3} \Rightarrow {x_L} = \frac{3}{4}L\alpha \Delta T$

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