JEE Advanced 2025 Paper 2 Physics Question 13 Solution

Detailed Solution

Because the piston is thermally conducting, in equilibrium the temperatures are equal: $T_L = T_R = T$ .
The piston is at $L/2$ , so the volumes are $V_L = A(L/2)$ and $V_R = A(L/2)$ .
Using the ideal gas law $PV = nRT$ : $P_L = \frac{n_L RT}{V_L} = \frac{(3/2)RT}{AL/2} = \frac{3RT}{AL}$ $P_R = \frac{n_R RT}{V_R} = \frac{1 \cdot RT}{AL/2} = \frac{2RT}{AL}$
Relationship between pressures: $P_L = 1.5 P_R$ .
Force balance on the piston: $P_L A = P_R A + F_{spring}$ .
The spring is compressed/extended by $\Delta x = L/2 - 2L/5 = L/10$ . Thus $F_{spring} = k(L/10)$ .
Substituting values into the force balance equation: $1.5 P_R A = P_R A + \frac{kL}{10}$ $0.5 P_R A = \frac{kL}{10}$ $P_R = \frac{2kL}{10A} = \frac{kL}{A} (0.2)$
Comparing with $P = \frac{kL}{A}\alpha$ , we find $\alpha = 0.2$ . Final Answer: 0.2

Question 13