Thermodynamics question for JEE Main (JEE Main 2025 (24 Jan, Shift 2)), with solution

Answer: (A) $5\pi$

For a cycle $\Delta U = 0$, so the heat exchanged equals the net work, which is the area enclosed by the cycle on the P–V diagram.

The path $A \to B \to C$ runs along the upper half of the circle and $C \to A$ returns along the horizontal diameter. The enclosed area is a semicircle (really a half-ellipse in SI units) with semi-axes
$$\Delta V = 100\ \text{cc} = 10^{-4}\ \text{m}^3, \qquad \Delta P = 100\ \text{kPa} = 10^5\ \text{Pa}$$

$$|Q| = \frac{1}{2}\pi(10^{-4})(10^5) = 5\pi\ \text{J}$$

Solution by Sreeraj P, M.Sc Physics