JEE Main 2025 Jan 24 Shift 2, Physics Q13: Laws of Thermodynamics

For a cyclic process, $\Delta U=0$, so $|Q|=|W|$, the magnitude of work being the area enclosed in the $P$-$V$ plane.

The cycle ABCA traces only the upper half of the circle (from A through B to C, and then back from C to A by a straight line along the V-axis at the diameter level).

The full circle has radius (in SI units) $R=100\text{cc}=100\times 10^{-6}{\text{m}}^3$ in volume and $R=100\text{kPa}=100\times 10^3\text{Pa}$ in pressure (the two semi-axes of the loop, both equal to 100 in their respective units in the plot).

Area of semicircle:

$W=\frac{1}{2}\pi R_V{R}_P=\frac{1}{2}\pi (100\times 10^{-6})(100\times 10^3)=\frac{1}{2}\pi (10)=5\pi \text{J}$

$\therefore |Q|=5\pi \text{J}$.

Correct option: (2)