Heat Engines, Refrigerators & Second Law question from JEE Main 2005 (AIEEE)

Answer: (d)

1. On a $T$-$S$ diagram, heat $=$ area under the path; net work $=$ area enclosed.
2. Heat absorbed along the slanted line ($S_0 \to 2S_0$): $Q_1 = \frac{(2T_0 + T_0)}{2} S_0 = \frac{3}{2}T_0S_0$.
3. Vertical path: $\Delta S = 0$, no heat. Bottom path rejects $Q_2 = T_0S_0$.
4. $W = Q_1 - Q_2 = \frac{1}{2}T_0S_0$, so $\eta = \frac{W}{Q_1} = \frac{1/2}{3/2} = \frac{1}{3}$.
5. So the answer is (d) $\frac{1}{3}$.