WBJEE 2026 Physics and Chemistry Paper

Concept:For a heat engine, efficiency is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_1$ is the source temperature and $T_2$ is the sink temperature.When the source temperature is fixed, the engine with the lowest sink temperature has the highest efficiency.Explanation:All three engines draw steam at the same source temperature, $T_1 = 130^{\circ}C = 403 \, K$.Convert each sink temperature from Celsius to Kelvin.Engine A: $T_2 = 20^{\circ}C = 293 \, K$Engine B: $T_2 = 40^{\circ}C = 313 \, K$Engine C: $T_2 = 50^{\circ}C = 323 \, K$Now compare their efficiencies.$\eta_A = 1 - \frac{293}{403} = 1 - 0.727 = 0.273$$\eta_B = 1 - \frac{313}{403} = 1 - 0.777 = 0.223$$\eta_C = 1 - \frac{323}{403} = 1 - 0.801 = 0.199$Clearly, $\eta_A > \eta_B > \eta_C$.This confirms that a lower sink temperature gives a higher efficiency.Answer:Engine A is the most efficient engine.Correct option: A.