Heat Engine Efficiency

Building on NCERT Class 11, Chapter 12 (Thermodynamics), the efficiency of a heat engine is defined as the ratio of useful work output to the heat absorbed from the hot reservoir, $\eta =\frac{W}{Q_H}$, where by energy conservation $W=Q_H-Q_C$. Here $Q_H=1000$ J and $Q_C=600$ J, so the work output per cycle is $W=1000-600=400$ J. The efficiency is therefore $\eta =\frac{400}{1000}=0.40$, or 40 percent. The option 60 percent is wrong because it mistakenly uses the rejected heat fraction $Q_C/Q_H$ rather than the work fraction. The option 100 percent is wrong because it would violate the Second Law, which forbids converting all absorbed heat into work in a cyclic engine. The option 25 percent is wrong because it divides work by the rejected heat instead of by the absorbed heat. A plausibility check: efficiency must lie between 0 and 1 for any real engine, and $0<0.40<1$ is consistent; also the heat rejected (600 J) exceeds the work (400 J), which is typical for low-efficiency engines.