NEET UG PHYSICS MCQ: THERMODYNAMICS

Option B (${T_2} = 0K$) is correct because the efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_2$ is the temperature of the cold reservoir and $T_1$ is the temperature of the hot reservoir. For $\eta$ to be 100%, $T_2$ must be 0K.