JEE Physics 2025 Question with Solution

Given: A Carnot engine $E$ works between $T_H = 473\,\text{K}$ and $T_C = 273\,\text{K}$. Two other engines work as $E_1: 473\,\text{K} \to 373\,\text{K}$ and $E_2: 373\,\text{K} \to 273\,\text{K}$.

Find: The correct relation among $\eta_{12}$, $\eta_1$ and $\eta_2$.

For a Carnot engine,

So for engine $E$,

For engine $E_1$,

For engine $E_2$,

Now compare $\eta_{12}$ with $\eta_1 + \eta_2$:

while

Hence,

Also, the series relation for two Carnot stages is

which gives

So the solution's listed option $\eta_{12} = \eta_1\eta_2$ is inconsistent with the working shown in the solution. Based on the solution derivation, the correct option is A.

Therefore, the correct option is A, i.e. $\eta_{12} < \eta_1 + \eta_2$.

Given: The engines operate in two stages from $473\,\text{K}$ to $373\,\text{K}$ and from $373\,\text{K}$ to $273\,\text{K}$.

Find: A quick relation between the overall efficiency and stage efficiencies.

Use the fact that for staged Carnot engines, the fractions of heat rejected multiply:

Therefore,

So,

which gives

Since $\eta_1\eta_2 > 0$, it follows immediately that

Therefore, the correct option is A.