Q 11-11-111JEE MainJEE Main 2022 (28 Jul, Shift 1)Medium
A Carnot engine has efficiency of $50\%$. If the temperature of sink is reduced by $40^\circ$C, its efficiency increases by $30\%$. The temperature of the source will be :
Answer: (C) $266.7$ K
Initially $1 - \dfrac{T_2}{T_1} = 0.5$, so $T_2 = 0.5\,T_1$.
The efficiency rises by $30\%$ of its value, to $0.5\times1.3 = 0.65$:
$$1 - \frac{T_2 - 40}{T_1} = 0.65 \Rightarrow 0.5\,T_1 - 40 = 0.35\,T_1$$
$$T_1 = \frac{40}{0.15} \approx 266.7\ \text{K}$$
Solution by Sreeraj P, M.Sc Physics