Q 11-11-027NEETJEE MainEasy
The efficiency of a Carnot engine is $\dfrac{1}{4}$. When the temperature of its sink is lowered by $50$ K, the efficiency becomes $\dfrac{1}{3}$. The temperature of the source is
Answer: (B) $600$ K
Concept. Write $\eta = 1 - \dfrac{T_2}{T_1}$ for both cases and subtract.
Step 1. First case. $T_2 = \dfrac34T_1$.
Step 2. Second case. $T_2 - 50 = \dfrac23T_1$.
Step 3. Subtract. $50 = \dfrac{T_1}{12} \Rightarrow T_1 = 600$ K.
Answer. $600$ K (Option B).
Solution by Sreeraj P, M.Sc Physics