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
$1 - \dfrac{T_2}{T_1} = \dfrac{1}{4} \Rightarrow T_2 = \dfrac{3}{4}T_1$.
After lowering the sink: $1 - \dfrac{T_2 - 50}{T_1} = \dfrac{1}{3} \Rightarrow T_2 - 50 = \dfrac{2}{3}T_1$.
Subtracting: $50 = \left(\dfrac{3}{4} - \dfrac{2}{3}\right)T_1 = \dfrac{T_1}{12} \Rightarrow T_1 = 600$ K (and $T_2 = 450$ K).
Solution by Sreeraj P, M.Sc Physics