Q 11-11-114JEE MainJEE Main 2021 (31 Aug, Shift 1)Easy
A reversible engine has an efficiency of $\dfrac14$. If the temperature of the sink is reduced by $58^\circ\text{C}$, its efficiency becomes double. Calculate the temperature of the sink:
Answer: (C) $174$ K
$1 - \dfrac{T_2}{T_1} = \dfrac14 \Rightarrow T_2 = \dfrac34T_1$.
$1 - \dfrac{T_2 - 58}{T_1} = \dfrac12 \Rightarrow T_2 - 58 = \dfrac{T_1}{2}$.
$\dfrac34T_1 - \dfrac12T_1 = 58 \Rightarrow T_1 = 232$ K, so $T_2 = 174$ K.
Solution by Sreeraj P, M.Sc Physics