JEE Main · 2010 · 11 Apr 2010 · Offline · Heat & Thermodynamics

Although the problem speaks of a car engine, the working relation is the Carnot efficiency $\eta =1-T_2/T_1$, so we only need the ratio of the two temperatures. The ends of the adiabatic expansion are linked by $T{V}^{\gamma -1}=\text{constant}$, and for a diatomic gas $\gamma =7/5$, so $\gamma -1=2/5$. With the volume growing from $V$ to $32V$, $\frac{T_1}{T_2}=(32{)}^{2/5}=(2^5{)}^{2/5}=2^2=4$. The cold end is thus at one quarter of the hot temperature, and $\eta =1-\frac{1}{4}=0.75$.