Q92 / 180 · AIIMS 2010 · Physics
A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by $62^{\circ} \mathrm{C}$, the efficiency of the engine is doubled. The temperatures of the source and sink are
- Correct answerA$99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$
- B$80^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$
- C$95^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$
- D$90^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}$
Solution
$$
\eta_1=1-\frac{T_L}{T_H}=\frac{W}{Q_1}=\frac{1}{6}
$$
or $5 T_H-6 T_L=0$
$\eta_2=1-\frac{T_L-62}{T_H}=2 \eta_1=\frac{1}{3}$
(Given)
$\Rightarrow 1-\frac{1}{3}=\frac{T_L-62}{T_H}$
or $2 T_H-3 T_L=-186$
Solving (i) and (ii), we get
$$
\begin{aligned}
& \therefore \quad T_H=372 \mathrm{~K}=99^{\circ} \mathrm{C} \\
& T_L=\frac{5}{6} T_H=\frac{5}{6} \times 372 \mathrm{~K}=310 \mathrm{~K}=37^{\circ} \mathrm{C}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.