Q115 / 180 · JIPMER 2018 · Physics
The efficiency of a heat engine is $1 / 6$. Its efficiency double when the temperature of sink decrease by $62^{\circ} \mathrm{C}$, its efficiency doubles. Then, what is the temperature of source?
- A$470 K$
- Correct answerB$372 K$
- C$542 K$
- D$1042 K$
Solution
$\because$ Efficiency of a heat engine,
$\eta=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1}$
or $\quad \frac{1}{6}=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1}$
$\frac{\mathrm{T}_2}{\mathrm{~T}_1}=1-\frac{1}{6}=\frac{5}{6} \Rightarrow \mathrm{T}_2=\frac{5 \mathrm{~T}_1}{6}$ ...(i)
According to question, it's efficiency is doubled when temperature of sink decreased by $62^{\circ} \mathrm{C}$.
$\therefore \quad 2 \eta=1-\frac{\mathrm{T}_2-62}{\mathrm{~T}_1}$
$\Rightarrow \quad 2\left(\frac{1}{6}\right)=1-\frac{\left(\frac{5 T_1}{6}-62\right)}{T_1} \quad$ [From Eq. (i)]
$\begin{array}{rlrl}\Rightarrow & \frac{5 \mathrm{~T}_1}{6}-62 & =\mathrm{T}_1\left(1-\frac{2}{6}\right)=\frac{2 \mathrm{~T}_1}{3} \\ \text { or } & \frac{5 \mathrm{~T}_1}{6}-\frac{2 \mathrm{~T}_1}{3} & =62 \Rightarrow \frac{\mathrm{T}_1}{6}=62 \\ \text { or } & \mathrm{T}_1=372 \mathrm{~K}\end{array}$
Hence, the correct answer is option: 'B'.