Q81 / 90 · JEE Main 2011 · Physics

A Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac{1}{6}$. When $T_2$ is lowered by 62 $\mathrm{K}$, its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ are, respectively :

  • A
    $372 \mathrm{~K}$ and $330 \mathrm{~K}$
  • B
    $330 \mathrm{~K}$ and $268 \mathrm{~K}$
  • C
    $310 \mathrm{~K}$ and $248 \mathrm{~K}$
  • Correct answer
    D
    $372 \mathrm{~K}$ and $310 \mathrm{~K}$

Solution

$\eta_1=\frac{T_1-T_2}{T_1}=\frac{1}{6}$ $\eta_2=\frac{T_1-\left(T_2-62\right)}{T_1}=\frac{1}{3}$ $\Rightarrow \frac{T_1-T_2}{T_1}+\frac{62}{T_1}=\frac{1}{3}$ $\frac{1}{6}+\frac{62}{T_1}=\frac{1}{3}$ $\frac{62}{T_1}=\frac{1}{6}$ $\therefore T_1=62 \times 6=372 \mathrm{~K}$ $\frac{T_1-T_2}{T_1}=\frac{1}{6}$ $1-\frac{T_2}{T_1}=\frac{1}{6}$ $\frac{T_2}{372}=\frac{5}{6}$ $\Rightarrow T_2=310 \mathrm{~K}$

Hence, the correct answer is option: 'D'.

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