Q137 / 150 · MHT CET 2024 (15 May Shift 2) · Physics

Heat engine operating between temperature $T_1$ and $T_2$ has efficiency $\frac{1}{6}$. When $T_2$ is lowered by 62 K , its efficiency increases to $\frac{1}{3}$. Then $\mathrm{T}_1$ and $\mathrm{T}_2$ respectively are

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

Solution

Given, $\eta_1=\frac{1}{6}=1-\frac{T_2}{T_1}=\Rightarrow T_2=\frac{5}{6} T_1$
Also, $\eta_2=\frac{1}{3}=1-\frac{\left(T_2-62\right)}{T_1}$
Substituting (i), $\begin{array}{ll} & \frac{\left(\frac{5}{6} T_1-62\right)}{T_1}=1-\frac{1}{3} \\ & \frac{5}{6} T_1-62=T_1 \frac{2}{3} \\ \therefore \quad & T_1=372 \mathrm{~K} \\ \therefore \quad & T_2=\frac{5}{6} \times 372=310 \mathrm{~K} \end{array}$

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

Relevant Questions