Q138 / 160 · AP EAMCET 2024 (21 May Shift 1) · Physics

In a Carnot engine, when the temperatures are $T_2=$ $0^{\circ} \mathrm{C}$ and $T_1=200^{\circ} \mathrm{C}$, its efficiency is $\eta_1$ and when the temperatures are $T_1=0^{\circ} \mathrm{C}$ and $T_2=-200^{\circ}$, its efficiency is $\eta_2$. Then the value $\frac{\eta_1}{\eta_2}$ is

  • Correct answer
    A
    0.58
  • B
    0.73
  • C
    0.64
  • D
    0.42

Solution

For carnot engine, $\begin{aligned} & \text {At } T_1=200^{\circ} \mathrm{C}, \mathrm{~T}_2=0^{\circ} \mathrm{C} \\ & \eta_1=1-\frac{T_2}{T_1}=1-\frac{273+0}{273+200}=\frac{200}{473} \\ & \text { At } T_1=0^{\circ} \mathrm{c}, \mathrm{~T}_2=-200^{\circ} \mathrm{c} \\ & \eta_2=1-\frac{T_2}{\mathrm{~T}_1}=1-\frac{273+200}{273+0}=\frac{200}{273} \\ & \therefore \frac{\eta_1}{\eta_2}=\frac{200}{473} \times \frac{273}{200}=0.58 \end{aligned}$

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

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