AP EAMCET · PHYSICS · Thermodynamics
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
- A 0.58
- B 0.73
- C 0.64
- D 0.42
Answer & Solution
Correct Answer
(A) 0.58
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