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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2022
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  • MCQ

Let \(\eta_{1}\) is the efficiency of an engine at \(T _{1}=447^{\circ}\,C\) and \(T _{2}=147^{\circ}\,C\) while \(\eta_{2}\) is the efficiency at \(T _{1}=947^{\circ}\,C\) and \(T _{2}=47^{\circ}\,C\). The ratio \(\frac{\eta_{1}}{\eta_{2}}\) will be.

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Answer & Solution

Correct Answer

(B)  \(0.56\)

Step-by-step Solution

Detailed explanation

Efficiency \(\eta=1-\frac{ T _{ L }}{ T _{ H }}\)

\(\eta_{1}=1-\frac{147+273}{447+273}=1-\frac{420}{720}\)

\(\eta_{1}=\frac{300}{720}\)

\(\eta_{2}=1-\frac{47+273}{947+273}=1-\frac{320}{1220}\)

\(\eta_{2}=\frac{900}{1220}\)

\(\frac{\eta_{1}}{\eta_{2}}=\frac{300}{720} \times \frac{1220}{900}=\frac{122}{72 \times 3}\)

\(\frac{\eta_{1}}{\eta_{2}}=0.56\)

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