JEE Mains · Physics · STD 11 - 11. thermodynamics
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.
- A \(0.41\)
- B \(0.56\)
- C \(0.73\)
- D \(0.70\)
Answer & Solution
Correct Answer
(B) \(0.56\)
Step-by-step Solution
Detailed explanation
Efficiency \(\eta=1-\frac{ T _{ L }}{ T _{ H }}\) \(\eta_{1}=1-\frac{147+273}{447+273}=1-\frac{420}{720}\) \(\eta_{1}=\frac{300}{720}\) \(\eta_{2}=1-\frac{47+273}{947+273}=1-\frac{320}{1220}\) \(\eta_{2}=\frac{900}{1220}\)…
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