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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.

  1. A \(0.41\)
  2. B  \(0.56\)
  3. C \(0.73\)
  4. D \(0.70\)
Verified Solution

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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