Let η₁ is the efficiency of an engine at T₁ = 447 and T₂ = 147 while η₂ is the efficiency at T₁ = 947 and T₂ = 47. The ratio (η₁)/(η₂) will be: (a) 0.41 (b) 0.56 (c) 0.73 (d) 0.70

Answer: (b). Convert to kelvin: 720 K, 420 K and 1220 K, 320 K. η₁ = 1 - (420)/(720) = (300)/(720) ≈ 0.417 η₂ = 1 - (320)/(1220) = (900)/(1220) ≈ 0.738 (η₁)/(η₂) ≈ 0.56. So the answer is (b) 0.56.

Let η₁ is the efficiency of an engine at T₁ = 447 and T₂ = 147 while… | JEE Main 2022 (25 Jul, Shift 2)

Q34JEE Main 2022 · 25 JUL · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

Let $\eta_1$ is the efficiency of an engine at $T_1 = 447^\circ\text{C}$ and $\mathrm{T}_2 = 147^\circ\text{C}$ while $\eta_2$ is the efficiency at $\mathrm{T}_1 = 947^\circ\text{C}$ and $\mathrm{T}_2 = 47^\circ\text{C}$ . The ratio $\frac{\eta_1}{\eta_2}$ will be:

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Answer: (b)

Convert to kelvin: 720 K, 420 K and 1220 K, 320 K.
$\eta_1 = 1 - \frac{420}{720} = \frac{300}{720} \approx 0.417$
$\eta_2 = 1 - \frac{320}{1220} = \frac{900}{1220} \approx 0.738$
$\frac{\eta_1}{\eta_2} \approx 0.56$ . So the answer is (b) 0.56.

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