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MHT CET · Physics · Thermodynamics

An engine operating between temperatures \(T_1\) and \(T_2\) has efficiency \(\frac{1}{5}\). When \(\mathrm{T}_2\) is lowered by 45 K , its efficiency becomes \(\frac{1}{2}\). Temperatures \(\mathrm{T}_1\) and \(\mathrm{T}_2\) are respectively

  1. A \(100 \mathrm{~K}, 70 \mathrm{~K}\)
  2. B \(160 \mathrm{~K}, 120 \mathrm{~K}\)
  3. C \(140 \mathrm{~K}, 110 \mathrm{~K}\)
  4. D \(150 \mathrm{~K}, 120 \mathrm{~K}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(150 \mathrm{~K}, 120 \mathrm{~K}\)

Step-by-step Solution

Detailed explanation

\(\eta_1 = 1 - \frac{T_2}{T_1} \implies \frac{1}{5} = 1 - \frac{T_2}{T_1} \implies \frac{T_2}{T_1} = \frac{4}{5}\) \(\eta_2 = 1 - \frac{T_2 - 45}{T_1} \implies \frac{1}{2} = 1 - \frac{T_2 - 45}{T_1} \implies \frac{T_2 - 45}{T_1} = \frac{1}{2}\)
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