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
- A \(100 \mathrm{~K}, 70 \mathrm{~K}\)
- B \(160 \mathrm{~K}, 120 \mathrm{~K}\)
- C \(140 \mathrm{~K}, 110 \mathrm{~K}\)
- D \(150 \mathrm{~K}, 120 \mathrm{~K}\)
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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