JEE Main 2021PhysicsThermodynamicsMediumNumerical

JEE Main 2021 — Thermodynamics Question with Solution

JEE Main 2021 (25 Feb Shift 2)

Question

A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52 K, its efficiency is doubled. The temperature in Kelvin of the source will be ___ .

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Show full solutionCorrect answer: 208
Correct answer
208

Step-by-step explanation

Efficiency of reversible engine is η=14=WQin=1-T2 T1

From above, T2 T1=34

When the temperature of the sink is reduced by 52 K then its efficiency is doubled.

Then, T2-52 T1=12

⇒T2 T1-52 T1=12⇒34-52 T1=12⇒52 T1=34-12=14⇒T1=52×4=208 K

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About this question

This is a previous-year question from JEE Main 2021, covering the Thermodynamics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.