JEE Main 2022PhysicsThermodynamicsMediumMCQ

JEE Main 2022 — Thermodynamics Question with Solution

JEE Main 2022 (28 Jul Shift 1)

Question

A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40°C, its efficiency increases by 30%. The temperature of the source will be :

Choose an option

Show full solutionCorrect option: C
Correct answer
C266.7 K

Step-by-step explanation

Initial efficiency η0=50%, the increase in efficiency is 30%. Therefore, new efficiency η=501+30100=65%.

Efficiency of the Carnot engin is given by η=1-TLTH

⇒η0=12=1-TLTH

When the temperature of the sink is decreased by 40°C=40 K.

η=121·3=1-TL-40 TH

⇒121·3=12+40TH

⇒TH=266.7 K

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Thermodynamics chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2022, 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.