JEE Main 2018PhysicsThermodynamicsMediumMCQ

JEE Main 2018 — Thermodynamics Question with Solution

JEE Main 2018 (15 Apr)

Question

A Carnot's engine works like a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at a lower temperature. The amount of work done in each cycle to operate the refrigerator is,

Choose an option

Show full solutionCorrect option: B
Correct answer
B420 J

Step-by-step explanation

η=1− T 2 T 1 , efficiency =1-T2T1=WQ2+W⇒1-250300=WQ2+W =420 J

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About this question

This is a previous-year question from JEE Main 2018, 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.