JEE Main 2020PhysicsCapacitanceHardNumerical

JEE Main 2020 — Capacitance Question with Solution

JEE Main 2020 (02 Sep Shift 1)

Question

A 5 μF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 μF capacitor. If the energy change during the charge redistribution is X100 J then value of X to the nearest integer is :

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Show full solutionCorrect answer: 4
Correct answer
4

Step-by-step explanation

C1=5μF  V1=220Volt

C2=2.5μF  V2=0

Heat loss;

ΔH=U1-Ut=12C1C2C1+C2v1-v22

=12×5×2.5(5+2.5)(220-0)2μJ

=52×3×22×22×100×10-6J

=5×11×223×10-4J=55×223×10-4J

=12103×10-4J=12103×10-3J=4×10-2

According to questions

x100=4×10-2

so, x=4

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About this question

This is a previous-year question from JEE Main 2020, covering the Capacitance 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.