JEE Main 2021PhysicsElectrostaticsHardNumerical

JEE Main 2021 — Electrostatics Question with Solution

JEE Main 2021 (25 Feb Shift 1)

Question

512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is V in Volt.

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Show full solutionCorrect answer: 128
Correct answer
128

Step-by-step explanation

Q=512 q

 Volume i= Volume f

512×43πr3=43πR3

29r3=R3

R=8 r

2=kqr

V=kQR=k512q8r

V=128 V.

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About this question

This is a previous-year question from JEE Main 2021, covering the Electrostatics 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.