JEE Main 2021PhysicsCapacitanceMediumMCQ

JEE Main 2021 — Capacitance Question with Solution

JEE Main 2021 (01 Sep Shift 2)

Question

A capacitor is connected to a 20 V battery through a resistance of 10Ω. It is found that the potential difference across the capacitor rises to 2 V in 1 μs. The capacitance of the capacitor is ________μF.
Given In 109=0.105

Choose an option

Show full solutionCorrect option: A
Correct answer
A0.95

Step-by-step explanation

V=V01-e-t/RC

2=201-e-1μs10C

110=1-e-1×10-610C

e-1×10-610C=910

1×10-610C=ln109

C=1×10-610×ln109=10-71.05=11.05μF=100105μF=0.95μF

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Capacitance 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 2021, 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.