JEE Main 2019PhysicsElectrostaticsMediumMCQ

JEE Main 2019Electrostatics Question with Solution

JEE Main 2019 (10 Apr Shift 2)

Question

A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field  E , as shown in figure. Its bob has mass m and charge  q . The time period of the pendulum is given by:

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Show full solutionCorrect option: A
Correct answer
A2πLg2+qEm2

Step-by-step explanation


Two forces will be acting on the mass m, gravity force and force by the electric field.

Thus, the effective force =  qE2+mg2
From force equation, geff=g2+qEm2
Time period of the pendulum is T=2πLgeff
=2π Lg2+qEm2

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About this question

This is a previous-year question from JEE Main 2019, 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.