JEE Main 2022PhysicsGravitationMediumMCQ

JEE Main 2022 — Gravitation Question with Solution

JEE Main 2022 (25 Jul Shift 2)

Question

The length of a seconds pendulum at a height h=2R from earth surface will be:

(Given: R= Radius of earth and acceleration due to gravity at the surface of earth g=π2 m s-2)

Choose an option

Show full solutionCorrect option: D
Correct answer
D19m

Step-by-step explanation

The acceleration due to gravity at height 2R from the surface of the Earth will be,

g'=gR2R+h2⇒g'=gR2R+2R2⇒g'=g9

Length of the second's pendulum at the surface of the Earth is 1 m.

Since, time period of a pendulum is given by,

T=2πlg⇒T'T=l'l×gg'⇒1=l'l×gg'⇒l'=l9=19 m

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About this question

This is a previous-year question from JEE Main 2022, covering the Gravitation 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.