JEE Main 2024PhysicsOscillationsMediumNumerical

JEE Main 2024 — Oscillations Question with Solution

JEE Main 2024 (30 Jan Shift 2)

Question

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be _________s.

[take g=π2 m s-2]

Enter your answer

Show full solutionCorrect answer: 8
Correct answer
8

Step-by-step explanation

Acceleration due to gravity at surface of earth, g=GMR2.

Now, at distance R from the surface of the earth, g'=GMR+R2=g4

Therefore,
T=2πlg4

⇒T=2π4×4 g

⇒T=2π4π=8 s

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About this question

This is a previous-year question from JEE Main 2024, covering the Oscillations 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.