JEE Main 2024PhysicsLaws of MotionMediumNumerical

JEE Main 2024 — Laws of Motion Question with Solution

JEE Main 2024 (29 Jan Shift 2)

Question

A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential components of its acceleration are equal. If its speed at t=0 is 4 m s-1, the time taken to complete the first revolution will be 1α1-e-2π s, where α=______.

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Show full solutionCorrect answer: 8
Correct answer
8

Step-by-step explanation

Given:

a→c=a→t

⇒v2r=dvdt

⇒∫4vdvv2=∫0tdtr

⇒-1v4v=tr

⇒-1v+14=2t

⇒v=41-8t=dsdt

⇒4∫0tdt1-8t=∫0sds

r=0.5 m, distance covered s=2πr=π

⇒4×[ln(1-8t)]0t-8=π

⇒ℓn(1-8t)=-2π

⇒1-8t=e-2π

⇒t=1-e-2π18 s

So, α=8.

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About this question

This is a previous-year question from JEE Main 2024, covering the Laws of Motion 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.