JEE Main 2022PhysicsLaws of MotionMediumMCQ

JEE Main 2022 — Laws of Motion Question with Solution

JEE Main 2022 (27 Jun Shift 2)

Question

One end of a massless spring of spring constant k and natural length l0 is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity ω about an axis passing trough fixed end, then the elongation of the spring will be

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Show full solutionCorrect option: C
Correct answer
Cmω2l0k-mω2

Step-by-step explanation

The particle is moving in a horizontal circle, so it is accelerated towards the centre with magnitude v2r. The horizontal force on the particle is due to the spring and is given by kx, where x is the elongation and k is the spring constant.

kx=mv2r=mω2r=mω2l0+x

⇒k-mω2x=mω2l0

⇒x=mω2l0k-mω2

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About this question

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