JEE Main 2020PhysicsLaws of MotionMediumMCQ

JEE Main 2020 — Laws of Motion Question with Solution

JEE Main 2020 (08 Jan Shift 1)

Question

A particle of mass m is fixed to one end of a light spring having force constant k and unstretched length l. The other end is fixed. The system is given an angular speed ω about the fixed end of the spring such that it rotates in a circle in gravity free space. Then the stretch in the spring is:

Choose an option

Show full solutionCorrect option: B
Correct answer
Bmlω2k-mω2

Step-by-step explanation

As we know, the spring force will give the necessary centripetal force for rotation.

So, 

mω2l+x=kx
lx+1=kmω2

 Thus, the stretch in the spring
x=lmω2k-mω2

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About this question

This is a previous-year question from JEE Main 2020, 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.