JEE Main 2022PhysicsLaws of MotionEasyMCQ

JEE Main 2022 — Laws of Motion Question with Solution

JEE Main 2022 (25 Jun Shift 2)

Question

A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity ω. The coefficient of static friction between the bottom of the beaker and the surface of the disc is μ. The beaker will revolve with the disc if :

Choose an option

Show full solutionCorrect option: B
Correct answer
BR≤μgω2

Step-by-step explanation

As the disc is revolving in circular motion about an axis passing through centre and perpendicular to its plane, thus centrifugal force must act on it. The forces acting on disc is shown below.

 From FBD

For circular motion, frictional force must balance the required centripetal force.

So, f=mω2R

⇒μN≥mω2R

⇒μmg≥mω2R

⇒R≤μgω2

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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.