JEE Main 2023PhysicsRotational MotionHardNumerical

JEE Main 2023 — Rotational Motion Question with Solution

JEE Main 2023 (30 Jan Shift 2)

Question

A uniform disc of mass 0.5 kg and radius r is projected with velocity 18 m s-1 at t=0 s on a rough horizontal surface. It starts off with a purely sliding motion at t=0 s. After 2 s it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after 2 s will be ______ J.

(given, coefficient of friction is 0.3 and g=10 m s-2).

 

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Show full solutionCorrect answer: 54
Correct answer
54

Step-by-step explanation

Acceleration of the disc, a=-μkg=-3 m s-2.

Now, velocity of the centre of mass

v=u+at=18-3×2

⇒v=12 m s-1

Now total kinetic energy during pure rolling(v=ωr) will become,

KE=12mv2+12Iω2=12mv2+12mr22v2r2

⇒KE=34mv2

=3×18=54 J

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About this question

This is a previous-year question from JEE Main 2023, covering the Rotational 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.