JEE Main 2023PhysicsRotational MotionHardNumerical

JEE Main 2023 — Rotational Motion Question with Solution

JEE Main 2023 (01 Feb Shift 1)

Question

A solid cylinder is released from rest from the top of an inclined plane of inclination 30° and length 60 cm. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is ______ m s-1. (Given g=10 m s-2)

 

Enter your answer

Show full solutionCorrect answer: 2
Correct answer
2

Step-by-step explanation

Let speed of cylinder upon reaching the bottom of the inclined plane be v.

Here, gain in kinetic energy=loss in potential energy

So, 12Iω2+12mv2=mglsinθ

⇒12mr22ω2+12mv2=mglsin30°

For pure rolling, v=rω

⇒12mr22vr2+12mv2=mglsin30°

⇒14mv2+12mv2=mglsin30°⇒34mv2=mgl2⇒v=4gl6

Putting the values, we have

⇒v=4×10×0.66=2 m s-1

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Rotational Motion chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

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.