JEE Main 2019PhysicsRotational MotionEasyMCQ

JEE Main 2019 — Rotational Motion Question with Solution

JEE Main 2019 (09 Apr Shift 2)

Question

Moment of inertia of a body about a given axis is 1.5 kg m2. Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s2 must be applied about the axis for a duration of:

Choose an option

Show full solutionCorrect option: B
Correct answer
B2 s

Step-by-step explanation

We know that, 

Rotational kinetic energy, K=12Iω2 

⇒12Iω2=1200
⇒12×1.5×ω2=1200
⇒ω=40 rad s-1

We also know that, ω=αt
⇒t=4020=2 s

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About this question

This is a previous-year question from JEE Main 2019, 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.