JEE Main 2024PhysicsRotational MotionMediumNumerical

JEE Main 2024 — Rotational Motion Question with Solution

JEE Main 2024 (30 Jan Shift 2)

Question

Two discs of moment of inertia I1=4 kg m2 and I2=2 kg m2 about their central axes & normal to their planes, rotating with angular speeds 10 rad s-1 & 4 rad s-1 respectively are brought into contact face to face with their axe of rotation coincident. The loss in kinetic energy of the system in the process is _________J.

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Show full solutionCorrect answer: 24
Correct answer
24

Step-by-step explanation

Applying conservation of angular momentum, we get

I1ω1+I2ω2=I1+I2ω0⇒4×10+2×4=4+2ω0⇒ω0=8 rad s-1

Now, initial rotational kinetic energy is E1=12I1ω12+12I2ω22=216 J

and final rotational kinetic energy is

E2=12I1+I2ω02=192 J

Required, energy loss ΔE=E1-E2=24 J

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About this question

This is a previous-year question from JEE Main 2024, 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.