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JEE Main Physics question

PhysicsSystems of Particles and Rotational MotionJEE Main

A disc of moment of inertia (9.8/π²) kg·m² is rotating at 600 rpm. If the frequency of rotation is changed from 600 rpm to 300 rpm, then what is the work done?

Answer: (A)

Solution

Rotational KE = ½Iω² = ½I(2πn)². Work done = ½I(2π)²(n₁²−n₂²), with n₁ = 10

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Question ID VK-153565