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MHT CET202521 Apr 2025Evening ShiftPhysicsRotational MotionActual

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is I. It is rotating with angular velocity . Another identical ring is gently placed on it so that their centres coincide. If both the rings are rotating about the same axis then loss in kinetic energy is

Options

  1. AI ^2 16
  2. BI ^2 8
  3. CI ^2 4
  4. DI ^2 2

Correct answer

C. I ^2 4

Step-by-step solution

The moment of inertia I and initial angular velocity characterize the first ring. The system's initial angular momentum is conserved since no external torque acts upon it. The initial angular momentum is L_i = I . When a second identical ring is placed concentrically, the total moment of inertia becomes I_f = I + I = 2I . Conservation of angular momentum gives I = 2I _f , so the final angular velocity is _f = / 2 . The initial kinetic energy is KE_i = 1 2 I ^2 . The final kinetic energy is KE_f = 1 2 (2I)( /2)^2 =

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