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MHT CET202611 April 2026Evening 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 rings are rotating about the same axis then loss in kinetic energy is

Options

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

Correct answer

C. I ^2 4

Step-by-step solution

Initial angular momentum of the system is L_i = I Initial kinetic energy is K_i = 1 2 I ^2 When an identical ring is placed on it, the new moment of inertia becomes I_f = I + I = 2I By conservation of angular momentum, I_f _f = I_i _i 2I _f = I _f = 2 Final kinetic energy is K_f = 1 2 (2I) ( 2 )^2 = 1 4 I ^2 Loss in kinetic energy is K = K_i - K_f = 1 2 I ^2 - 1 4 I ^2 = I ^2 4 Answer: I ^2 4

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