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
- AI ^2
- BI ^2 2
- CI ^2 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