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 ω. 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 (a) Iω² (b) Iω²2 (c) Iω²4 (d) Iω²8

Answer: (c). No external torque: Iω=(2I)ω', so ω'=(ω)/(2). Ki=12Iω², Kf=12(2I)(ω²)/(4)=(Iω²)/(4) Loss =(Iω²)/(2)-(Iω²)/(4)=(Iω²)/(4) So the answer is (c) (Iω²)/(4).

The moment of inertia of a ring about an axis passing through its… | MHT CET 2026 (11 Apr, Shift 2)

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Angular Momentum & its Conservation

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 $\omega$ . 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

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Answer: (c)

No external torque: $I\omega=(2I)\omega'$ , so $\omega'=\frac{\omega}{2}$ .
$K_i=\frac12I\omega^2$ , $K_f=\frac12(2I)\frac{\omega^2}{4}=\frac{I\omega^2}{4}$
Loss $=\frac{I\omega^2}{2}-\frac{I\omega^2}{4}=\frac{I\omega^2}{4}$
So the answer is (c) $\frac{I\omega^2}{4}$ .

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