The moment of inertia of a ring about an axis passing though the centre and perpendicular to its plane is l. It is rotating ring is gently placed `omega` 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 isA. `(lomega^(2))/(2)`B. `(lomega^(2))/(4)`C. `(lomega^(2))/(6)`D. `(lomega^(2))/(8)`

The moment of inertia of a ring about an axis passing though the centr

1.	The moment of inertia of a ring about an axis passing though the centre and perpendicular to its plane is l. It is rotating ring is gently placed `omega` 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 isA. `(lomega^(2))/(2)`B. `(lomega^(2))/(4)`C. `(lomega^(2))/(6)`D. `(lomega^(2))/(8)`
Answer» Correct Answer - B According to the law of conservation of angular momentum, `lomega=` constant Now, according to the question `l_(1)omega_(1)=l_(2)omega_(2)orlomega=(2l)omega_(2)` `omega_(2)=(omega)/(2)` New kinetic energy `=[(1)/(2)l_(2)omega_(2)^(2)]=(1)/(2)(2l)xx((omega)/(2))^(2)=(lomega^(2))/(4)` Loss in kinetic energy `(K_(L))=K_(i)-K_(j)` `=(1)/(2)lomega^(2)-(lomega^(2))/(4)=(lomega^(2))/(4)`