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 is

Options

A. `(lomega^(2))/(2)`
B. `(lomega^(2))/(4)`
C. `(lomega^(2))/(6)`
D. `(lomega^(2))/(8)`

Answer

B

Explanation

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)`