The moment of inertia of a ring about an axis passing through the 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 the rings are rotating about the same axis,then the loss in kinetic energy is

$A$ uniform bar of length $12 \text{ cm}$ and mass $20m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ are moving in opposite directions with the same speed $v$ and in the same plane as the bar. These masses strike the bar simultaneously and get stuck to it. After the collision, the entire system is rotating with an angular frequency $\omega$. The ratio of $v$ and $\omega$ is:

$A$ uniform cylinder of radius $R$ is spun with angular velocity $\omega$ about its axis and then placed into a corner. The coefficient of friction between the cylinder and the planes is $\mu$. The number of turns taken by the cylinder before stopping is given by

$A$ pendulum consists of a bob of mass $m=0.1 \ kg$ and a massless inextensible string of length $L=1.0 \ m$. It is suspended from a fixed point at height $H=0.9 \ m$ above a frictionless horizontal floor. Initially,the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. $A$ horizontal impulse $P=0.2 \ kg \cdot m/s$ is imparted to the bob at some instant. After the bob slides for some distance,the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is $J \ kg \cdot m^2/s$. The kinetic energy of the pendulum just after the lift-off is $K$ Joules. $(1)$ The value of $J$ is. . . . . . $(2)$ The value of $K$ is. . . . . Give the answers of the questions $(1)$ and $(2)$.

$A$ uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $J$ is applied to the end $B$,perpendicular to the rod in the horizontal direction. Find the speed of particle $P$ at a distance $\frac{l}{6}$ from the centre towards $A$ of the rod after time $t = \frac{\pi m l}{12 J}$.

The angular momentum of a rotating body is $L$. When the frequency of the rotating body is tripled and its kinetic energy is made one-third, the new angular momentum becomes: