The angular momentum of a rigid body of mass $m$ about an axis is $n$ times the linear momentum $(P)$ of the body. The total kinetic energy of the rigid body is:

Two discs $A$ and $B$ are mounted coaxially on a vertical axle. The discs have moments of inertia $I$ and $2I$ respectively about the common axis. Disc $A$ is imparted an initial angular velocity $2\omega$ using the entire potential energy of a spring compressed by a distance $x_1$. Disc $B$ is imparted an angular velocity $\omega$ by a spring having the same spring constant and compressed by a distance $x_2$. Both the discs rotate in the clockwise direction.
$1.$ The ratio of $x_1/x_2$ is
$(A)$ $2$ $(B)$ $1/2$ $(C)$ $\sqrt{2}$ $(D)$ $1/\sqrt{2}$
$2.$ When disc $B$ is brought in contact with disc $A$,they acquire a common angular velocity in time $t$. The average frictional torque on one disc by the other during this period is
$(A)$ $\frac{2I\omega}{3t}$ $(B)$ $\frac{9I\omega}{2t}$ $(C)$ $\frac{9I\omega}{4t}$ $(D)$ $\frac{3I\omega}{2t}$
$3.$ The loss of kinetic energy during the above process is
$(A)$ $\frac{I\omega^2}{2}$ $(B)$ $\frac{I\omega^2}{3}$ $(C)$ $\frac{I\omega^2}{4}$ $(D)$ $\frac{I\omega^2}{6}$

$A$ particle performs rotational motion with an angular momentum $L$. If the frequency of rotation is doubled and its kinetic energy becomes one-fourth,the new angular momentum becomes:

The angular momentum of a particle performing uniform circular motion is $L$. If the kinetic energy of the particle is doubled and the frequency is halved,then the new angular momentum becomes:

$A$ particle is moving in a uniform circular motion with angular momentum $L$. If its angular frequency is doubled and its kinetic energy is halved,then the new angular momentum of the particle will be .......

$A$ cubical block of side $a = 30\,cm$ is moving with velocity $v = 2\,m/s$ on a smooth horizontal surface. The surface has a small bump at a point $O$ as shown in the figure. The angular velocity (in $rad/s$) of the block immediately after it hits the bump is: