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:

$A$ thin uniform bar lies on a frictionless horizontal surface and is free to move in any way on the surface. Its mass is $0.300 \, kg$ and length is $2 \, m$. Two particles,each of mass $0.100 \, kg$,are moving on the same surface towards the two ends of the bar in a direction perpendicular to the bar. One particle moves with a velocity of $10 \, m/s$ towards one end,and the other moves with a velocity of $5 \, m/s$ towards the other end. If the collision between the particles and the bar is completely elastic and both particles strike the bar simultaneously,find the velocity of the centre of mass of the bar after the collision in $m/s$.

In the figure shown,the plank is being pulled to the right with a constant speed $v$. If the cylinder does not slip,then:

$A$ disc of radius $R$ is rolling purely on a flat horizontal surface with a constant angular velocity $\omega$. The angle between the velocity and acceleration vectors of point $P$ (which is at the same horizontal level as the center $C$) is

$A$ ring rolls without slipping on the ground. Its centre $C$ moves with a constant speed $u$. $P$ is any point on the ring. The speed of $P$ with respect to the ground 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
$(A)$ $\frac{I\omega^2}{2}$ $(B)$ $\frac{I\omega^2}{3}$ $(C)$ $\frac{I\omega^2}{4}$ $(D)$ $\frac{I\omega^2}{6}$