$A$ uniform bar of length $6l$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ moving in the same horizontal plane with speed $2v$ and $v$ respectively,strike the bar (as shown in the figure) and stick to the bar after collision. The total rotational kinetic energy about the centre of mass $c$ will be:

$A$ 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

$A$ particle performs uniform circular motion with an angular momentum $L$. If the angular frequency of the particle is doubled and its kinetic energy is halved,what will be its new angular momentum?

Two rings of the same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass of the ring $= m$,radius $= r$)

$A$ sphere of mass $M$ and radius $R$ is attached by a light rod of length $l$ to a point $P$. The sphere rolls without slipping on a circular track as shown. It is released from the horizontal position. The angular momentum of the system about $P$ when the rod becomes vertical is:

$A$ thin ring of mass $2 \ kg$ and radius $1 \ m$ is rolling without slipping on a horizontal plane with velocity $1 \ m/s$. $A$ small ball of mass $1 \ kg$,moving with velocity $2 \ m/s$ in the opposite direction,hits the ring at a height of $1.8 \ m$ and goes vertically up with velocity $1 \ m/s$. Immediately after the collision:
$(A)$ the ring has pure rotation about its stationary $CM$.
$(B)$ the ring comes to a complete stop.
$(C)$ friction between the ring and the ground is to the left.
$(D)$ there is no friction between the ring and the ground.