$A$ thin hollow cylinder slides with velocity $v$ without rotating. It then rolls without slipping with the same velocity $v$. Find the ratio of the kinetic energies in the two cases.

$A$ uniform rod of mass $m$ and length $l$ hinged at its end is released from rest when it is in the horizontal position. The normal reaction at the hinge when the rod becomes vertical is:

$A$ solid sphere of mass $M$,radius $R$ and having moment of inertia about an axis passing through the centre of mass as $I$,is recast into a disc of thickness $t$,whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$. Then,the radius of the disc will be

$A$ block of mass $m = 1 \, kg$ slides with velocity $v = 6 \, m/s$ on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about $O$ and swings as a result of the collision,making an angle $\theta$ before momentarily coming to rest. If the rod has mass $M = 2 \, kg$ and length $l = 1 \, m$,the value of $\theta$ is approximately (Take $g = 10 \, m/s^2$) (in $^{\circ}$)

This question has Statement $1$ and Statement $2$. Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement $1$ : When the moment of inertia $I$ of a body rotating about an axis with angular speed $\omega$ increases,its angular momentum $L$ remains unchanged,but the kinetic energy $K$ decreases if no external torque is applied.
Statement $2$ : $L = I\omega$ and the rotational kinetic energy $K = \frac{1}{2}I\omega^2 = \frac{L^2}{2I}$.

$A$ thin rod of mass $M$ and length $a$ is free to rotate in a horizontal plane about a fixed vertical axis passing through point $O$. $A$ thin circular disc of mass $M$ and radius $a/4$ is pivoted on this rod with its center at a distance $a/4$ from the free end so that it can rotate freely about its vertical axis,as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity $\Omega$ and the disc rotating about its vertical axis with angular velocity $4\Omega$. The total angular momentum of the system about the point $O$ is $\left(\frac{Ma^2\Omega}{48}\right) n$. The value of $n$ is. . . . .