$A$ uniform rod of mass $m$ and length $\ell$ hinged at end $A$ is released from the horizontal position shown in the figure. Just after the rod is released:

Column $I$	Column $II$
$(A)$ Angular acceleration of $C$	$(P)$ $\frac{3g}{2}$
$(B)$ Angular acceleration of $B$	$(Q)$ $\frac{3g}{2\ell}$
$(C)$ Acceleration of $C$	$(R)$ $\frac{3g}{4}$
$(D)$ Acceleration of $B$	$(S)$ $\frac{3g}{\ell}$

$A$ particle falls freely near the surface of the earth. Consider a fixed point $O$ (not vertically below the particle) on the ground.

$A$ uniform sphere of mass $m$ and radius $R$ is placed on a rough horizontal surface. The sphere is struck horizontally at a height $h$ from the floor. Match the following:

$(a)$ $h = \frac{R}{2}$	$(i)$ Sphere rolls without slipping with a constant velocity and no loss of energy.
$(b)$ $h = R$	$(ii)$ Sphere spins clockwise,loses energy by friction.
$(c)$ $h = \frac{3R}{2}$	$(iii)$ Sphere spins anti-clockwise,loses energy by friction.
$(d)$ $h = \frac{7R}{5}$	$(iv)$ Sphere has only a translational motion,loses energy by friction.

$A$ 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:

One twirls a circular ring (of mass $M$ and radius $R$) near the tip of one's finger as shown in Figure $1$. In the process,the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone,shown by the dotted line. The radius of the path traced out by the point where the ring and the finger are in contact is $r$. The finger rotates with an angular velocity $\omega_0$. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger are in contact (Figure $2$). The coefficient of friction between the ring and the finger is $\mu$ and the acceleration due to gravity is $g$.
$(1)$ The total kinetic energy of the ring is
$[A]$ $M \omega_0^2 R^2$ $[B]$ $\frac{1}{2} M \omega_0^2(R-r)^2$ $[C]$ $M \omega_0^2(R-r)^2$ $[D]$ $\frac{3}{2} M \omega_0^2(R-r)^2$
$(2)$ The minimum value of $\omega_0$ below which the ring will drop down is
$[A]$ $\sqrt{\frac{g}{\mu(R-r)}}$ $[B]$ $\sqrt{\frac{2 g}{\mu(R-r)}}$ $[C]$ $\sqrt{\frac{3 g}{2 \mu(R-r)}}$ $[D]$ $\sqrt{\frac{g}{2 \mu(R-r)}}$
Given the answers to questions $(1)$ and $(2)$:

Two identical rings $A$ and $B$ of same mass $M$ and radius $R$ are revolving. Ring $A$ revolves around its own diameter, and ring $B$ revolves about a tangential axis in its own plane. Both rings $A$ and $B$ have the same rotational kinetic energy. The ratio of the angular velocity of ring $B$ $(\omega_B)$ to that of ring $A$ $(\omega_A)$ is: