$A$ cube of side $a$ is moving with velocity $v$ on a smooth horizontal surface. It hits a linear raised obstacle $O$ on the horizontal surface (as shown in the figure). The angular speed of the block after hitting $O$ will be

$A$ 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 $B$	$(S)$ $\frac{3g}{\ell}$

On a solid sphere lying on a horizontal surface,a force $F$ is applied at a height of $R/2$ from the centre of mass. The initial acceleration of a point at the top of the sphere is (there is no slipping at any point).

$A$ thin rod $AB$ of length $L$ and mass $m$ is held horizontally so that it can freely rotate in a vertical plane about the end $A$ as shown in the figure. The potential energy of the rod when it hangs vertically is taken to be zero. The end $B$ of the rod is released from rest from a horizontal position. At the instant the rod makes an angle $\theta$ with the horizontal:

$A$ homogeneous disc of mass $2 \ kg$ and radius $15 \ cm$ is rotating about its axis (which is fixed) with an angular velocity $4 \ rad/s$. The linear momentum of the disc is

$A$ uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $J$ is applied to the end $B$,perpendicular to the rod in the horizontal direction. Find the speed of particle $P$ at a distance $\frac{l}{6}$ from the centre towards $A$ of the rod after time $t = \frac{\pi m l}{12 J}$.