$A$ uniform body of mass $M$ and radius $R$ has a small mass $m$ attached at its edge as shown in the figure. The system is placed on a perfectly rough horizontal surface such that mass $m$ is at the same horizontal level as the centre of the body. It is assumed that there is no slipping at point $A$. If $I_A$ is the moment of inertia of the combined system about the point of contact $A$,then the normal reaction at point $A$ just after the system is released from rest is ........ $N$. ($M = 6 \ kg$,$m = 2 \ kg$,$I_A = 4 \ kg \ m^2$,$R = 1 \ m$,$g = 10 \ m/s^2$)

$A$ 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:

Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is $I$. If the same rod is bent into a ring and its moment of inertia about its diameter is $I^{\prime}$,then the ratio $\frac{I}{I^{\prime}}$ is

$A$ thin uniform circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane about an axis passing through its centre and perpendicular to the plane with angular velocity $\omega$. Another disc of same mass but half the radius is gently placed over it coaxially. The angular speed of the composite disc will be :

$A$ cubical block of side $a = 30\,cm$ is moving with velocity $v = 2\,m/s$ on a smooth horizontal surface. The surface has a small bump at a point $O$ as shown in the figure. The angular velocity (in $rad/s$) of the block immediately after it hits the bump 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 $(2)$: