$A$ square lamina $OABC$ of side length $10 \ cm$ is pivoted at $O$. Forces act on the lamina as shown in the figure. If the lamina remains stationary,then the magnitude of $F$ is:

$A$ uniform rod of length $2L$ has one end on a horizontal floor. It is inclined at an angle $\alpha$ to the horizontal floor. It falls without slipping,rotating about the point of contact. Its angular velocity when it hits the horizontal floor will be:

$A$ ring of mass $m = 1 \ kg$ and radius $R = 1.25 \ m$ is kept on a rough horizontal ground. $A$ small body of same mass $m = 1 \ kg$ is stuck to the top of the ring. When it is given a slight push forward,the ring starts rolling purely on the ground. What is the maximum speed of the centre of the ring (in $m/s$)?

An object has a moment of inertia of $3 \ kg \cdot m^2$. It is rotating with an angular velocity of $2 \ rad/s$. If a mass of $12 \ kg$ is moving with a velocity of $v \ m/s$,at what value of $v$ will their kinetic energies be equal?

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 $(2)$:

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).