$A$ wheel of moment of inertia $2 \ kg \ m^2$ is rotating about an axis passing through its centre and perpendicular to its plane at a speed of $60 \ rad \ s^{-1}$. Due to friction, it comes to rest in $5$ minutes. The angular momentum of the wheel three minutes before it stops rotating is:

The figure shows a system consisting of $(i)$ a ring of outer radius $3R$ rolling clockwise without slipping on a horizontal surface with angular speed $\omega$ and $(ii)$ an inner disc of radius $2R$ rotating anti-clockwise with angular speed $\omega/2$. The ring and disc are separated by frictionless ball bearings. The system is in the $x-z$ plane. The point $P$ on the inner disc is at distance $R$ from the origin,where $OP$ makes an angle of $30^{\circ}$ with the horizontal. Then with respect to the horizontal surface,
$(A)$ the point $O$ has linear velocity $3R\omega\hat{i}$.
$(B)$ the point $P$ has a linear velocity $\frac{11}{4}R\omega\hat{i} + \frac{\sqrt{3}}{4}R\omega\hat{k}$.
$(C)$ the point $P$ has linear velocity $\frac{13}{4}R\omega\hat{i} - \frac{\sqrt{3}}{4}R\omega\hat{k}$.
$(D)$ The point $P$ has a linear velocity $(3 - \frac{\sqrt{3}}{4})R\omega\hat{i} + \frac{1}{4}R\omega\hat{k}$.

Fill in the blanks:
$(1)$ If the velocity of the center of mass of a body is $v_{cm} = 0$ and the angular velocity is $\omega = 0$,the body is said to be in ............. equilibrium.
$(2)$ Angular momentum is generated in a body when a ............. acts on it.
$(3)$ If a barrel is filled half with water,its center of gravity will move ............. .
$(4)$ The point at which the entire mass of a body is assumed to be concentrated is called the ............. .

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 $(2)$:

$A$ particle performs rotational motion with an angular momentum $L$. If the frequency of rotation is doubled and its kinetic energy becomes one-fourth,the new angular momentum becomes:

$A$ uniform cylinder of radius $1 \,m$, mass $1 \,kg$ spins about its axis with an angular velocity $20 \,rad/s$. At a certain moment, the cylinder is placed into a corner as shown in the figure. The coefficient of friction between the horizontal wall and the cylinder is $\mu$, whereas the vertical wall is frictionless. If the number of rounds made by the cylinder is $5$ before it stops, then the value of $\mu$ is (acceleration due to gravity, $g=10 \,m/s^2$)