The angular momentum of a rigid body of mass $m$ about an axis is $n$ times the linear momentum $(P)$ of the body. The total kinetic energy of the rigid body is:

Match Column-$I$ with Column-$II$.

Column-$I$	Column-$II$
$(1)$ $SI$ unit of torque	$(a)$ $m$
$(2)$ $SI$ unit of radius of gyration	$(b)$ $N\,m$
	$(c)$ $Js^{-2}$

$A$ solid sphere of mass $M$ and radius $R$ has a moment of inertia $I$ about its diameter. It is recast into a disc of thickness $t$ whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains $I$. The radius of the disc will be:

$A$ 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, $g=10 \,m/s^2$)

The position vector $\vec{r}$ of a particle of mass $m$ is given by the following equation:
$\vec{r}(t) = \alpha t^3 \hat{i} + \beta t^2 \hat{j}$
where $\alpha = 10/3 \ m \ s^{-3}$,$\beta = 5 \ m \ s^{-2}$ and $m = 0.1 \ kg$. At $t = 1 \ s$,which of the following statement$(s)$ is(are) true about the particle?
$(A)$ The velocity $\vec{v}$ is given by $\vec{v} = (10 \hat{i} + 10 \hat{j}) \ m \ s^{-1}$
$(B)$ The angular momentum $\vec{L}$ with respect to the origin is given by $\vec{L} = -(5/3) \hat{k} \ N \ m \ s$
$(C)$ The force $\vec{F}$ is given by $\vec{F} = (2 \hat{i} + 1 \hat{j}) \ N$
$(D)$ The torque $\vec{\tau}$ with respect to the origin is given by $\vec{\tau} = -(20/3) \hat{k} \ N \ m$

$A$ 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: