$A$ particle performs uniform circular motion with an angular momentum $L$. If the angular frequency of the particle is doubled and its kinetic energy is halved,what will be its new angular momentum?

$A$ 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:

$A$ circular hoop of mass $m$ and radius $R$ rests flat on a horizontal frictionless surface. $A$ bullet,also of mass $m$ and moving with a velocity $v$,strikes the hoop and gets embedded in it. The thickness of the hoop is much smaller than $R$. The angular velocity with which the system rotates after the bullet strikes the hoop is

The torque acting on a body about a point is equal to $\overrightarrow{A} \times \overrightarrow{L}$,where $\overrightarrow{A}$ is a constant vector and $\overrightarrow{L}$ is the angular momentum about that point. This implies that:

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

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

$A$ uniform cylinder of radius $R$ is spun with angular velocity $\omega$ about its axis and then placed into a corner. The coefficient of friction between the cylinder and the planes is $\mu$. The number of turns taken by the cylinder before stopping is given by