Match the linear motion formulas in Column-$I$ with their corresponding rotational motion formulas in Column-$II$.

Column-$I$	Column-$II$
$(1)$ $W = F \Delta x$	$(a)$ $P = \tau \omega$
$(2)$ $P = Fv$	$(b)$ $W = \tau \Delta \theta$
	$(c)$ $L = I \omega$

$A$ uniform bar of length $6l$ and mass $8m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ moving in the same horizontal plane with speeds $2v$ and $v$ respectively,strike the bar (as shown in the figure) and stick to the bar after the collision. The total energy after the collision (about the center of mass,$c$) will be:

The power $(P)$ is supplied to a rotating body having moment of inertia '$I$' and angular acceleration '$\alpha$'. Its instantaneous angular velocity is

$A$ uniform rod of length $2L$ is placed with one end in contact with a horizontal surface. The other end is released from an angle $\alpha$ with the horizontal,such that the end in contact does not slip. What will be its angular velocity when it becomes horizontal?

$A$ tangential force $F$ is applied on a disc of radius $R$,due to which it deflects through an angle $\theta$ from its initial position. The work done by this force would be

Write the formula for power and angular momentum in rotational motion.