$A$ tangential force $F$ is applied on the rim of a ring of radius $R$. If the ring rotates by an angle $\theta$,what is the work done by the force?

The $M.I.$ of a body about the given axis is $1.2 \, kg \cdot m^2$ and initially the body is at rest. In order to produce a rotational kinetic energy of $1500 \, J$,an angular acceleration of $25 \, rad/s^2$ must be applied about that axis for a duration of ........ $s$.

$A$ 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:

$A$ thin rod $MN$,free to rotate in the vertical plane about the fixed end $N$,is held horizontal. When the end $M$ is released,the speed of this end,when the rod makes an angle $\alpha$ with the horizontal,will be proportional to (see figure):

$A$ thin uniform metal rod of mass $M$ and length $L$ is swinging about a horizontal axis passing through its end. Its maximum angular velocity is $\omega$. Its centre of mass rises to a maximum height of (where $g$ is the acceleration due to gravity):

$A$ thin uniform circular disc of mass $\frac{10}{\pi^2} \,kg$ and radius $2 \,m$ is rotating about an axis passing through its centre and perpendicular to its plane. The work done to increase the angular speed of the disc from $90 \,rev/min$ to $120 \,rev/min$ is (in $\,J$)