$A$ thin rod of mass $m$ and length $l$ is oscillating about a horizontal axis through its one end. Its maximum angular speed is $\omega$. Its centre of mass will rise up to a maximum height of:

$A$ disc of mass $1\,kg$ and radius $R$ is free to rotate about a horizontal axis passing through its centre and perpendicular to the plane of the disc. $A$ body of the same mass as that of the disc is fixed at the highest point of the disc. Now the system is released. When the body comes to the lowest position,its angular speed will be $4 \sqrt{\frac{x}{3 R}} \text{ rad s}^{-1}$ where $x=$ (Given $g = 10 \text{ m s}^{-2}$)

The Moon revolves around the Earth in an orbit of radius $R$ with a time period of revolution $T$. It also rotates about its own axis with a time period $T$. If the mass of the Moon is $M$ and its radius is $r$,the total kinetic energy of the Moon is:

$A$ 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):

$A$ 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 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 $\,J$)