$A$ body with a moment of inertia of $3 \ kg \cdot m^2$ rotating with an angular speed of $2 \ rad/s$ has the same kinetic energy as a mass of $12 \ kg$ moving with a speed of ......... $m/s$.

The angular momentum of a particle performing uniform circular motion is $L$. If the kinetic energy of the particle is doubled and the frequency is halved,then the new angular momentum becomes:

$A$ sphere of mass $M$ and radius $R$ is attached by a light rod of length $l$ to a point $P$. The sphere rolls without slipping on a circular track as shown. It is released from the horizontal position. The angular momentum of the system about $P$ when the rod becomes vertical is:

Consider a disc rotating in the horizontal plane with a constant angular speed $\omega$ about its centre $O$. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown,two pebbles $P$ and $Q$ are simultaneously projected at an angle towards $R$. The velocity of projection is in the $y-z$ plane and is same for both pebbles with respect to the disc. Assume that $(i)$ they land back on the disc before the disc completes $\frac{1}{8}$ rotation,$(ii)$ their range is less than half disc radius,and $(iii)$ $\omega$ remains constant throughout. Then

If the earth suddenly contracts to $(1/3)^{rd}$ of its present size without any change in its mass, the ratio of the kinetic energy of the earth after and before contraction will be (Earth is assumed to be a rotating sphere about itself).

$A$ roller is made by joining together two cones at their vertices $O$. It is kept on two rails $AB$ and $CD$,which are placed asymmetrically (see figure),with its axis perpendicular to $CD$ and its centre $O$ at the centre of the line joining $AB$ and $CD$ (see figure). It is given a light push so that it starts rolling with its centre $O$ moving parallel to $CD$ in the direction shown. As it moves,the roller will tend to