$A$ thin rod of length $L$ is suspended from one end and rotates with $n$ revolutions per second. What will be the rotational kinetic energy of the rod?

$A$ body is rotating about its own axis. Its rotational kinetic energy is $x$ and its angular momentum is $y$. Hence,its moment of inertia about its own axis is:

$A$ solid sphere of mass $M$ and radius $R$ is rotating about its diameter. $A$ solid cylinder of the same mass and radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation ($K_{\text{sphere}}$ to $K_{\text{cylinder}}$) will be:

An energy of $484\,J$ is spent in increasing the speed of a flywheel from $60\,rpm$ to $360\,rpm$. The moment of inertia of the flywheel is $.............\,kg\cdot m^2$.

$A$ thin uniform rod of mass $m$ and length $l$ is suspended from one end,which can oscillate in a vertical plane about the point of suspension. It is pulled to one side and then released. It passes through the equilibrium position with angular speed $\omega$. The kinetic energy while passing through the mean position is

If the angular velocity of a body rotating about a given axis increases by $20 \%$,then its kinetic energy of rotation will increase by: (in $\%$)