The moments of inertia of two freely rotating bodies $A$ and $B$ are $I_{A}$ and $I_{B}$ respectively. Given $I_{A} > I_{B}$ and their angular momenta are equal. If $K_{A}$ and $K_{B}$ are their kinetic energies,then:

$A$ 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:

$A$ flywheel rotating about a fixed axis has a kinetic energy of $500 \text{ J}$. If its angular frequency is $5 \text{ Hz}$, calculate the moment of inertia of the wheel about the axis of rotation. (Take $\pi^2 = 10$) (in $\text{ kg m}^2$)

Two bodies $A$ and $B$ have their moments of inertia $I_1$ and $I_2$ respectively about their axis of rotation. If their kinetic energies of rotation are equal and their angular momenta $L_1$ and $L_2$ respectively are in the ratio $1 : \sqrt{3}$,then $I_2$ will be

$A$ spherical solid ball of $1\,kg$ mass and radius $30\,cm$ is rotating about an axis passing through its centre with an angular velocity of $50\,rad/s$. The kinetic energy of rotation is ......... $J$.

$A$ solid cylinder of mass $M$ and radius $R$ is rotating about its geometrical axis. $A$ solid sphere of the same mass and same radius is also rotating about its diameter with an angular speed half that of the cylinder. The ratio of the kinetic energy of rotation of the sphere to that of the cylinder will be