$A$ 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 the 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:

Two rotating bodies $A$ and $B$ of masses $m$ and $2m$ with moments of inertia $I_A$ and $I_B$ $(I_B > I_A)$ have equal kinetic energy of rotation. If $L_A$ and $L_B$ are their angular momenta respectively,then:

If the angular momentum of a rotating body is increased by $200\ \%$,then its kinetic energy of rotation will be increased by .......... $\%$

$A$ thin uniform rod of length $L = 2 \text{ m}$, cross-sectional area $A$, and density $d$ is rotated about an axis passing through its center and perpendicular to its length with angular velocity $\omega$. If the value of $\omega$ in terms of its rotational kinetic energy $E$ is $(\alpha E/Ad)^{1/2}$, then the value of $\alpha$ 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 $\%$)