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

$A$ 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 $P$ and $Q$ of masses $m$ with moment of inertia $I_P$ and $I_Q$ $(I_Q > I_P)$ have equal kinetic energy of rotation. If $L_P$ and $L_Q$ are their angular momenta respectively,then:

The moment of inertia of a body about a given axis,rotating with an angular velocity of $1 \ rad/s$,is numerically equal to '$P$' times its rotational kinetic energy. The value of '$P$' is:

$A$ 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

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: