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:

$A$ body having a moment of inertia about its axis of rotation equal to $3 \ kg \ m^{2}$ is rotating with an angular velocity of $3 \ rad \ s^{-1}$. The kinetic energy of this rotating body is the same as that of a body of mass $27 \ kg$ moving with velocity $v$. The value of $v$ is: (in $m \ s^{-1}$)

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

Two bodies have their moments of inertia $I$ and $2I$ respectively about their axis of rotation. If their kinetic energies of rotation are equal,their angular momentum will be in the ratio

$A$ 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:

$A$ thin circular disc of mass $12 \,kg$ and radius $0.5 \,m$ rotates with an angular velocity of $100 \,rad/s$. The rotational kinetic energy of the disc is (in $\,kJ$)