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

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:

$A$ uniform thin rod of length $l$ and mass $m$ is suspended from one of its ends and is rotated at $f$ rotations per second. The rotational kinetic energy of the rod will be

$A$ solid cylinder of mass $M$ and radius $R$ is rotating about its geometrical axis. $A$ solid sphere of 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:

Three identical square plates rotate about the axes shown in the figure in such a way that their kinetic energies are equal. Each of the rotation axes passes through the centre of the square. Then the ratio of angular speeds $\omega_{1}: \omega_{2}: \omega_{3}$ is