If a body completes one revolution in $\pi \ s$,then the moment of inertia is related to the rotational kinetic energy $(K)$ by which of the following expressions?

An energy of $484\,J$ is spent in increasing the speed of a flywheel from $60\,rpm$ to $360\,rpm$. The moment of inertia of the flywheel is $.............\,kg\cdot m^2$.

$A$ 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 identical particles each of mass $m$ are separated by a distance $d$. The axis of rotation passes through the midpoint of $d$ and is perpendicular to the length $d$. If $K$ is the rotational kinetic energy of the system,then the angular frequency $\omega$ is:

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:

$A$ flywheel rotating about a fixed axis has a kinetic energy of $360 \ J$ when its angular speed is $30 \ rad/s$. The moment of inertia of the wheel about the axis of rotation is ...... $kg \ m^2$.