$A$ 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 $m \ s^{-1}$)

$A$ 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:

The angular momentum of a body rotating about a fixed axis is increased by $10\%$. What is the percentage increase in its rotational kinetic energy (in $\%$)?

The ratio of rotational and translatory kinetic energies of a sphere is

$A$ solid sphere of mass $M$ and radius $R$ is rotating about its diameter. $A$ disc of the same mass and radius is also rotating about an axis passing through its centre and perpendicular to the plane,but its angular speed is twice that of the sphere. The ratio of the kinetic energy of the disc to that of the sphere is: (in $: 1$)

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: