Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is $I$. If the same rod is bent into a ring and its moment of inertia about its diameter is $I^{\prime}$,then the ratio $\frac{I}{I^{\prime}}$ is

$A$ body with a moment of inertia of $3 \ kg \cdot m^2$ rotating with an angular speed of $2 \ rad/s$ has the same kinetic energy as a mass of $12 \ kg$ moving with a speed of ......... $m/s$.

$A$ solid cylinder of mass $3 \ kg$ is rolling on a horizontal surface with velocity $4 \ m/s$. It collides with a horizontal spring whose one end is fixed to a rigid support. The force constant of the spring is $200 \ N/m$. The maximum compression produced in the spring will be (assume the collision between the cylinder and the spring is elastic). (in $m$)

The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is $I$. It is rotating with angular velocity $\omega$. Another identical ring is gently placed on it so that their centres coincide. If both rings are rotating about the same axis, then the loss in kinetic energy is

$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that:

$A$ solid metallic sphere of radius '$R$' having moment of inertia '$I$' about its diameter is melted and recast into a solid disc of radius '$r$' of uniform thickness. The moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is also equal to '$I$'. The ratio $\frac{r}{R}$ is