$A$ 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 ......... $m/s$.

$A$ uniform bar of length $12 \text{ cm}$ and mass $20m$ lies on a smooth horizontal table. Two point masses $m$ and $2m$ are moving in opposite directions with the same speed $v$ and in the same plane as the bar. These masses strike the bar simultaneously and get stuck to it. After the collision,the entire system is rotating with an angular frequency $\omega$. The ratio of $v$ and $\omega$ is:

Two masses of $0.3 \, kg$ and $0.7 \, kg$ are attached to the ends of a rod of length $1.4 \, m$ and negligible mass. The rod is rotated about an axis perpendicular to its length with a constant angular speed. The point on the rod through which the axis should pass so that the work required to rotate the rod is minimum is:

$A$ metal sphere of radius $r$ and specific heat $S$ is rotated about an axis passing through its centre at a speed of $n$ rotations per second. It is suddenly stopped and $50 \%$ of its energy is used in increasing its temperature. Then,the rise in temperature of the sphere is

$A$ uniform circular disc of radius $R$,lying on a frictionless horizontal plane,is rotating with an angular velocity $\omega$ about its own axis. Another identical circular disc is gently placed on top of the first disc coaxially. The loss in rotational kinetic energy due to friction between the two discs,as they acquire a common angular velocity,is ($I$ is the moment of inertia of the disc).

This question has Statement $1$ and Statement $2$. Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement $1$ : When the moment of inertia $I$ of a body rotating about an axis with angular speed $\omega$ increases,its angular momentum $L$ remains unchanged,but the kinetic energy $K$ decreases if no external torque is applied.
Statement $2$ : $L = I\omega$ and the rotational kinetic energy $K = \frac{1}{2}I\omega^2 = \frac{L^2}{2I}$.