$A$ thin hollow sphere of mass $m$ is completely filled with a liquid of mass $m$. When the sphere rolls with a velocity $v$, the kinetic energy of the system is (neglect friction):

An object has a moment of inertia of $3 \ kg \cdot m^2$. It is rotating with an angular velocity of $2 \ rad/s$. If a mass of $12 \ kg$ is moving with a velocity of $v \ m/s$,at what value of $v$ will their kinetic energies be equal?

$A$ ring of mass $m$ and radius $R$ has three particles attached to the ring as shown in the figure. The centre of the ring has a speed $v_0$. The kinetic energy of the system is: (Slipping is absent) (in $, mv_0^2$)

$A$ thin ring of mass $2 \ kg$ and radius $1 \ m$ is rolling without slipping on a horizontal plane with velocity $1 \ m/s$. $A$ small ball of mass $1 \ kg$,moving with velocity $2 \ m/s$ in the opposite direction,hits the ring at a height of $1.8 \ m$ and goes vertically up with velocity $1 \ m/s$. Immediately after the collision:
$(A)$ the ring has pure rotation about its stationary $CM$.
$(B)$ the ring comes to a complete stop.
$(C)$ friction between the ring and the ground is to the left.
$(D)$ there is no friction between the ring and the ground.

$A$ rotating body has angular momentum $L$. If its frequency is doubled and its kinetic energy is halved,what will be its new angular momentum?

$A$ small particle of mass $m$ is projected at an angle $\theta$ with the $x$-axis with an initial velocity $v_{0}$ in the $x-y$ plane as shown in the figure. For time $t < \frac{v_{0} \sin \theta}{g}$,the angular momentum of the particle is (where $\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along the $x, y$ and $z$ axes respectively):