$A$ disc of mass $1 \, kg$ and radius $0.1 \, m$ is rotating with angular velocity $20 \, rad/s$. What is the angular velocity (in $rad/s$) if a mass of $0.5 \, kg$ is placed on the circumference of the disc?

$A$ uniform disc of mass $M$ and radius $R$ is rotating about a horizontal axis passing through its centre with angular velocity $\omega$. $A$ piece of mass $m$ breaks from the disc and flies off vertically upwards. The angular speed of the disc will be

$A$ person is rotating on a rotating table with an angular speed of $\omega$. He holds two identical masses in his outstretched hands. Without moving his hands,he simply drops the masses. How will his angular velocity change?

Given below are two statements: One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A)$ : In a conical pendulum,angular momentum about its vertical axis remains constant.
Reason $(R)$ : Net torque about the vertical axis of a conical pendulum is not zero.
In the light of the above statements,choose the most appropriate answer from the options given below:

$A$ smooth uniform rod of length $L$ and mass $M$ has two identical beads of negligible size,each of mass $m$,which can slide freely along the rod. Initially,the two beads are at the center of the rod and the system is rotating with angular velocity ${\omega _0}$ about an axis perpendicular to the rod and passing through the midpoint of the rod (see figure). There are no external torques. When the beads reach the ends of the rod,the angular velocity of the system is

$A$ uniform disc of radius $R$ is placed on a smooth horizontal surface. $A$ particle of mass $m$ is moving with velocity $v$ on the surface and hits the disc at the periphery and sticks to it. The angular velocity of the system after the collision is: