$Assertion$ : For a system of particles under a central force field,the total angular momentum is conserved.
$Reason$ : The torque acting on such a system is zero.

State and explain the law of conservation of angular momentum.

$A$ thin and circular disc of mass $M$ and radius $R$ is rotating in a horizontal plane about its geometrical axis with an angular velocity $\omega$. If another disc of same dimensions but of mass $M/4$ is placed gently on the first disc coaxially,then the new angular velocity of the system is:

$A$ particle of mass $1 \ kg$ is subjected to a force which depends on the position as $\vec{F} = -k(x \hat{i} + y \hat{j}) \ N$ with $k = 1 \ kg \ s^{-2}$. At time $t = 0$,the particle's position is $\vec{r} = (\frac{1}{\sqrt{2}} \hat{i} + \sqrt{2} \hat{j}) \ m$ and its velocity is $\vec{v} = (-\sqrt{2} \hat{i} + \sqrt{2} \hat{j} + \frac{2}{\pi} \hat{k}) \ m \ s^{-1}$. Let $v_x$ and $v_y$ denote the $x$ and $y$ components of the particle's velocity,respectively. Ignore gravity. When $z = 0.5 \ m$,the value of $(x v_y - y v_x)$ is . . . . . $m^2 \ s^{-1}$.

$A$ circular disc has a moment of inertia $I_2$ about an axis passing through its center and perpendicular to its plane. Another disc with a moment of inertia $I_1$ is placed on it. If the system rotates about the same axis with an initial angular velocity $\omega$,what will be the final angular velocity of the combined system?

$A$ small bead of mass $m$ moving with velocity $v$ gets threaded on a stationary semicircular ring of mass $m$ and radius $R$ kept on a horizontal table. The ring can freely rotate about its centre. The bead comes to rest relative to the ring. What will be the final angular velocity of the system?