When torque applied on a system is zero,which of the following will be constant?

Assertion $:-$ When no external torque is applied and moment of inertia of a rotating body changes,then its angular momentum remains conserved,but its kinetic energy changes.
Reason $:-$ Angular momentum does not depend upon moment of inertia of the body.

$A$ mass is whirled in a circular path with constant angular velocity and its linear velocity is $v$. If the string is now halved keeping the angular momentum same,the linear velocity 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 . . . . . $m^2 \ s^{-1}$.

If the torque applied on a particle is zero,then its angular momentum will be:

Two small beads are initially at $A$ on an equilateral triangle made of uniform wires. The triangle is rotated about the vertical axis $AO$. The beads are released from rest simultaneously and slide down along $AB$ and $AC$ respectively (as shown in the figure). Which quantities are conserved as the beads slide down? (Friction is negligible)