According to the principle of conservation of angular momentum, angular momentum:

$A$ man is standing on the edge of a circular platform. The platform is rotating with a constant angular velocity about an axis passing through its center and perpendicular to its plane. If the man walks along the radius towards the axis, the angular velocity will ........ .

$A$ solid sphere with uniform density and radius $R$ is rotating initially with constant angular velocity $\omega_1$ about its diameter. After some time,it starts losing mass at a uniform rate,with no change in its shape. The angular velocity of the sphere when its radius becomes $R/2$ is $x\omega_1$. The value of $x$ is . . . . . . .

Consider a uniform rod of mass $M=4m$ and length $\ell$ pivoted about its centre. $A$ mass $m$ moving with velocity $v$ making an angle $\theta=\frac{\pi}{4}$ to the rod's long axis collides with one end of the rod and sticks to it. The angular speed of the rod-mass system just after the collision is

$A$ 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

$A$ solid cube of wood of side $2a$ and mass $M$ is resting on a horizontal surface as shown in the figure. The cube is free to rotate about a fixed axis $AB$. $A$ bullet of mass $m$ $(m << M)$ and speed $v$ is shot horizontally at the face opposite to $ABCD$ at a height of $4a/3$ from the surface to impart the cube an angular speed $\omega$. It strikes the face and embeds in the cube. Then,$\omega$ is close to (Note: The moment of inertia of the cube about an axis perpendicular to the face and passing through the centre of mass is $2Ma^2/3$).