$A$ thin uniform rod of length $l$ and mass $m$ is swinging freely about a horizontal axis passing through its end. Its maximum angular speed is $\omega$. Its centre of mass rises to a maximum height of:

Two rings of the same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass of the ring $= m$,radius $= r$)

$Assertion$ : $A$ helicopter must necessarily have two propellers.
$Reason$ : Two propellers are provided in a helicopter in order to conserve linear momentum.

$A$ rod of mass $m$ and length $l$ is hinged at one end to a horizontal floor and stands vertically. If it is allowed to fall,the velocity with which its upper end strikes the floor is:

$A$ thin rod of mass $M$ and length $a$ is free to rotate in a horizontal plane about a fixed vertical axis passing through point $O$. $A$ thin circular disc of mass $M$ and radius $a/4$ is pivoted on this rod with its center at a distance $a/4$ from the free end so that it can rotate freely about its vertical axis,as shown in the figure. Assume that both the rod and the disc have uniform density and they remain horizontal during the motion. An outside stationary observer finds the rod rotating with an angular velocity $\Omega$ and the disc rotating about its vertical axis with angular velocity $4\Omega$. The total angular momentum of the system about the point $O$ is $\left(\frac{Ma^2\Omega}{48}\right) n$. The value of $n$ is. . . . .

In the figure shown,a ring $A$ is initially rolling without sliding with a velocity $v$ on the horizontal surface of the body $B$ (of the same mass as $A$). All surfaces are smooth. $B$ has no initial velocity. What will be the maximum height reached by $A$ on $B$?