$A$ 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:

$A$ thin ring of mass $2 \ kg$ and radius $1 \ m$ is rolling without slipping on a horizontal plane with velocity $1 \ m/s$. $A$ small ball of mass $1 \ kg$,moving with velocity $2 \ m/s$ in the opposite direction,hits the ring at a height of $1.8 \ m$ and goes vertically up with velocity $1 \ m/s$. Immediately after the collision:
$(A)$ the ring has pure rotation about its stationary $CM$.
$(B)$ the ring comes to a complete stop.
$(C)$ friction between the ring and the ground is to the left.
$(D)$ there is no friction between the ring and the ground.

$A$ uniform cube of mass $m$ and side $a$ is placed on a frictionless horizontal surface. $A$ vertical force $F$ is applied to the edge as shown in the figure. Match the following (most appropriate choice):

$(a)$ $\frac{mg}{4} < F < \frac{mg}{2}$	$(i)$ Cube will move up
$(b)$ $F > \frac{mg}{2}$	$(ii)$ Cube will not exhibit motion
$(c)$ $F > mg$	$(iii)$ Cube will begin to rotate about $A$
$(d)$ $F = \frac{mg}{4}$	$(iv)$ Normal reaction effectively at $a/3$ from $A$,no motion

$A$ bar of length $L$ carrying a small mass $m$ at one of its ends rotates with a uniform angular speed $\omega$ in a vertical plane about the midpoint of the bar. During the rotation,at some instant of time when the bar is horizontal,the mass is detached from the bar but the bar continues to rotate with the same $\omega$. The mass moves vertically up,comes back,and reaches the bar at the same point. At that place,the acceleration due to gravity is $g$.

$Assertion :$ $A$ hollow shaft is found to be stronger than a solid shaft made of the same material and having the same mass per unit length.
$Reason :$ The torque required to produce a given twist in a hollow cylinder is greater than that required to twist a solid cylinder of the same length and material.

$A$ rod of mass $m$ and length $L$,pivoted at one of its ends,is hanging vertically. $A$ bullet of the same mass moving at speed $v$ strikes the rod horizontally at a distance $x$ from its pivoted end and gets embedded in it. The combined system now rotates with angular speed $\omega$ about the pivot. The maximum angular speed $\omega_M$ is achieved for $x=x_M$. Then
$(A)$ $\omega=\frac{3 v x}{ L ^2+3 x^2}$
$(B)$ $\omega=\frac{12 v x}{L^2+12 x^2}$
$(C)$ $x_M=\frac{L}{\sqrt{3}}$
$(D)$ $\omega_M=\frac{v}{2 L} \sqrt{3}$