$A$ 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. . . . .

$A$ thin and uniform rod of mass $M$ and length $L$ is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement$(s)$ is/are correct,when the rod makes an angle $60^{\circ}$ with vertical? [$g$ is the acceleration due to gravity]
$(1)$ The radial acceleration of the rod's center of mass will be $\frac{3g}{4}$
$(2)$ The angular acceleration of the rod will be $\frac{3\sqrt{3}g}{4L}$
$(3)$ The angular speed of the rod will be $\sqrt{\frac{3g}{2L}}$
$(4)$ The normal reaction force from the floor on the rod will be $\frac{Mg}{16}$

$A$ rigid uniform bar $AB$ of length $L$ is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time,the angle made by the bar with the vertical is $\theta$. Which of the following statements about its motion is/are correct?
$[A]$ The midpoint of the bar will fall vertically downward
$[B]$ The trajectory of the point $A$ is a parabola
$[C]$ Instantaneous torque about the point in contact with the floor is proportional to $\sin \theta$
$[D]$ When the bar makes an angle $\theta$ with the vertical,the displacement of its midpoint from the initial position is proportional to $(1-\cos \theta)$

$A$ uniform rod of mass $m$ and length $\ell$ hinged at end $A$ is released from the horizontal position shown in the figure. Just after the rod is released:

Column $I$	Column $II$
$(A)$ Angular acceleration of $C$	$(P)$ $\frac{3g}{2}$
$(B)$ Angular acceleration of $B$	$(Q)$ $\frac{3g}{2\ell}$
$(C)$ Acceleration of $C$	$(R)$ $\frac{3g}{4}$
$(D)$ Acceleration of $B$	$(S)$ $\frac{3g}{\ell}$

$A$ ring,a solid sphere,and a thin disc of different masses rotate with the same kinetic energy. Equal torques are applied to stop them. Which will make the least number of rotations before coming to rest?