$A$ 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:

$A$ body kept on a smooth horizontal surface is pulled by a constant horizontal force applied at the top point of the body. If the body rolls purely on the surface,its shape can be:

Which of the following statements are correct?

$A$ 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 $\frac{Mg}{16}$

$A$ wheel of moment of inertia $2 \ kg \ m^2$ is rotating about an axis passing through its centre and perpendicular to its plane at a speed of $60 \ rad \ s^{-1}$. Due to friction, it comes to rest in $5$ minutes. The angular momentum of the wheel three minutes before it stops rotating is:

$A$ uniform rod of length $l$ is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed $\omega$,the rod makes an angle $\theta$ with it (see figure). To find $\theta$,equate the rate of change of angular momentum (direction going into the paper) $\frac{m l^{2}}{12} \omega^{2} \sin \theta \cos \theta$ about the centre of mass $(CM)$ to the torque provided by the horizontal and vertical forces $F_{H}$ and $F_{V}$ about the $CM$. The value of $\theta$ is then such that: