$A$ 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:

$A$ uniform disc is acted upon by two equal forces of magnitude $F$. One of them acts tangentially to the disc,while the other acts at the central point of the disc. The friction between the disc surface and the ground surface is $nF$. If $r$ is the radius of the disc,then the value of $n$ would be:

$A$ 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:

$A$ man is sitting in a smooth groove on a horizontal circular table at the edge by holding a rope joined to the centre. The moment of inertia of the table is $I$. The mass of the man is $M$. The man now pulls the rope so that he comes to the centre. The angular velocity of the table:

$A$ solid cube of wood of side $2a$ and mass $M$ is resting on a horizontal surface as shown below. The cube is free to rotate about the fixed axis $AB$. $A$ bullet of mass $m (< M)$ and speed $v$ is shot horizontally at the face opposite to $ABCD$ at a height $h$ above the surface to impart the cube an angular speed $\omega_{c}$,so that the cube just topples over. Then,$\omega_{c}$ is (Note: the moment of inertia of the cube about an axis passing through the centre of mass and parallel to the edge is $2Ma^{2}/3$)

$A$ ring starting from rest rotates under a constant angular acceleration of $8 \ rad \ s^{-2}$ due to an applied torque. How many revolutions will the ring complete in $5 \ s$? How many revolutions will it complete in the $6^{th}$ second? If the torque becomes zero after $6 \ s$,how many revolutions will the ring complete in the $7^{th}$ second?