$A$ ring and a disc have the same mass and the same radius. The ratio of the moment of inertia of a ring about a tangent in its plane to that of the disc about its diameter is: (in $: 1$)

$A$ disc of circumference $s$ is at rest at a point $A$ on a horizontal surface when a constant horizontal force begins to act on its centre. Between $A$ and $B$ there is sufficient friction to prevent slipping,and the surface is smooth to the right of $B$. $AB = s$. The disc moves from $A$ to $B$ in time $T$. To the right of $B$,

$A$ uniform thin wooden plank $AB$ of length $L$ and mass $M$ is kept on a table with its $B$ end slightly outside the edge of the table. When an impulse $J$ is given to the end $B$,the plank moves up with the centre of mass rising a distance $h$ from the surface of the table. Then,

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 $B$?

$A$ ring of radius $R$ rolls without sliding with a constant velocity $u$. The radius of curvature of the path followed by any particle of the ring at the highest point of its path will be

$A$ block of mass $m = 1 \, kg$ slides with velocity $v = 6 \, m/s$ on a frictionless horizontal surface and collides with a uniform vertical rod and sticks to it as shown. The rod is pivoted about $O$ and swings as a result of the collision,making an angle $\theta$ before momentarily coming to rest. If the rod has mass $M = 2 \, kg$ and length $l = 1 \, m$,the value of $\theta$ is approximately (Take $g = 10 \, m/s^2$) (in $^{\circ}$)