$A$ uniform rod $AB$ of mass $m$ and length $l$ is at rest on a smooth horizontal surface. An impulse $J$ is applied to the end $B$,perpendicular to the rod in the horizontal direction. Find the speed of particle $P$ at a distance $\frac{l}{6}$ from the centre towards $A$ of the rod after time $t = \frac{\pi m l}{12 J}$.

$ABC$ is an equilateral triangle with $O$ as its centre. $\vec F_1, \vec F_2$ and $\vec F_3$ represent three forces acting along the sides $AB, BC$ and $AC$ respectively. If the total torque about $O$ is zero,then the magnitude of $\vec F_3$ is

$A$ ring rolls without slipping on the ground. Its centre $C$ moves with a constant speed $u$. $P$ is any point on the ring. The speed of $P$ with respect to the ground is:

Three solid spheres each of mass $m$ and diameter $d$ are stuck together such that the lines connecting the centres form an equilateral triangle of side length $d$. The ratio $I_{0} / I_{A}$ of the moment of inertia $I_{0}$ of the system about an axis passing through the centroid and perpendicular to the plane of the triangle,to the moment of inertia $I_{A}$ about an axis passing through the center of any one of the spheres and perpendicular to the plane of the triangle,is:

If the earth suddenly contracts to $(1/3)^{rd}$ of its present size without any change in its mass, the ratio of the kinetic energy of the earth after and before contraction will be (Earth is assumed to be a rotating sphere about itself).

The graph between $\log_e L$ and $\log_e P$ will be (where $L$ is angular momentum and $P$ is linear momentum):