Radius of gyration of a uniform thin rod of length $L$ about an axis passing normally through its centre of mass is

A particle of mass $m$ moves in the $XY$ plane with a velocity $V$ along the straight line $AB$ . If the angular momentum of the particle with respect to origin $O$ is $L_A$ when it is at $A$ and $L_B$ when it is at $B$ , then

Two loops $P$ and $Q$ are made from a uniform wire. The radii of $P$ and $Q$ are $r_1$ and $r_2$ respectively, and their moments of inertia are $I_1$ and $I_2$ respectively. If $I_2/I_1=4$ then $\frac{{{r_2}}}{{{r_1}}}$ equals

Particles of masses $m, 2m, 3m, ...... nm$ $grams$ are placed on the same line at distances $l, 2l, 3l,...., nl\, cm$ from a fixed point. The distance of the centre of mass of the particles from the fixed point (in centimetres) is

A circular disk of moment of inertia $I_t$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed ${\omega _i}$. Another disk of moment of inertia $I_b$ is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed ${\omega _f}$. The energy lost by the initially rotating disc to friction is

One circular ring and one ciruclar disc, both are having the same mass and radius. the ratio of their moments of inertia about the axis passing through their centres and perpendicular to their planes, will be