A disc and a ring of same mass are rolling and if their kinetic energies are equal, then the ratio of their velocities will be

A uniform thin wooden plank $A B$ 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 centre of mass rising a distance $h$ from the surface of the table. Then,

A circular disc of moment of inertia $I_t$, is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega_i$ . Another disc of moment of inertia $l_b$ is dropped coaxially onto the rotating disc. Initially the second disc has zero angular speed. Eventually both the discs rotate with a constant angular speed $\omega_f$. The energy lost by the initially rotating disc to friction is 

A circular disc of mass $M$ and radius $R$ is rotating about its axis with angular speed $\omega_{1}$ If another stationary disc having radius $\frac{ R }{2}$ and same mass $M$ is dropped co-axially on to the rotating disc. Gradually both discs attain constant angular speed $\omega_{2}$. The energy lost in the process is $p \%$ of the initial energy. Value of $p$ is

Two discs of moment of inertia $I_1$ and $I_2$ and angular speeds ${\omega _1}\,{\rm{and }}{\omega _2}$ are rotating along collinear axes passing through their centre of mass and perpendicular to their plane. If the two are made to rotate together along the same axis the rotational $KE$ of system will be