Two rotating bodies $A$ and $B$ of masses $m$ and $2\,m$ with moments of inertia $I_A$ and $I_B (I_B> I_A)$ have equal kinetic energy of rotation. If $L_A$ and $L_B$ be their angular momenta respectively, then

Two coaxial discs, having moments of inertia $I_1$ and $\frac{I_1}{2}$ are a rotating with respectively angular velocities $\omega_1$ and $\frac{\omega_1}{2}$, about their common axes. They are brought in contact with each other and thereafter they rotate with a common angular velocity. If $E_f$ and $E_i$ are the final and initial total energies, then $(E_f -E_i)$ is

A fly wheel of moment of inertia $I$ is rotating at $n$ revolutions per $sec$. The work needed to double the frequency would be

Two bodies have their moments of inertia $I$ and $2 I$ respectively about their axis of rotation. If their kinetic energies of rotation are equal, their angular momentum will be in the ratio

The moment of inertia of a body about a given axis is $2.4\ kg-m^2$.  To produce a rotational kinetic energy of $750\ J$, an angular acceleration of $5\ rad/s^2$ must be applied about that axis for.......... $\sec$

If the rotational kinetic energy of a body is increased by $300\ \%$ then the percentage increase in its angular momentum will be .......... $\%$