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

Solution :
As we know that, the kinetic energy of a rotating body,
$$KE = \frac{1}{2}I{\omega ^2} = \frac{1}{2}\frac{{{I^2}{\omega ^2}}}{I} = \frac{{{L^2}}}{{2I}}$$
Also, angular momentum, $$L = I\omega $$
Thus, $${K_A} = {K_B}$$
$$\eqalign{ & \Rightarrow \frac{1}{2}\frac{{L_A^2}}{{{I_A}}} = \frac{1}{2}\frac{{L_B^2}}{{{I_B}}} \cr & \Rightarrow {\left( {\frac{{{L_A}}}{{{L_B}}}} \right)^2} = \frac{{{I_A}}}{{{I_B}}} \Rightarrow \frac{{{L_A}}}{{{L_B}}} = \sqrt {\frac{{{I_A}}}{{{I_B}}}} \cr & L \propto \sqrt I \cr & \therefore {L_A} < {L_B}\,\,\left[ {\because {I_B} > {I_A}} \right] \cr} $$