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

Solution :
As for linear motion $$KE = \frac{{{p^2}}}{{2m}}$$
Similarly, for rotational motion $$K{E_{rot}} = \frac{{{L^2}}}{{2I}}$$
As said, $${\left( {KE} \right)_{rot}}$$   remains same.
i.e. $$\frac{1}{2}{I_1}\omega _1^2 = \frac{1}{2}{I_2}\omega _2^2$$
$$\eqalign{ & \Rightarrow \frac{1}{{2{I_1}}}{\left( {{I_1}{\omega _1}} \right)^2} = \frac{1}{{2{I_2}}}{\left( {{I_2}{\omega _2}} \right)^2} \cr & \Rightarrow \frac{{L_1^2}}{{{I_1}}} = \frac{{L_2^2}}{{{I_2}}} \cr & \Rightarrow \frac{{{L_1}}}{{{L_2}}} = \sqrt {\frac{{{I_1}}}{{{I_2}}}} \cr & {\text{but}}\,{I_1} = I,\,{I_2} = 2I \cr & \therefore \frac{{{L_1}}}{{{L_2}}} = \sqrt {\frac{I}{{2I}}} = \frac{1}{{\sqrt 2 }} \cr & {\text{or}}\,\,{L_1}:{L_2} = 1:\sqrt 2 \cr} $$