A particle of mass $2\;kg$ is on a smooth horizontal table and moves in a circular path of radius $0.6\;m$ The height of the table from the ground is $0.8\;m$ If the angular speed of the particle is $12 rad\;s^{-1}$ the magnitude of its angular momentum about a point on the ground right under the center of the circle is :

Question

$\begin{array}{1 1} 8.64\;kg\;m^2s^{-1} \\ 11.52\;kg\;m^2s^{-1} \\ 14.4\;kg\;m^2s^{-1} \\ 20.16\;kg\;m^2s^{-1} \end{array}$