A uniform rod of length 'l' is pivoted at one of itd ends on a vertical shaft of negligible radius When the shaft rotates at angular speed `omega` the rod makes an angle `theta` with it (see figure) . To find `theta` equate the rate of change of angular momentum (direction going into the paper) `(ml^2)/(12) omega^2 sin theta cos theta` about the centre of mass (CM) to the torque provided by the horizontal and vertical forces `F_H` and `F_V` about the CM. The value of `theta` is then such that :

Options

A. `cos theta = (g)/(2lomega^2)`
B. `cos theta = (3g)/(2lomega^2)`
C. `cos theta = (g)/(lomega^2)`
D. `cos theta = (2g)/(3lomega^2)`