Correct Answer - B
(b) We start with the particle under a net force model in the `x` and `y` directions :
`sum F_x = ma_x : T sin theta = (mv^2)/( r)`
`sum F_y = ma_y : T cos theta =mg`
So `(sin theta)/(cos theta) = (v^2)/(rg)` and `v = sqrt(rg((sin theta)/(cos theta)))`
then `L = rmv sin 90^@ = rm sqrt(rg((sin theta)/(cos theta))) = sqrt(m^2 gr^3 (sin theta)/(cos theta))`
and since `r = l sin theta`,
`L = sqrt(m^2 gl^3 (sin^4 theta)/(cos theta))`.