Correct Answer - A
(a) work done by the torque
`Delta W = int tau d theta = int_0^(theta) F l cos theta d theta`
`Delta W = F l sin theta`
Now using work energy theorem,
`Delta W = Delta k`
:. `F l sin theta = [(1)/(2)((m l^2)/(3)) omega^2 - 0]`
Which gives, `omega = sqrt((6 F sin theta)/(m l))`.
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