A ball of mass 0 . 5kg is attached to a string of length 50cm . The...

Let the string makes angle $\theta$ with the vertical. Weight will be balanced by the vertical component of tension, which is $Tcos\theta = mg and the horizontal component towards the centre will provide the required centripetal force for circular motion. Therefore,</p><p>Tsin\theta = m{\omega}^{2}\left(lsin\theta\right)$, where $l =$length of the string.

$⇒ 400sin\theta = 0 .5 {\omega}^{2} \times 0 .5 sin\theta$

$⇒ \omega = 40rad{s}^{- 1}$