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

Let the string makes angle $\theta$ with the vertical. Weight will be balanced by the vertical component of tension, which is $T\cos \theta =mg$ and the horizontal component towards the centre will provide the required centripetal force for circular motion. Therefore,

$T\sin \theta =m{\omega }^2\left(\mathcal{l}\sin \theta \right)$, where $\mathcal{l}=$length of the string.

$\Rightarrow 400\sin \theta =0.5{\omega }^2\times 0.5\sin \theta$

$\Rightarrow \omega =40 \mathrm{rad} {\mathrm{s}}^{-1}$