A small block is connected to one end of two identical massless strings of length $16\frac{2}{3}cm$  each with their other ends fixed to a vertical rod. If the ratio of tensions  $\frac{T_1}{T_2} be 4:1$  then the angular velocity $\omega$ of the block will be

For horizontal equilibrium of the block,
$T_1\sin {60}^0+T_2\sin {60}^0=m{\omega }^{2}r=m{\omega }^2\left(1\sin {60}^0\right)\Rightarrow T_1+T_2=m{\omega }^{2}1\mathrm{........}\left(i\right)$For vertical equilibrium of the block
$T_1\cos {60}^0=T_2\cos {60}^0+mg$
$\Rightarrow T_1-T_2=\frac{mg}{\cos {60}^0}=2mg\mathrm{..............}\left(ii\right)$
$Dividing\left(i\right)by\left(ii\right)$


$\frac{T_1+T_2}{T_1-T_2}=\frac{{\omega }^{2}1}{2g}or\frac{\frac{T_A}{T_2}+1}{\frac{T_A}{T_2}-1}=\frac{{\omega }^{2}1}{2g}$
$i.e.,\frac{4+1}{4-1}=\frac{{\omega }^{2}1}{2g}\Rightarrow {\omega }^2=\frac{10g}{31}=\frac{10\times 9.8\times 3}{3\times 50\times {10}^{-2}}=196$
$\Rightarrow \omega =14rad{s}^{-1}$