In the figure, a ball of mass $m$ is tied with two strings of equal length $l$ whose other ends are connected to a rod as shown in the figure. If $T_1$ and $T_2$ are the tensions in the upper and lower strings respectively when the rod is rotated with angular velocity $\omega$, then

The correct option is A $T_1>T_2$
Applying the condition of vertical equilibrium on the ball

$T_1\cos \theta =mg+T_2\cos \theta$

Dividing with $\cos \theta$ both sides,

$\frac{T_1\cos \theta }{\cos \theta }=\frac{mg}{\cos \theta }+\frac{T_2\cos \theta }{\cos \theta }$

$T_1=\frac{mg}{\cos \theta }+T_2$

$\because \theta <{90}^{\circ }$, means $\cos \theta >0$

Hence, $\frac{mg}{\cos \theta }$ is $+ve$

$\therefore T_1>T_2$

Hence, option $\left(a\right)$ is the correct choice.