For the rotating container shown in figure, differences in the heights of the fluid between the centre and the sides (in $\text{cm}$) is-
[Take ${\pi }^2=10\&g=10{\text{m/s}}^2$ for calculation]

The correct option is A $12.5\text{cm}$
Let the difference in the heights between the centre and the sides be $h$.


For a rotating fluid, the pressure difference can be given as
$\Delta p=\frac{1}{2}\rho {\omega }^2{x}^2$
where $x$ is the distance between two points in a liquid, measured $\perp$ from the axis of rotation.
The pressure will increase in a direction away from the axis of rotation.
i.e $p_B>p_O$
$\Rightarrow \Delta p=p_B-p_0=\frac{1}{2}\rho {\omega }^2{x}^2...\left(1\right)$
The pressure variation along vertical direction is,
$p_B=p_A+\rho gh$
$\Rightarrow p_B-p_0=\rho gh...\left(2\right)$
[$p_A=p_0=$ atmospheric pressure]

From Eq.$\left(1\right)$ and $\left(2\right)$, on equating, we get
$\rho gh=\frac{1}{2}\rho {\omega }^2{x}^2$
$\Rightarrow h=\frac{{\omega }^2{x}^2}{2g}$
$\Rightarrow h=\frac{(5\pi {)}^2\times (0.1{)}^2}{2\times 10}$
$(\because x=10\text{cm}=0.1\text{m}\omega =5\pi \text{rad/s})$

$\therefore h=0.125\text{m}=12.5\text{cm}$