A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the centre of the circular base. If the radius of the vessel is r and angular velocity of rotation is w, then the different in the heights of the liquid at the centre of the vessel and the edge is

The maximum liquid velocity is determined by $v_s=r\omega$ when the cylindrical vessel is rotated at an angle of around its axis.
By using Bernoulli's theorem at the vessel's sides and in the center, we have

$\mathrm{P}+\frac{1}{2}{\mathrm{\rho v}}^2=\mathrm{constant}$

${\mathrm{P}}_{\mathrm{s}}+\frac{1}{2}\mathrm{\rho }{{\mathrm{v}}_{\mathrm{s}}}^2={\mathrm{P}}_{\mathrm{c}}+\frac{1}{2}\mathrm{\rho }{{\mathrm{v}}_{\mathrm{c}}}^2$

${\mathrm{P}}_{\mathrm{c}}-{\mathrm{P}}_{\mathrm{s}}=\frac{1}{2}\mathrm{\rho }{{\mathrm{v}}_{\mathrm{s}}}^2=\frac{1}{2}{\mathrm{\rho r}}^2{\mathrm{\omega }}^2\mathrm{\_\_\_\_\_}\left(1\right)$

The liquid rises at the sides of the vessel because $P_c$is greater than $P_s$. Using h as the difference between the liquid levels at the sides and the centre, we can find

${\mathrm{P}}_{\mathrm{c}}-{\mathrm{P}}_{\mathrm{s}}=\mathrm{\rho gh\_\_\_\_\_\_}\left(2\right)$

from equations 1 and 2

$\mathrm{\rho gh}=\frac{1}{2}{\mathrm{\rho r}}^2{\mathrm{\omega }}^2\Rightarrow \mathrm{h}=\frac{{\mathrm{r}}^2{\mathrm{\omega }}^2}{2\mathrm{g}}$

Hence the correct answer is $\frac{{\mathrm{r}}^2{\mathrm{\omega }}^2}{2\mathrm{g}}.$