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

Solution:

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 $\frac{{\mathrm{r}}^2{\mathrm{\omega }}^2}{2\mathrm{g}}.$