A liquid kept in a cylindrical vessel is rotated about vertical axis through the centre
of circular base. The difference in the heights of the liquid at the centre of vessel and
its edge is \((\mathrm{R}=\) radius of vessel, \(\omega=\) angular velocity of rotation, \(\mathrm{g}=\) acceleration
due to gravity)

When the cylindrical vessel is rotated at angular speed \(\omega\) about its axis, the velocity of the liquid at the sides is maximum, given by \(\mathrm{v}_{\mathrm{s}}=\mathrm{r} \omega\)
Applying Bernoulli's theorem at the sides and at the center of the vessel, we have
\(\mathrm{P}+\frac{1}{2} \rho \mathrm{v}^{2}=\) constant \(\mathrm{P}_{\mathrm{s}}+\frac{1}{2} \rho \mathrm{v}_{\mathrm{s}}^{2}=\mathrm{P}_{\mathrm{c}}+\frac{1}{2} \rho \mathrm{v}_{\mathrm{c}}{ }^{2}\)
where
\(\mathrm{P}_{\mathrm{s}}=\) pressure at the sides
\(\mathrm{v}_{\mathrm{s}}=\) velocity of the liquid at the sides
\(\mathrm{P}_{\mathrm{c}}=\) pressure at the center
\(\mathrm{v}_{\mathrm{c}}=\) velocity of the liquid at the center
\(\mathrm{P}_{\mathrm{c}}-\mathrm{P}_{\mathrm{s}}=\frac{1}{2} \rho \mathrm{v}_{\mathrm{s}}^{2}=\frac{1}{2} \rho \mathrm{r}^{2} \omega^{2} \ldots(\mathrm{I}) \quad \because \mathrm{v}_{\mathrm{c}}=0\)
Since \(\mathrm{P}_{\mathrm{c}}\) is greater than \(\mathrm{P}_{\mathrm{s}}\), the liquid rises at the sides of the vessel. Let \(\mathrm{h}\) be the difference in the levels of the liquids at the sides and at the center, so we have
\(\mathrm{P}_{\mathrm{c}}-\mathrm{P}_{\mathrm{s}}=\rho g h \ldots(\mathrm{II})\)
from (I) and (II) we have
\(\rho g h=\frac{1}{2} \rho r^{2} \omega^{2}\)
\(\Rightarrow h=\frac{r^{2} \omega^{2}}{2 g}\)