Physics - Fluid Question with Solution | TestHub

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 ${\mathrm{P}}_{\mathrm{c}}-{\mathrm{P}}_{\mathrm{s}}=\frac{1}{2}\rho {\mathrm{v}}_{\mathrm{s}}^2=\frac{1}{2}\rho {\mathrm{r}}^2{\omega }^2\dots (\mathrm{I})\because {\mathrm{v}}_{\mathrm{c}}=0$ 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 ${\mathrm{P}}_{\mathrm{c}}-{\mathrm{P}}_{\mathrm{s}}=\rho gh\dots (\mathrm{II})$ from (I) and (II) we have $\rho gh=\frac{1}{2}\rho r^2{\omega }^2$ $\Rightarrow h=\frac{r^2{\omega }^2}{2g}$