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

# 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

1. A

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

2. B

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

3. C

$\sqrt{2\mathrm{gr\omega }}$

4. D

$\frac{{\mathrm{\omega }}^{2}}{2{\mathrm{gr}}^{2}}$

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### 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

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

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

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