Question

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 $\omega$, then the difference in the heights of the liquid at the centre of the vessel and the edge is:

• A. $\frac{r\omega }{2g}$
• B. $\frac{r^2{\omega }^2}{2g}$
• C. $\sqrt{2gr\omega }$
• D. $\frac{{\omega }^2}{2g{r}^2}$

Answer 1

Answer: (B)

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

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 $v_s=r\omega$
Applying Bernoulli's theorem at the sides and at the center of the vessel, we have
$P+\frac{1}{2}\rho v^2=$ constant
$P_s+\frac{1}{2}\rho {v_s}^2=P_c+\frac{1}{2}\rho {v_c}^2$
where
$P_s=$ pressure at the sides
$v_s=$ velocity of the liquid at the sides
$P_c=$ pressure at the center
$v_c=$ velocity of the liquid at the center
$P_c-P_s=\frac{1}{2}\rho {v_s}^2=\frac{1}{2}\rho r^2{\omega }^2$ ...$\left(I\right)$ $\because v_c=0$
Since $P_c$ is greater than $P_s$, the liquid rises at the sides of the vessel. Let $h$ be the difference in the levels of the liquids at the sides and at the center, so we have
$P_c-P_s=\rho gh$ ...$\left(II\right)$
from $\left(I\right)$ and $\left(II\right)$ we have
$\rho gh=\frac{1}{2}\rho r^2{\omega }^2\Rightarrow h=\frac{r^2{\omega }^2}{2g}$