A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the center of the circular base. If the radius of the vessel is r and angular velocity of rotation is $\alpha$ , then the difference in the heights of the liquid at the center of the vessel and the edge is
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From Bernoulli's theorem,
$P_A + \frac{1}{2} dv_A^2 + dgh_A = P_B + \frac{1}{2} dv_B^2 + dgh_B$ Here, $h_A = h_B$ $\therefore P_A + \frac{1}{2} dv_A^2 = P_B + \frac{1}{2} dv_B^2$ $\Rightarrow P_A - P_B = \frac{1}{2} d [v_B^2 - v_A^2]$ Now, $v_A = 0$, $v_B = r \omega$ and $P_A - P_B = hdg$ $\therefore hdg = \frac{1}{2} d r^2 \omega^2$ or $h = \frac{r^2 \omega^2}{2g}$
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