First slide

Fluid dynamics

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 $ω$ , then the difference in the heights of the liquid at the centre of the vessel and the edge is

Moderate

A

$\frac{rω}{2 g}$
B

$\frac{r^{2} ω^{2}}{2 g}$
C

$\sqrt{2 grω}$
D

$\frac{ω^{2}}{2 {gr}^{2}}$

Solution

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, $h_{A} = h_{B}$
$\begin{array}{l} ∴ P_{A} + \frac{1}{2} {dv}_{A}^{2} = P_{B} + \frac{1}{2} {dv}_{B}^{2} \\ \Rightarrow P_{A} - P_{B} = \frac{1}{2} d [v_{B}^{2} - v_{A}^{2}] \end{array}$
$Now, v_{A} = 0, v_{B} = rω and P_{A} - P_{B} = hdg$
$∴ hdg = \frac{1}{2} {dr}^{2} ω^{2} or h = \frac{r^{2} ω^{2}}{2 g}$

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