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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:

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$\frac{r ω}{2 g}$

$\frac{r^{2} ω^{2}}{2 g}$

$\sqrt{2 g r ω}$

$\frac{ω^{2}}{2 g r^{2}}$

answer is B.

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Detailed Solution

From Bernoulli’s theorem,

$P_{A} + \frac{1}{2} d v_{A}^{2} + d g h_{A}$ $= P_{B} + \frac{1}{2} d v_{B}^{2} + d g h_{B}$

Here, $h_{A} = h_{B}$

$∴$ $P_{A} + \frac{1}{2} d v_{A}^{2} = P_{B} + \frac{1}{2} d v_{B}^{2}$

$P_{A} - P_{B} = \frac{1}{2} d [v_{B}^{2} - v_{A}^{2}]$

Now, $v_{A} = 0$ , $v_{B} = r ω$ and $P_{A} - P_{B} = h d g$

$∴$ $h d g = \frac{1}{2} d r^{2} ω^{2}$ or $h = \frac{r^{2} ω^{2}}{2 g}$

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