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A liquid is kept in a cylindrical vessel which is rotated along its axis. The liquid rises at the sides. If the radius of the vessel is r and the speed of revolution is n rotations/second, then find the difference in height of the liquid at the centre of vessel and its sides.

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$h = \frac{2 π^{2} n^{2} r^{2}}{g}$

$h = \frac{4 π^{2} n^{2} r^{2}}{g}$

$h = \frac{π^{2} n^{2} r^{2}}{g}$

$h = \frac{π^{2} n^{2} r^{2}}{2 g}$

answer is A.

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Detailed Solution

When the cylindrical vessel rotates, the velocity at the sides is higher. Hence pressure at the sides is lower. Equation of surface liquid is

$y = \frac{ω^{2} x^{2}}{2 g}$

$Here ω = 2 π n,$

x = R, y = h

$∴ \frac{(2 π n)^{2} (R)^{2}}{2 g} = \frac{2 π^{2} n^{2} R^{2}}{g}$

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