Liquid in a cylindrical vessel of radius 0.3 m is spun at 2 revolutions per second. What is the height difference of the liquid surface between the vessel's centre and its rim?

0.01 m
0.02 m
0.04 m
0.72 m

Correct answer: 0.72 m

Solution

For a liquid rotating as a rigid body, the free surface is a paraboloid and the rise at the wall relative to the centre is h = ω² r² / (2g). With ω = 4π rad/s, r = 0.3 m: h = (4π)² * 0.09 / (2*9.8) ≈ 157.9*0.09/19.6 ≈ 0.72 m.

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