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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:

r2ω2/2g

√2grω

rω2/g

ω2/2gr2

Solution:

Along the radius pressure is related with radius as according to the following relation,
dP/dr = ρω²r
PB − PA = ρω²r²/2
Similarly, PB − PC = ρgh. (ii)
From (i) (ii), PA − PC = 1/2ρr²ω² − ρgh
But PA = PC = Po
=> h = r²ω²/2g