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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:

A
rω2g
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B
r2ω22g
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C
2grω
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D
ω22gr2
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Solution

The correct option is A r2ω22g
When the cylindrical vessel is rotated at angular speed ω about its axis, the velocity of the liquid at the sides is maximum, given by vs=rω
Applying Bernoulli's theorem at the sides and at the center of the vessel, we have
P+12ρv2= constant
Ps+12ρvs2=Pc+12ρvc2
where
Ps= pressure at the sides
vs= velocity of the liquid at the sides
Pc= pressure at the center
vc= velocity of the liquid at the center
PcPs=12ρvs2=12ρr2ω2 ...(I) vc=0
Since Pc is greater than Ps, the liquid rises at the sides of the vessel. Let h be the difference in the levels of the liquids at the sides and at the center, so we have
PcPs=ρgh ...(II)
from (I) and (II) we have
ρgh=12ρr2ω2h=r2ω22g

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