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Question

Length of horizontal arm of a uniform cross section U-tube is l=21cm and ends of both of the vertical arms are open to surrounding pressure of 10×500N/m2. A liquid of density ρ=103kg/m3 is poured into the tube such that liquid just fills the horizontal part of the tube. Now, one of the open ends is sealed and the tube is then rotated about a vertical axis passing through the other vertical arm with angular velocity ω0=10rad/s. If the length of each vertical arm is a=6cm, calculate the length of air column in the sealed arm. (in cm)
157526_6393650609b9445abe6702c1da211abc.png

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Solution

Initial volume of right arm V=sa ,s=cross section . P=10500 pa
let the mercury rises to x height Vf=s(ax)
let Pf be the final pressure on top of mercury on right tube
using PV=PfVf
Pf=Paax,
now pressure at the B
PB=Paax+ρgx
centripetal will be provided by the pressure difference between A and B
F1=(PBPA)s .......(1), also PA=10500N/m2
mass of liquid in AB m=ρs(lx)
center of mass of liquid in AB from axis of rotation r=(l+x)/2
F2=mω2r
now F1=F2
on solving we get x=1 cm
so column of air =6-1=5 cm

401992_157526_ans_0d5ffd07b88847a487a8b6199bbddfca.png

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