Length of horizontal arm of a uniform cross section `U`-tube is `l= 21 cm` and ends of both of the vertical arms are open to surrounding of pressure `10500 N//m^(2)`. A liquid of density `rho = 10^(3) kg//m^(3)` 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 `omega_(0) = 10 rad//s`. If length of each vertical arm is `a = 6 cm`, calculate the length of air column in the sealed arm. (in cm)

Answer

5

Explanation

Due to rotation let the shift of the liqu\id be `x cm`.
Let the cross sectoinal area of the tube of `A`.

In the right limb for compressed air,
`p_(1)v_(1)=p_(2)v_(2)`
`p_(0)Axx6=p_(2)a(6-x)`
`p_(2)=(6p_(1))/(6-x)`
Force at te corver `C` of right limb due to liquid above,
`p_(1)=[(6p_(1))/(6-x)+xrhog]A`
Mass of the liquid in the horizontal arm `M=rho(l-x)A`
It is rotated about left limb. Then the cenripetal force is
`F_(2)=momega_(0)^(2)r=rho(l-x)Aomega_(0)^(2)(1+x)/2(l^(2)-x^(2))`
`F_(2)=momega_(0)^(2)r=rho(1=x)Aomega_(0)^(2)(1+x)/2=(rhoAomega_(0)^(2))/2(1^(2)-x^(2))`
But `F_(1)=F_(2)`
`implies(rhoAomega_(0)^(2)(1^(2)-x^(2)))/2=[(deltarho_(0))/(delta-x)+xrhox]A`
`=(10^(5)xx100x(21^(2)-x^(2))xx10^(-4))/2`
`=[(deltaxx10500)/((delta-x))+x xx 10^(5)xx10xx10^(-2)]`
On solving we get `x=0.01m=1cm`
The length of the air column `=6-1=5cm`