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 $10\times 500N/m^2$. 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 ${\omega }_0=10rad/s$. If the length of each vertical arm is $a=6cm$, calculate the length of air column in the sealed arm. (in cm)

Answer 1

Initial volume of right arm V=sa ,s=cross section . P=10500 pa

let the mercury rises to x height $V_f=s(a-x)$

let $P_f$ be the final pressure on top of mercury on right tube

using $PV=P_f{V}_f$

$P_f=\frac{Pa}{a-x}$,

now pressure at the B

$P_B=\frac{Pa}{a-x}+\rho gx$

centripetal will be provided by the pressure difference between A and B

$F_1=(P_B-P_A)s$ .......(1), also $P_A=10500N/m^2$

mass of liquid in AB $m=\rho s(l-x)$

center of mass of liquid in AB from axis of rotation $r=(l+x)/2$

$F_2=m{\omega }^{2}r$

now $F_1=F_2$

on solving we get x=1 cm

so column of air =6-1=5 cm