Length of horizontal arm of a uniform cross section U-tube is $l=21 \mathrm{cm}$ and ends of both of the vertical arms are open to surrounding of pressure $10500\mathrm{N}/{\mathrm{m}}^2$ A liquid of density $\rho ={10}^3\mathrm{kg}/{\mathrm{m}}^3$ 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 ${\omega }_0=10\mathrm{rad}/\mathrm{s}$ If length of each vertical arm is a=6 cm, calculate the length of air column in the sealed arm. (in cm)

Due to rotation, let the shift of the liquid be x cm. Let the cross-sectional area of the tube be A

In the right limb for compressed air,

${\mathrm{p}}_1{\mathrm{v}}_1={\mathrm{p}}_2{\mathrm{v}}_2\Rightarrow {\mathrm{p}}_0\mathrm{A}\times 6={\mathrm{p}}_0\mathrm{A}(6-\mathrm{x})$

$\text{ or }{p}_2=\frac{6p_1}{6-x}\text{ (i) }$

Force at the corner C of right limb climb due to liquid above,

${\mathrm{F}}_1=\left[\frac{6{\mathrm{p}}_1}{6-\mathrm{x}}+\mathrm{x\rho g}\right]\mathrm{A}$

Mass of the liquid in the horizontal arm, $M=\rho (I-x)A$ 

it is rotated about left limb. Then the centripetal force is

${\mathrm{F}}_2={\mathrm{M\omega }}_0^2\mathrm{r}=\mathrm{\rho }(l-\mathrm{x}){\mathrm{A\omega }}_0\frac{l+\mathrm{x}}{2}$

$\Rightarrow F_2={\mathrm{A\omega }}_0^2\frac{l+\mathrm{x}}{2}=\frac{{\mathrm{\rho A\omega }}_0^2}{2}\left(l^2-{\mathrm{x}}^2\right)$

$But F_1 = F_2$

$\Rightarrow \frac{{\mathrm{\rho A\omega }}_0^2\left(l^2-{\mathrm{x}}^2\right)}{2}=\left[\frac{6{\mathrm{p}}_0}{6-\mathrm{x}}+\mathrm{x\rho g}\right]\mathrm{A}$

$\Rightarrow \frac{{10}^5\times 100\times \left({21}^2-x^2\right)\times {10}^{-4}}{2}=\left[\frac{6\times 10500}{(6-\mathrm{x})}+\mathrm{x}\times {10}^5\times 10\times {10}^{-2}\right]$

on solving, we get x = 0.01m=1 cm
The length of the air column = 6–1 = 5 cm.