Question

Length of horizontal arm of a uniform cross section $U-$tube is $l=21cm$ and ends of both of the vertical arm are open to surrounding of pressure $10\times 500N/m^2$. 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 than rotated about a vertical axis passing through the other vertical arm with angular velocity ${\omega }_0=10rad/s$. If length of each vertical arm is $a=6cm$, calculated the length of air column in the sealed arm. (in $cm$)

Answer 1

Due to rotation , let the shift of liquid be X cm

Let the cross sectional area of the tube I.A

In the right limb for compressed ar

$P_1{V}_1=P_2{V}_2$

$P_{1}A\times 6=P_{2}A(6-x)$

$P_2=\frac{6P_1}{6-x}$

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

$F_1=[\frac{6P_1}{6-x}+x\rho g]A$

Mass of liquid in $=1M=\rho (l-x)n$

horizontal arm

It is rotated about left limb. Then the centre petal force is

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

$=\frac{\rho A{\omega }_0^2}{2}(l^2-x^2)$

But $F_1=F_2$

$\frac{\rho A{\omega }_0^2(l^2-x^2)}{2}=(\frac{\delta {\rho }_0}{\delta -x}+x\rho g)A$

$\Rightarrow x=1cm$

length of arm column $=5cm$