Question

Length of the horizontal arm of a uniform cross-section U-tube is $l=5 \, cm$ and both ends of the vertical arms are open to the surrounding pressure of $7700 \, N \, m^{- 2}$ . A liquid of density $\rho =10^{3} \, kg \, m^{- 3}$ is poured into the tube such that the 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 $ due to which the liquid rises up to half the length of the vertical arm. If the length of each vertical arm is $a=6 \, cm$ . What is the value of $\omega $ (in $rad \, s^{- 1}$ )?

Answer 1

Let cross-section area of the tube $=A$
In the right limb for compressed air
$x=3 \, cm$
$p_{1}V_{1}=p_{2}V_{2}$
$p_{0}A\times 6=p_{2}A\times 3$
$p_{2}=2p_{0}........(1)$

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Force at the corner of the right limb due to liquid above,
$F_{1}=\left[2 p_{0} + x \rho g\right]A$
Mass of the liquid in the horizontal arm, $m=\rho \left(\right.l-x\left.\right)A$
It is rotated about the left limb, then centripetal force,
$m\left(\omega \right)^{2}r=\rho \left(l - x\right)A\left(\omega \right)^{2}\times \frac{l + x}{2}=\frac{\rho A \left(\omega \right)^{2}}{2}\times \left(l^{2} - x^{2}\right)$
$\frac{\rho A \left(\omega \right)^{2} \left(l^{2} - x^{2}\right)}{2}=\left[2 p_{0} + x \rho g\right]A-p_{0}A$
$\omega =\sqrt{\frac{2 \left(p_{0} + x \rho g\right)}{\rho \left(l^{2} - x^{2}\right)}}=100 \, rad \, s^{- 1}$