Question

Figure shows a three arm tube in which a liquid is filled upto levels of height $l$. It is now rotated at an angular frequency $\omega$ about an axis passing through arm $B$. The angular frequency $\omega$ at which level of liquid in arm $B$ becomes zero is

• A. $\sqrt{\frac{2g}{3l}}$
• B. $\sqrt{\frac{g}{l}}$
• C. $\sqrt{\frac{3g}{l}}$
• D. $\sqrt{\frac{3g}{2l}}$

Answer 1

The correct option is C $\sqrt{\frac{3g}{l}}$

Consider a liquid element of thickness $dx$ at a distance of $x$ from axis of rotation. Let $A$ be the area of the element.
For equilbrium of this element, centripetal force should be balanced by pressure force
i.e $dp.A=\rho Adx.{\omega }^{2}x$
Integrating both sides, taking the limits of $x$ from 0 to $l$ and pressure from point B to C,
${\int }_{P_B}^{P_C}dp={\int }_0^l\rho {\omega }^{2}xdx$
$\Rightarrow P_C-P_B=\frac{\rho {\omega }^2{l}^2}{2}$

When water level in arm B becomes zero, $P_B$ is atmospheric,
i.e $P_C=P_{atm}+\frac{\rho {\omega }^2{l}^2}{2}-\left(1\right)$

Now, if level of water in arm B is zero, then original water columns of height $l$ should be distributed among arms A and C.

So, height of water column in $A$ and $C$ $=l+\frac{l}{2}=\frac{3l}{2}$
Excess pressure in arm C due to water column is $P_C-P_{atm}=\frac{3l\rho g}{2}$
Using $\left(1\right)$,
$\frac{\rho {\omega }^2{l}^2}{2}=\frac{3l\rho g}{2}$
$\Rightarrow \omega =\sqrt{\frac{3g}{l}}$