Question

A thin uniform tube is bent into a circle of radius $r$ in the vertical plane. Equal volumes of two immiscible liquids, whose densities are ${\rho }_1$ and ${\rho }_2$ $({\rho }_1>{\rho }_2)$ fill half the circle. The angle $\theta$ between the radius vector passing through the common interface and the vertical is:

• A. $\theta ={\tan }^{-1}\left[\frac{\pi }{2}\left(\frac{{\rho }_1-{\rho }_2}{{\rho }_1+{\rho }_2}\right)\right]$
• B. $\theta ={\tan }^{-1}\frac{\pi }{2}\left(\frac{{\rho }_1-{\rho }_2}{{\rho }_1+{\rho }_2}\right)$
• C. $\theta ={\tan }^{-1}\pi \left(\frac{{\rho }_1}{{\rho }_2}\right)$
• D. None of above

Answer 1

Answer: (D)

None of above

Let us find the pressure at the lowest point 1. Since the liquid has density ${\rho }_1$ and height $h_1$ on the left hand side of point 1, we have

$P_1={\rho }_{1}g{h}_1$.............(1)

Since two liquid columns of height $h_{2}and{h}_2^{\prime }$ and densities ${\rho }_{1}and{\rho }_2$ are situated above point 1, on the right hand side, we have

$P_2={\rho }_{1}g{h}_2+{\rho }_{2}g{h}_2^{\prime }$..............(2)

Equating $P_{1}and{P}_2$, we get

${\rho }_1{h}_2+{\rho }_2{h}_2^{\prime }={\rho }_1{h}_1$

Substituting

$h_2^{\prime }=Rsin\theta +Rcos\theta ,h_2=R(1-cos\theta )and{h}_1=R(1-sin\theta )$

${\rho }_{1}R(1-cos\theta )+{\rho }_{2}R(sin\theta +cos\theta )={\rho }_{1}R(1-sin\theta )$

This gives

$tan\theta =\frac{{\rho }_1-{\rho }_2}{{\rho }_1+{\rho }_2}$