Question

A small uniform tube is bent into a circular tube of radius $R$ and kept in the vertical plane. Equal volume of two liquids of densities $\rho$ and $\sigma (\rho >\sigma )$ fill half of the tube as shown . $\theta$ is the angle which the radius passing through the interface makes with the vertical. Then

• A. $\theta ={\tan }^{-1}\left(\frac{\rho -\sigma }{\rho +\sigma }\right)$
• B. $\theta ={\tan }^{-1}\left(\frac{\sigma -\rho }{\rho +\sigma }\right)$
• C. $\theta ={\tan }^{-1}\left(\frac{\rho }{\rho +\sigma }\right)$
• D. $\theta ={\tan }^{-1}\left(\frac{\rho }{\rho -\sigma }\right)$

Answer 1

The correct option is A $\theta ={\tan }^{-1}\left(\frac{\rho -\sigma }{\rho +\sigma }\right)$

Pressure at A from both side must balance. Figure is self-explanatory.
$\sigma h_{2}g=\rho h_{1}g\sigma R\sqrt{2}\sin (45+\theta )=\rho R\left(\cos \theta -\sin \theta \right]\sigma \left(\cos \theta +\sin \theta \right]=\rho \left(\cos \theta -\sin \theta \right]\tan \theta =\frac{\rho -\sigma }{\rho +\sigma }$