question_answer

A small uniform tube is bent into a circular tube of radius R and kept in the vertical plane. Equal volumes 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:

A) \[\theta =ta{{n}^{-1}}\left( \frac{\rho -\sigma }{\rho +\sigma } \right)\]

B) \[\theta =ta{{n}^{-1}}\left( \frac{\sigma -\rho }{\sigma +\rho } \right)\]

C) \[\theta =ta{{n}^{-1}}\left( \frac{\rho }{\rho +\sigma } \right)\]

D) \[\theta =ta{{n}^{-1}}\left( \frac{\rho }{\rho -\sigma } \right)\]

Correct Answer: A

Solution :

Pressure at ?A? from both side must balance. Figure is self-explanatory,

\[\sigma {{h}_{2}}g=\rho {{h}_{1}}g\]

\[\sigma \,\sin \,\,(45{}^\circ +\theta )=\rho R\,\,[\cos \theta -\sin \theta ]\]

\[\sigma \,\,[\cos \theta +\sin \theta ]=\rho \,\,[\cos \theta -\sin \theta ]\]

\[\tan \theta =\frac{\rho -\sigma }{\rho +\sigma }\]