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

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

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

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

Correct Answer: A

Solution :

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

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

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

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

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