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, |
\[\sigma {{h}_{2}}g=\rho {{h}_{1}}g\] |
\[\sigma \,\sin \,\,(45{}^\circ +\theta )=\rho R\,\,[\cos \theta -\sin \theta ]\] |
\[\sigma \,\,[\cos \theta +\sin \theta ]=\rho \,\,[\cos \theta -\sin \theta ]\] |
\[\tan \theta =\frac{\rho -\sigma }{\rho +\sigma }\] |
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