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. |
\[\sigma \,{{h}_{2}}g=\rho {{h}_{1}}g\] |
\[\sigma \sin (45{}^\circ +\theta )=\rho R\,[cos\theta -sin\theta ]\] |
\[\sigma \,[cos\theta +sin\theta ]=\rho \,[cos\theta -sin\theta ]\] |
\[\tan \theta =\frac{\rho -\sigma }{\rho +\sigma }\] |
You need to login to perform this action.
You will be redirected in
3 sec