question_answer

A thin uniform tube is bent into a circle of radius in the vertical plane. Equal volumes of two immiscible liquids, whose densities are and fill half the circle. The angle between the radius vector passing through the common interface and the vertical is:                                                                                               [JEE Online 15-04-2018]

A)  \[\theta ={{\tan }^{-1}}\left[ \frac{\pi }{2}\left( \frac{{{\rho }_{1}}-{{\rho }_{2}}}{{{\rho }_{1}}+{{\rho }_{2}}} \right) \right]\]

B)  \[\theta ={{\tan }^{-1}}\frac{\pi }{2}\left( \frac{{{\rho }_{1}}-{{\rho }_{2}}}{{{\rho }_{1}}+{{\rho }_{2}}} \right)\]

C)  \[\theta ={{\tan }^{-1}}\pi \left( \frac{{{\rho }_{1}}}{{{\rho }_{2}}} \right)\]

D)  None of above

Correct Answer: D

Solution :

[d] Let us find the pressure at the lowest point 1. Since the liquid has density

\[{{\rho }_{2}}\] and height

\[h_{2}^{'}\] on the right hand side of point 1, we have

\[{{p}_{1}}={{\rho }_{1}}g{{h}_{1}}................(1)\]

Since two liquid columns of height \[{{h}_{1}}\] and \[{{h}_{2}}\] densities \[{{\rho }_{1}}\] are \[{{\rho }_{2}}\] are situated above point 1, on the left-hand side, we have

\[{{P}_{2}}={{\rho }_{1}}g{{h}_{2}}+{{\rho }_{2}}gh_{2}^{'}..............(2)\]

Equating \[{{P}_{1}}\] and \[{{P}_{2}}\], we get

\[{{\rho }_{1}}{{h}_{2}}+{{\rho }_{2}}h_{2}^{'}={{\rho }_{1}}{{h}_{1}}\]

Substituting

\[h_{2}^{'}=R\sin \theta +R\cos \theta ,\,\,{{h}_{2}}=R(1-\cos \theta )\]and \[{{h}_{1}}=R(1-\sin \theta )\]

\[{{\rho }_{1}}R(1-\cos \theta )+{{\rho }_{2}}R(\sin \theta +\cos \theta )={{\rho }_{1}}R(1-sin\theta )\]

This gives

\[\tan \theta =\frac{{{\rho }_{1}}-{{\rho }_{2}}}{{{\rho }_{1}}+{{\rho }_{2}}}\]