A thin uniform tube is bent into a circle of radius $r$ in the virtical plane. Equal volumes of two immiscible liquids, whose densities are ${\rho _1}$ and ${\rho _2}\left( {{\rho _1} > {\rho _2}} \right)$ fill half the circle. The angle $\theta$ between the radius vector passing through the common interface and the vertical is
JEE MAIN 2018, Diffcult
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Pressure at interface $A$ must be same from both the sides to be in equilibrium.

$\therefore \left( {R\cos \theta  + R\sin \theta } \right){\rho _2}g$

$ = \left( {R\cos \theta  - R\sin \theta } \right){\rho _1}g$

$ \Rightarrow \frac{{{\rho _1}}}{{{\rho _2}}} = \frac{{\cos \theta  + \sin \theta }}{{\cos \theta  - \sin \theta }} = \frac{{1 + \tan \theta }}{{1 - \tan \theta }}$

$ \Rightarrow {\rho _1} - {\rho _1}\,\tan \,\theta  = {\rho _2} + {\rho _2}\,\tan \,\theta $

$ \Rightarrow \left( {{\rho _1} + {\rho _2}} \right)\tan \theta  = {\rho _1} - {\rho _2}$

$\therefore \theta  = {\tan ^{ - 1}}\left( {\frac{{{\rho _1} - {\rho _2}}}{{{\rho _1} + {\rho _2}}}} \right),so\,closest\,answer\,is\left( a \right)$

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