$A$ thin uniform tube is bent into a circle of radius $r$ in the vertical plane. Equal volumes of two immiscible liquids,whose densities are ${\rho _1}$ and ${\rho _2}$ $({\rho _1} > {\rho _2})$,fill half the circle. The angle $\theta$ between the radius vector passing through the common interface and the vertical is

• A
  $\theta = {\tan ^{ - 1}}\left[ {\frac{\pi }{2}\left( {\frac{{{\rho _1} - {\rho _2}}}{{{\rho _1} + {\rho _2}}}} \right)} \right]$
• B
  $\theta = {\tan ^{ - 1}}\left[ \frac{\pi }{2}\left( {\frac{{{\rho _1} + {\rho _2}}}{{{\rho _1} - {\rho _2}}}} \right) \right]$
• C
  $\theta = {\tan ^{ - 1}}\left[ \pi \left( {\frac{{{\rho _1}}}{{{\rho _2}}}} \right) \right]$
• D
  $\theta = {\tan ^{ - 1}}\left[ \frac{\pi }{2}\left( {\frac{{{\rho _2}}}{{{\rho _1}}}} \right) \right]$