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

$$\theta = \tan^{-1}\frac{\pi}{2}\left(\frac{\rho_1 + \rho_2}{\rho_1 - \rho_2}\right)$$

C

$$\theta = \tan^{-1}\pi\left(\frac{\rho_1}{\rho_2}\right)$$

D

$$\theta = \tan^{-1}\frac{\pi}{2}\left(\frac{\rho_2}{\rho_1}\right)$$