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:
Answer: (B) $\theta = \tan^{-1}\left(\dfrac{\rho_1-\rho_2}{\rho_1+\rho_2}\right)$
Each liquid fills a quarter of the circle, i.e. an arc of $90^\circ$. Measure positions by the angle from the lowest point of the circle; a point at angle $\varphi$ is at height $-r\cos\varphi$ relative to the centre.
The heavier liquid lies lower. Let it occupy the arc from $\varphi = -a$ to $\varphi = 90^\circ - a$, and the lighter liquid the arc from $90^\circ - a$ to $180^\circ - a$. The interface is at $\varphi = 90^\circ - a$, at height $-r\sin a$. Both free surfaces are at atmospheric pressure.
Pressure at the interface, through the heavy liquid (free surface at height $-r\cos a$):
$$p - p_0 = \rho_1 g\,r(\sin a - \cos a)$$
Through the light liquid (free surface at height $+r\cos a$):
$$p - p_0 = \rho_2 g\,r(\sin a + \cos a)$$
Equating:
$$\tan a\,(\rho_1 - \rho_2) = \rho_1 + \rho_2$$
The radius to the interface makes angle $\theta = 90^\circ - a$ with the vertical, so
$$\tan\theta = \cot a = \frac{\rho_1 - \rho_2}{\rho_1 + \rho_2}$$
Solution by Sreeraj P, M.Sc Physics