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 ${\rho }_2\left({\rho }_1>{\rho }_2\right)$ fill half the circle. The angle $\theta$  between the radius vector passing through the common interface and the vertical is

Let us find the pressure at the lowest point 1. Since the liquid has density ρ₁​ and height h₁​ on the left hand side of point 1, we have

P₁​=ρ₁​gh_1​.............(1)

Since two liquid columns of height h2​ and h₂′​ and densities ρ₁​ and ρ₂​ are situated above point 1, on the right hand side, we have

P₂​=ρ₁​gh₂​+ρ2​gh₃​..............(2)

Equating P1​andP2​, we get

ρ₁​h₂​+ρ2​h₃​=ρ₁​h_1​

Substituting

h₃​=Rsinθ+Rcosθ, h₂​=R(1−cosθ) and h₁​=R(1−sinθ)

ρ₁​R(1−cosθ)+ρ₂​R(sinθ+cosθ)=ρ₁​R(1−sinθ) 

This gives$\tan \theta =\frac{{\rho }_1+{\rho }_2}{{\rho }_1-{\rho }_2}$