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 ρ₁ and ρ₂ (ρ₁ > ρ₂), fill half the circle. The angle θ 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 = ρ₁gh₁ (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 = ρ₁gh₂ + ρ₂gh₂'. (2)
Equating P₁ and P₂, we get
ρ₁h₂ + ρ₂h₂' = ρ₁h₁
Substituting h₂' = Rsinθ + Rcosθ, h₂ = R(1−cosθ) and h₁ = R(1−sinθ)
ρ₁R(1−cosθ) + ρ₂R(sinθ + cosθ) = ρ₁R(1−sinθ)
This gives
tanθ = (ρ₁−ρ₂)/(ρ₁+ρ₂)