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Question:

A solid sphere of volume V and density ρ floats at the interface of two immiscible liquids of densities ρ₁ and ρ₂ respectively. If ρ₁ < ρ < ρ₂, then the ratio of volume of the parts of the sphere in upper and lower liquid is:

ρ₂−ρ/ρ−ρ₁

ρ+ρ₂/ρ+ρ₁

ρ−ρ₁/ρ₂−ρ

ρ+ρ₁/ρ+ρ₂

Solution:

V=Volume of solid sphere.
Let V₁=Volume of the part of the sphere immersed in a liquid of density ρ₁ and V₂=Volume of the part of the sphere immersed in liquid of density ρ₂.
According to law of floatation, Vρg=V₁ρ₁g+V₂ρ₂g (i)
and V=V₁+V₂ (ii)
Hence from eqns (i) and (ii), V₁ρg+V₂ρg=V₁ρ₁g+V₂ρ₂g
or V₁(ρ−ρ₁)g=V₂(ρ₂−ρ)g
or V₁/V₂=(ρ₂−ρ)/(ρ−ρ₁)