A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil is: (specific gravity of kerosene oil = 0.8, specific gravity of ice = 0.9)

Question

G1 Physics · Accepted Answer

Image Let: V₁ = volume immersed in water V₂ = volume immersed in kerosene oil For floating equilibrium: V₁ρ(w)g + V₂ρ(o)g = (V₁ + V₂)ρ(ice)g Using relative densities: ρ(w) = 1 ρ(o) = 0.8 ρ(ice) = 0.9 Therefore: V₁ + 0.8V₂ = 0.9V₁ + 0.9V₂ 0.1V₁ = 0.1V₂ Hence: V₁ : V₂ = 1 : 1