A uniform body of density 900 kg/m³ floats completely within two immiscible liquids at their horizontal interface. The upper liquid has density 800 kg/m³ and the lower liquid has density 1000 kg/m³. If N percent of the body's volume lies in the lower liquid, find N.
Physics · Fluid Mechanics · pascal law and buoyant force
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Question

A uniform body of density 900 kg/m³ floats completely within two immiscible liquids at their horizontal interface. The upper liquid has density 800 kg/m³ and the lower liquid has density 1000 kg/m³. If N percent of the body's volume lies in the lower liquid, find N.

Physics · Fluid Mechanics · pascal law and buoyant force

G1 Physics · Accepted Answer

Let fraction f of the body volume be in the lower liquid. Then fraction 1−f is in the upper liquid. For floating equilibrium, weight = total buoyant force. Therefore, ρ(body)Vg = ρ(lower)fVg +ρ(upper)(1−f)Vg. Cancel Vg: 900 = 1000f+800(1−f). Thus, 900 = 1000f+800−800f = 800+200f. Hence, f = 100/200 = 0.50. Therefore, N = 50.