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

$\frac{ρ - ρ_{1}}{ρ_{2} - ρ}$

$\frac{ρ_{2} - ρ}{ρ - ρ_{1}}$

$\frac{ρ + ρ_{1}}{ρ + ρ_{2}}$

$\frac{ρ + ρ_{2}}{ρ + ρ_{1}}$

Solution

$\frac{ρ_{2} - ρ}{ρ - ρ_{1}}$

V = Volume of solid sphere

Let V₁ = Volume of the part of the sphere immersed in a liquid of density p₁ and

V₂ = Volume of the part of the sphere immersed in liquid of density p₂

The law of flotation says that for a floating object the weight of the object equals the weight of the fluid.

According to law of flotation,

Vρ g = V₁ p₁g + V₂ p₂g ....(i)

V₁ ( ρ $-$ ρ₁ ) g = V₂ ( p₂ $-$ ρ )

⇒ $\frac{V_{1}}{V_{2}} = \frac{ρ_{2} - ρ}{ρ - ρ_{1}}$

Some More Questions From Mechanical Properties of Fluids Chapter

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