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A solid uniform ball having volume V and density r floats at the interface of two immiscible liquids as shown in figure.
The densities of the upper and the lower liquids are $ρ_{1}$ and $ρ_{2}$ respectively, such that $ρ_{1} < ρ < ρ_{2} .$ What fraction of the volume of the ball will be in the lower liquid :

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$\frac{ρ_{1} - ρ}{ρ_{1} - ρ_{2}}$

$\frac{ρ_{1}}{ρ_{1} - ρ_{2}}$

$\frac{ρ - ρ_{2}}{ρ_{1} - ρ_{2}}$

$\frac{ρ_{1} - ρ_{2}}{ρ_{2}}$

answer is C.

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Detailed Solution

In floating condition

W = Th

Let V₁ volume immersed in lower liquid then

$Vρg = (V - V_{1}) ρ_{1} g + V_{1} ρ_{2} g$

$V (ρ - ρ_{1}) = V_{1} (ρ_{2} - ρ_{1})$

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

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