A solid uniform ball having volume V and density $ρ$ floats at the interface of two 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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$\frac{ρ - ρ_{2}}{ρ_{1} - ρ_{2}}$

$\frac{ρ_{1}}{ρ_{1} - ρ_{2}}$

$\frac{ρ_{1} - ρ}{ρ_{1} - ρ_{2}}$

$\frac{ρ_{1} - ρ_{2}}{ρ_{2}}$

answer is C.

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Detailed Solution

Let x - fraction of volume is in the lower liquid.

Then,

Weight = Upthrust

$W = U_{ρ_{1}} + U_{ρ_{2}} or Vρg = (1 - x) {Vρ}_{1} g + {xVρ}_{2} g Solving this equation, we get x = \frac{ρ - ρ_{1}}{ρ_{2} - ρ_{1}} o r, x = \frac{ρ_{1} - ρ}{ρ_{1} - ρ_{2}}$

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