A solid sphere having volume V and density $ρ$ floats at the interface of two immiscible liquids of densities $ρ_{1} and ρ_{2}$ respectively. lf $ρ_{1} < ρ < ρ_{2}$ , then the ratio of volume of the parts of the sphere in upper and lower liquid is

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$\frac{ρ - ρ_{1}}{ρ_{2} - ρ}$

$\frac{ρ + ρ_{2}}{ρ + ρ_{1}}$

$\frac{ρ + ρ_{1}}{ρ + ρ_{2}}$

$\frac{ρ_{2} - ρ}{ρ - ρ_{1}}$

answer is B.

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Detailed Solution

According to law of flotation,
$Vρg = V_{1} ρ_{1} g + V_{2} ρ_{2} g - - - - - - (i)$

and $V = V_{1} + V_{2} - - - - (ii)$

Hence from Eqs. (i) and (ii),
$V_{1} ρg + V_{2} ρg = V_{1} ρ_{1} g + V_{2} ρ_{2} g$

Or $V_{1} (ρ - ρ_{1}) g = V_{2} (ρ_{2} - ρ) or \frac{V_{1}}{V_{2}} = \frac{ρ_{2} - ρ}{ρ - ρ_{1}}$

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