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Q.

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

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a

$\frac{ρ - ρ_{1}}{ρ_{2} - ρ}$

b

$\frac{ρ + ρ_{2}}{ρ + ρ_{1}}$

c

$\frac{ρ + ρ_{1}}{ρ + ρ_{2}}$

d

$\frac{ρ_{2} - ρ}{ρ - ρ_{1}}$

answer is B.

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Detailed Solution

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$(V_{1} + V_{2}) ρg = V_{1} ρ_{1} g + V_{2} ρ_{2} g$
$V_{1} (ρ - ρ_{1}) = V_{2} (ρ_{2} - ρ) \frac{V_{1}}{V_{2}} = \frac{ρ_{2} - ρ}{ρ - ρ_{1}}$

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A solid sphere of volume v and density ρ floats at the interface of two immiscible liquids of densities ρ1and ρ2 respectively. If ρ1<ρ<ρ2,then the ratio of volume.of the parts of the sphere in upper and lower liquid is: