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Q.

A sphere of volume V and density 'd' floats at interface of two immiscible liquids of densities $d_{1}$ and $d_{2}$ respectively. If $d_{1} > d > d_{2}$ , then the ratio of volume of parts of sphere in upper and lower liquids is

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a

$\frac{(d_{1} - d)}{(d_{2} + d)}$

b

$\frac{(d - d_{1})}{(d - d_{2})}$

c

$\frac{(d_{1} - d)}{(d_{2} - d)}$

d

$\frac{(d_{2} - d)}{(d - d_{1})}$

answer is D.

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Detailed Solution

$(v_{1} + v_{2}) d g = v_{1} d_{2} g + v_{2} d_{1} g (v_{1} + v_{2}) d = v_{1} d_{2} + v_{2} d_{1} (\frac{v_{1}}{v_{2}} + 1) d = (\frac{v_{1}}{v_{2}}) d_{2} + d_{1} \frac{v_{1}}{v_{2}} d + d = \frac{v_{1}}{v_{2}} d_{2} + d_{1} \frac{v_{1}}{v_{2}} (d - d_{2}) = d_{1} - d \frac{v_{1}}{v_{2}} = \frac{d_{1} - d}{d - d_{2}} \frac{v_{2}}{v_{1}} = \frac{d_{2} - d}{d - d_{1}}$

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