A solid sphere of volume V and density - floats at the interface of two immiscible liquids of densities -1 and-2 respectively- If -1 - - - -2- then find out the ratio of volume of the parts of the sphere in upper and lower liquids is

The correct option is A ρ #160; _ 2 ρ #160; _ 1 ρ ρ #160; _ 1 Let v n #160; ^ th part of the sphere is inside the liquid with density ...

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Byju's Answer

Standard VIII

Physics

Law of Floatation

A solid spher...

Question

A solid sphere of volume V and density $ρ$ floats at the interface of two immiscible liquids of densities $ρ_{1} a n d ρ_{2}$ respectively. If $ρ_{1} < ρ < ρ_{2}$ , then find out the ratio of volume of the parts of the sphere in upper and lower liquids is

\frac{ρ_{2} - ρ_{1}}{ρ - ρ_{1}}

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\frac{ρ + ρ_{1}}{ρ + ρ_{2}}

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\frac{ρ + ρ_{2}}{ρ + ρ_{1}}

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\frac{\sqrt{ρ_{1} ρ_{2}}}{ρ}

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Solution

The correct option is A $\frac{ρ_{2} - ρ_{1}}{ρ - ρ_{1}}$
Let ${\frac{v}{n}}^{t h}$ part of the sphere is inside the liquid with density $ρ_{2}$
$v ρ g = \frac{v}{n} ρ_{2} g + v (1 - \frac{1}{n}) ρ_{1} g ρ - \frac{ρ_{2}}{n} = (1 - \frac{1}{n}) ρ_{1} ρ - ρ_{1} = \frac{ρ_{2} - ρ_{1}}{n} n = \frac{ρ_{2} - ρ_{1}}{ρ - ρ_{1}}$

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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 find out the ratio of volume of the parts of the sphere in upper and lower liquids is

The correct option is A ρ2−ρ1ρ−ρ1Let vnth part of the sphere is inside the liquid with density ρ2 vρg=vnρ2g+v(1−1n)ρ1gρ−ρ2n=(1−1n)ρ1ρ−ρ1=ρ2−ρ1nn=ρ2−ρ1ρ−ρ1

A solid sphere of volume V and density $ρ$ floats at the interface of two immiscible liquids of densities $ρ_{1} a n d ρ_{2}$ respectively. If $ρ_{1} < ρ < ρ_{2}$ , then find out the ratio of volume of the parts of the sphere in upper and lower liquids is