An object of density - and volume V floats at the interface of two liquids 1 and 2 of densities -1 and -2 with volume V1 in liquid 1 and -2 in liquid 2- Then for equilibrium -1- - -2 value of V2V1 is-

For the upper half of the body applying Archimedes principle m_1g=ρ_1V_1g ............. (i) nbsp; nbsp; And for lower part again applying we get, m_2g=ρ_2 ...

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Byju's Answer

Standard XIII

Physics

Floating

An object of ...

Question

An object of density $ρ$ and volume $V$ floats at the interface of two liquids $1$ and $2$ of densities $ρ_{1}$ and $ρ_{2}$ with volume $V_{1}$ in liquid $1$ and $ρ_{2}$ in liquid $2$ . Then for equilibrium $(ρ_{1} < ρ < ρ_{2})$ value of $\frac{V_{2}}{V_{1}}$ is:

\frac{(ρ_{1} - ρ_{2})}{(ρ - ρ_{1})}

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\frac{(ρ_{2} - ρ_{1})}{(ρ - ρ_{2})}

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\frac{(ρ_{1} - ρ)}{(ρ_{2} - ρ)}

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\frac{(ρ - ρ_{1})}{(ρ_{2} - ρ)}

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Solution

The correct option is D $\frac{(ρ - ρ_{1})}{(ρ_{2} - ρ)}$
For the upper half of the body applying Archimedes principle
$m_{1} g = ρ_{1} V_{1} g . . . . . . . . . . . . . (i)$

And for lower part again applying we get,
$m_{2} g = ρ_{2} V_{2} g . . . . . . . . . . . . . . . . (i i)$

Adding $(i)$ and $(i i)$

$(m_{1} + m_{2}) g = (ρ_{1} V_{1} + ρ_{2} V_{2}) g$

$ρ V = ρ_{1} V_{1} + ρ_{2} V_{2}$

Putting $V = V_{1} + V_{2}$

$V_{1} (ρ - ρ_{1}) = V_{2} (ρ_{2} - ρ)$

$∴ \frac{V_{2}}{V_{1}} = (\frac{ρ - ρ_{1}}{ρ_{2} - ρ})$

Hence, Option $(B)$ is the correct answer.

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