A layer of oil with density $724 k g / m^{3}$ floats on water of density $1000 k g / m^{2}$ . A block floats at the oil-water interface with $1 / 6$ of its volume in oil and $5 / 6$ of its volume in water, as shown in the figure. What is the density of the block?

1024 k g / m^{3}

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1276 k g / m^{3}

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776 k g / m^{3}

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954 k g / m^{3}

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Solution

The correct option is C $954 k g / m^{3}$
Given,
Density of oil $(ρ_{1}) = 724 k g / m^{3}$
Density of water $(ρ_{2}) = 1000 k g / m^{3}$
According to Archemedes' principle,
Upthrust $=$ Weight of the liquid displaced
$V_{ρ_{b} g} = V_{0 ρ_{1} g} + V_{w ρ_{2} g}$
( $V_{0}$ and $V_{w}$ are volume of oil and water displaced, respectively)
$V_{ρ_{b} g} = \frac{V}{6} (724) g + \frac{5 V}{6} (1000) g$
where, $ρ_{b} =$ density of block
and $V_{0} = \frac{V}{6}$ and $V_{w} = \frac{5 V}{6}$
$\Rightarrow V_{ρ_{b} g} = (\frac{724 V}{6} + \frac{5000 V}{6}) g$
$\Rightarrow V_{ρ_{b} g} = (\frac{724 V + 5000 V}{6}) g$
$\Rightarrow V_{ρ_{b} g} = \frac{5724}{6} V g$
$ρ_{b} = 954 k g / m^{3}$

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A layer of oil with density 724 kg/m3 floats on water of density 1000 kg/m2. A block floats at the oil-water interface with 1/6 of its volume in oil and 5/6 of its volume in water, as shown in the figure. What is the density of the block?

A layer of oil with density $724 k g / m^{3}$ floats on water of density $1000 k g / m^{2}$ . A block floats at the oil-water interface with $1 / 6$ of its volume in oil and $5 / 6$ of its volume in water, as shown in the figure. What is the density of the block?