A vessel contains oil $(ρ = 0.8 g m / c m^{3})$ over mercury $(ρ = 13.6 g m / c m^{3})$ . A homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. Then density of material of the sphere in $g m / c m^{3}$ is

3.3

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6.4

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7.2

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2.8

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Solution

The correct option is C 7.2
Half of sphere immersed in mercury and half in oil.
Thus there will be upthrust due to both liquids.
Lets $ρ_{d}$ be the density of sphere
Upthrust due to buoyancy = Mass of mercury displaced + mass of oil displaced
Upthrust = $\frac{v}{2} \times 13.6 \times g + \frac{v}{2} \times 0.8 \times g$
Downward gravitational force due to sphere = $v \times ρ_{d} \times g$
Downward gravitational force = Upthrust due to buoyancy
$v \times ρ_{d} \times g$ = $\frac{v}{2} \times 13.6 \times g + \frac{v}{2} \times 0.8 \times g$
$ρ_{d} = 7.2 g m / c m^{3}$

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Similar questions

Q. A vessel contains oil (density = 0.8

g m / c m^{3}

) over mercury (density = 13.6

g m / c m^{3}

). A uniform sphere floats with half its volume immersed in mercury and the other half in oil. The density of the material of sphere in

g m / c m^{3}

is:

A vessel contains oil (density = 0.8 gm / $c m^{3}$ ) Over mercury (density = 13.6 gm / $c m^{3}$ ). homogeneous sphere floats with half of its volume immersed in mercury and the other half in oil. The density of the material of the sphere in gm/ $c m^{3}$