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A drop of oil rises through water with a...

A drop of oil rises through water with an acceleration `alphag`. If `alpha` is a constant quantity and g is the acceleration due to gravity, find the specific gravity of the oil. Neglect the friction of water.

Text Solution

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Let d and D be the density of water and oil respectively, and mass of the oil drop is m.
`therefore` Volume of the oil drop = `m/D`
= volume of the displaced water by the oil drop or, mass of the displaced water = density of water `xx` volume of the displaced water
`d xx m/D`
Now, buoyancy = weight of the displaced water = `(dmg)/D`
`therefore` Net upward force acting on the oil drop
= buoyant force - weight of the oil drop
= `(dmg)/D-mg=(d/D-1)mg`
`therefore` Acceleration of the oil drop inside water,
`alphag=((d/D-1)mg)/m=(d/D-1)g`
`therefore` Specific gravity of oil = `D/d=1/(1+alpha)`
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