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

Explanation

Let the densities of oil and water be D and d respectively and the mass of the oil drop be m.
`therefore` Volume of the oil drop = `m/D` = volume of water displaced by the oil drop
`therefore` Mass of displaced water = density of water `xx` volume of displaced water = `(dm)/D`
`therefore` Buoyancy = weight of displaced water = `(dmg)/D`
`therefore` Net upward thrust 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)/mor,alpha=(d/D-1)`
or, `d/D=1+alpha`
`therefore` Specific gravity of the oil = `D/d=1/(1+alpha)`.