A piece of wood of relative density 0.25 floats in a pot containing oil of relative density 0.81. What will be the fraction of volume of the wood above the surface of the oil. (density of water = 1000 kg/m3)
0.69
If V1 is the volume of the wood in oil and V2 is the volume of the wood in air, then,
V1ρ1g=V2ρ2g
Where ρ1 and ρ2 are the relative densities of oil and wood respectively
V1V2 = 0.250.81 = 0.31
Fraction of the volume above the surface of oil = 1−V1V2 = 0.69