A block of wood floats in fresh water with 0.848 of its volume V submerged and in oil with 0.910 V submerged. Find the density of (a) the wood and (b) the oil.

(a) 848 kg/m³

(b) 931.86 kg/m³

Explanation:

From Archimedes principle,

Since the weight of the displaced water is equal to the weight of the block,

(a)

D₁V₁ = DV ... Equation 1

Making D The subject of the equation above,

D = D₁V₁/V ... Equation 2

Where D₁ = density of water, V₁ = submerged volume in water, D = Density of block, V = volume of block.

But, V₁ = 0.848V, and D = 1000 kg/m³

Substituting these values into equation 2

D₁ = 1000*0.848V/V

D₁ = 848 kg/m³

The Density of block = 848 kg/m³

(b)

Also, From Archimedes principle

DV = D₃V₃ ... Equation 3

Making D₃ the subject of the equation,

D₃ = DV/V₃

where D₃ = density of oil, V₃ = volume of oil,

Given: V₃ = 0.910V, D = 848 kg/m³.

Substituting these values into equation 3

D₃ = 848V/0.910V

D³ = 931.86 kg/m³

Therefore the Density of oil = 931.86 kg/m³