A block of wood floats in fresh water with 0.663 of its volume V submerged and in oil with 0.913 V submerged. Find the density of (a) the wood and (b) the oil.

(a). 663kg/m³.

(b). 726.18kg/m³

Explanation:

(a). Let V be the volume of the block of wood,

Then the submerged volume in fresh water is

V₍s₎=0.663V

Where V₍s₎ = submerged volume in water.

Since the block is floating, according to Archimedes principle,

The weight of the displaced fresh water is equal to the weight of block.

∴D₍w₎V₍s₎ = DV ... (1)

Where D₍w₎ = Density of fresh water, D = Density of Block of wood

Since V₍s₎=0.663V, and D₍w₎=1000kg/m³.

We substitute the value of V₍s₎ and D₍w₎ in equation (1) above.

⇒ 1000*0.663V = D * V

∴ D * V = 1000 * 0.663V

Dividing both side of the equation by V

∴ D = 663kg/m³.

(b). Let V be the volume of the wood block. and V₍o₎ is the submerged volume in oil.

V₍o₎ = 0.913V

Since the wood block is floating, according to Archimedes principle,

The weight of the displaced oil is equal to the weight of the wood block.

IF D₍o₎ = Density of oil.

D₍o₎V₍o₎ = DV ... (2)

D=663kg/m³, and V₍o₎ = 0.913V,

Substituting the value of D and V₍o₎ in equation (2)

D₍o₎ * 0.913V = 663 * V

Dividing both side of the equation by the coefficient of D₍o₎

∴D₍o₎ * 0.913V/0.913V = 663V/0.913V

∴ D₍o₎ = 663/0.913

D₍o₎ = 726.177

D₍o₎ ≈ 726.18kg/m³