A block of wood has a density of 900 kg/m3. It is placed in a fluid in which it floats with two-thirds of its volume submerged. What is the density of the fluid? Express your answer in kilograms per cubic meter to three significant figures.

The density of the fluid = 1350 kg/m³.

Explanation:

According to law of floatation,

The weight of fluid displayed is equal to the weight of block wood.

Weight of block wood (W₁) = D₁V₁g ... equation 1a

Where D₁ = density of the block, V₁ = volume of the block, g = acceleration due to gravity.

Weight of the displaced fluid (W₂) = D₂V₂g ... equation 1b

Where D₂ = density of the displaced fluid, V₂ = Volume of the displaced fluid and g = acceleration due to gravity.

Since W₁ = W₂ (from law of floatation)

∴ D₁V₁g = D₂V₂g ... Equation 2

But The volume of the block submerged = 2/3V₁

From Archimedes principle,

the volume of subject immersed in water = the volume of the displaced fluid.

V₂ = 2/3V₁

Substituting this value into equation 2

D₁V₁g = D₂2/3V₁g ... 3

Simplifying equation 3,

Therefore, D₂ = 3/2 (D₁) ... Equation 4

where D₁ = 900 kg/m³

D₂ = 3/2 (900)

D₂ = 1350 kg/m³.

The density of the fluid = 1350 kg/m³.

Answer: 1350kg/m3

Explanation:

Db = Mb/Vb

900 = Mb/Vb

(where Db is the density of the block of wood = 900kg/m3, Mb is its mass and Vb is its volume

Df = Mf/Vf (where Df, Mf and Vf are the density, mass and velocity of the fluid respectively

note that the same mass is involved because the block of wood can only displace its own mass of the fluid

It displaces two-thirds of its own volume in the fluid.

2/3Vf = Vb

Vf = 2Vb/3

Db = Mb/Vb

900 = 2Mb/3Vf

2700Vf = 2Mb

1350Vf = Mb

since Mb = Mf

Df = 1350Vf/Vf

Df = 1350Kg/m3