Question

A plastic sphere floats in water with 45.0% of its volume submerged. This same sphere floats in glycerin with 35.8% of its volume submerged.

(a) Determine the density of the glycerin.
kg/m³

(b) Determine the density of the sphere.
kg/m³

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Density of water = ₁ = 1000 kg/m³

Density of the glycerin = ₂

Density of the sphere =

Volume of the sphere = V

Weight of the sphere = W = Vg

The sphere floats in water with 45% of it's volume submerged.

Volume of the sphere submerged in water = V₁ = 0.45V

Buoyancy force on the sphere from the water = F₁ = ₁V₁g

When floating in water the weight of the sphere is supported by the buoyancy force on it from water.

W = F₁

Vg = ₁V₁g

V = ₁(0.45V)

= 0.45₁

= (0.45)(1000)

= 450 kg/m³

The sphere floats in glycerin with 35.8% of it's volume submerged.

Volume of the sphere submerged in glycerin = V₂ = 0.358V

Buoyancy force on the sphere from the glycerin = F₂ = ₂V₂g

When floating in glycerin the weight of the sphere is supported by the buoyancy force on it from the glycerin.

W = F₂

Vg = ₂V₂g

V = ₂(0.358V)

= 0.358₂

450 = 0.358₂

₂ = 1256.98 kg/m³

a) Density of the glycerin = 1256.98 kg/m³

b) Density of the sphere = 450 kg/m³