Question

A hollow spherical body of inner and outer radii 6 cm and 8 cm respectively floats half submerged in water. Find the density of the material of the sphere.

Answer 1

Let the density of the spherical body is ρ g/cm³.

By the Principle of the Flotation,

Weight of the spherical body = Weight of the fluid displaced.

∴ Volume of the spherical body × density of spherical body × g = Volume of body displaced in water × density of water × g

Volume of the spherical body  = 4/3 π (R³ - r³)

= 4/3 × 22/7 × (8³ - 6³)

= 4/3π(8³ - 6³)

Volume of the spherical body displaced in water = Outer volume/2

= 2/3π(8³)

Also, density of water = 1 g/cm³.

∴ 4/3 π (R³ - r³) × ρ = [4/3 π (R³)]/2 × 1

∴ 4/3π(8³ - 6³) ×  ρ = 2/3π(8³)

 2(8³ - 6³) ×  ρ = (8³)

  2 × 296ρ = 512

  ρ = 0.8648 g/cm³

 ∴ ρ = 864.8 kg/m³.

Hence, the density of the material of the spherical body = 864.8 kg/m³.

Hope it helps.

Answer 2

$<i>$External volume = (4/3)πr³ = (4/3)π x 8³ = 2145cm³

Volume of hollow interior = (4/3)πr³ = (4/3)π x 6³ = 905cm³

Volume of material = 2145 - 905 = 1240cm³

Since the object is half submerged, it displaces a volume of water equal to half its external volume, i.e. 2145/2 = 1072.5cm³.

The density of water = 1g/cm³, so the mass of water displaced equals 1072.5g.

Upthrust equals weight of water displaced. Since the object is floating in equilibrium, its weight equals the upthrust. So the object's weight (or mass) equals the weight (or mass) of water displaced

So the mass of water = 1072.5g

Density of material = mass / volume = 1072.5/1240 = 0.865g/cc