Question

The diameter and thickness of a hollow metals sphere are 12 cm and 0.01 m respectively. The density of the metal is 8.88 gm per cm³ . Find the outer surface area and mass of the sphere.

Answer 1

Outer radius of the sphere, R = $\frac{12}{2}$ = 6 cm

Thickness of the sphere = 0.01 m = 0.01 × 100 cm = 1 cm (1 m = 100 cm)

∴ Inner radius of the sphere, r = R − 1 = 6 − 1 = 5 cm

Outer surface area of the sphere = $4\mathrm{\pi }{R}^2=4\times 3.14\times {\left(6\right)}^2=4\times 3.14\times 36$ = 452.16 cm²

Volume of metal in the sphere = $\frac{4}{3}\mathrm{\pi }\left(R^3-r^3\right)=\frac{4}{3}\times 3.14\times \left(6^3-5^3\right)=\frac{{\displaystyle 4}}{{\displaystyle 3}}\times 3.14\times \left(216-125\right)=\frac{{\displaystyle 4}}{{\displaystyle 3}}\times 3.14\times 91$ = 380.97 cm³

Density of the metal = 8.88 g/cm³

∴ Mass of the sphere = Volume of metal in the sphere × Density of the metal = 380.97 × 8.88 = 3383.01 g

Thus, the outer surface area and mass of the sphere are 452.16 cm² and 3383.01 g, respectively.