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 = \[4\pi R^2 = 4 \times 3 . 14 \times \left( 6 \right)^2 = 4 \times 3 . 14 \times 36\]  = 452.16 cm²

Volume of metal in the sphere = \[\frac{4}{3}\pi\left( R^3 - r^3 \right) = \frac{4}{3} \times 3 . 14 \times \left( 6^3 - 5^3 \right) = \frac{4}{3} \times 3 . 14 \times \left( 216 - 125 \right) = \frac{4}{3} \times 3 . 14 \times 91\]  = 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.