Question

The diameter and thickness of a hollow metal 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, [π = 3.14]

Answer 1

Given: For the hollow sphere, 

diameter (D) =12 cm, thickness = 0.01 m 

density of the metal = 8.88 gm per cm³ 

To find: i. Outer surface area of the sphere 

ii. Mass of the sphere.

Diameter of the sphere (D) 

= 12 cm 

∴ Radius of sphere (R) 

= d/2 = 12/2 = 6 cm 

∴ Surface area of sphere = 4πR² 

= 4 × 3.14 × 6² 

= 452.16 cm² 

Thickness of sphere = 0.01 m 

= 0.01 × 100 cm …[∵ 1 m = 100 cm] 

= 1 cm 

∴ Inner radius of the sphere (r) = Outer radius – thickness of sphere 

= 6 – 1 

= 5 cm 

∴ Volume of hollow sphere = Volume of outer sphere - Volume of inner sphere

∴ The outer surface area and the mass of the sphere are 452.16 cm² and 3383.19 gm respectively.