The diameter and thickness of a hollow metals sphere are 12 cm and 0.0

1.	The diameter and thickness of a hollow metals sphere are 12 cm and 0.01 mrespectively. The density of the metal is 8.88 gm per cm3. Find the outer surfacearea and mass of the sphere.
Answer» Diameter of sphere from outside = 12 cm Radius from outside R = 6 cm Surface area = 4 πR² = 4 (22/7) (6²) = 452.57 cm² Now, Thickness of sphere = 0.01 m = 1 cm Diameter from inside = 12 - 2 = 10 cm [ -2 is done to reduce from both sides ] Radius from inside r = 5 cm Volume of this sphere V = Volume of outside sphere - Volume of inside sphere So, V = 4/3 πR³ - 4/3 πr³ = 4/3 π (R³ - r³) V = 4/3 (22/7) (6³ - 5³) V = 4/3 (22/7) (91) V = 381.33 cm³ Now, Density = Mass / Volume Mass = Density x Volume Mass = (8.88) x (381.33) Mass = 3386.24 g = 3.38624 kg

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

Answer»

Diameter of sphere from outside = 12 cm

Radius from outside R = 6 cm

Surface area = 4 πR²

= 4 (22/7) (6²) = 452.57 cm²

Now, Thickness of sphere = 0.01 m = 1 cm

Diameter from inside = 12 - 2 = 10 cm [ -2 is done to reduce from both sides ]

Radius from inside r = 5 cm

Volume of this sphere V = Volume of outside sphere - Volume of inside sphere

So, V = 4/3 πR³ - 4/3 πr³ = 4/3 π (R³ - r³)

V = 4/3 (22/7) (6³ - 5³)

V = 4/3 (22/7) (91)

V = 381.33 cm³

Now, Density = Mass / Volume

Mass = Density x Volume

Mass = (8.88) x (381.33)

Mass = 3386.24 g = 3.38624 kg