Sum

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 cm^{3}. Find the outer surface area and mass of the sphere.

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#### Solution

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 cm^{2}

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 cm

^{3}Density of the metal = 8.88 g/cm

∴ 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 cm

^{3}∴ 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 cm

^{2}and 3383.01 g, respectively. Concept: Surface Area of a Combination of Solids

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