# 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. - Geometry

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 cm3. Find the outer surface area and mass of the sphere.

#### 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 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.
Concept: Surface Area of a Combination of Solids
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#### APPEARS IN

Balbharati Mathematics 2 Geometry 10th Standard SSC Maharashtra State Board
Chapter 7 Mensuration
Practice set 7.4 | Q 6 | Page 161