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Question
Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted and recasted into a single solid sphere. Taking π = 3.1, find the surface area of the solid sphere formed.
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Solution
Let radius of the larger sphere be ‘R’
Volume of single sphere
= Vol. of sphere 1 + Vol. of sphere 2 + Vol. of sphere 3
`4/3piR^3 = 4/3pir_1^3 + 4/3pir_2^3 + 4/3pir_3^3`
`4/3piR^3 = 4/3pir6^3 + 4/3pir8^3 + 4/3pi10^3`
R3 = [63 + 83 + 103]
R3 = [216 + 512 + 1000]
R3 = 1728
R3 = 123
∴ R = 12
Surface area of the sphere
= 4πR2
= 4 × 3.1 × 122
= 4 × 3.1 × 12 × 12
= 1785.6 cm2
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