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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.
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 = 1728
`=>` R = 12
Surface area of the sphere
= 4πR2
= 4π122
= 1785.6 cm2
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