The surface area of a solid sphere is increased by 21% without changing its shape. Find the percentage increase in its - 1) radius , 2) volume
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Let the original and increased surface areas be A1 and A2.
A1/A2 = (r1/r2)2
1.21A1/A1 = (r2/r1)2
1.1 = r2/r1
∴ r2 = 1.1r1
∴ Increase percentage = (r2 – r1) / (r1) x 100 % = (1.1r1 – r1)/ (r1) x 100 % = 10
Hence, the percentage increase in diameter is 10 %
Let the original and increased volume be V1 and V2.
V1/V2 = (r1/r2)3
V2/V1 = (11/10)3
= 1.331
∴ V2 = 1.331 V1
Increase % = (V2 – V1) / (V1) x 100 % = (1.331 V1 – V1) / V1 x 100 %
= 33.1%
Hence, the percentage increase in volume is 33.1 %
A1/A2 = (r1/r2)2
1.21A1/A1 = (r2/r1)2
1.1 = r2/r1
∴ r2 = 1.1r1
∴ Increase percentage = (r2 – r1) / (r1) x 100 % = (1.1r1 – r1)/ (r1) x 100 % = 10
Hence, the percentage increase in diameter is 10 %
Let the original and increased volume be V1 and V2.
V1/V2 = (r1/r2)3
V2/V1 = (11/10)3
= 1.331
∴ V2 = 1.331 V1
Increase % = (V2 – V1) / (V1) x 100 % = (1.331 V1 – V1) / V1 x 100 %
= 33.1%
Hence, the percentage increase in volume is 33.1 %
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