Solution :
Let r be the radius of a sphere and δr be the error in measuring the radius. Then, r = 9 cm and δr = 0.03 cm.
Let V be the volume of the sphere. Then,
V = 43πr3 ⟹ dVdr = 4πr2
⟹ (dVdr)r=9 = 4π×92 = 324π
Let δV be the error in V due to error in V due to error δr in r. Then,
δV = dVdr δr ⟹ δV = 324π×0.03 = 9.72πcm3
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