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If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be

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a
8%
b
2%
c
4%
d
6%

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detailed solution

Correct option is D

V=43πR3;ln⁡V=ln⁡43π+ln⁡R3Differentiating,dVV=3dRRError in the determination of the volume= 3 × 2% = 6%

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