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Q. 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

AIPMTAIPMT 2008Physical World, Units and Measurements

Solution:

$V=\frac{4}{3} \pi R^{3} ; \ln V=\ln \left(\frac{4}{3} \pi\right)+\ln R^{3}$
Differentiating, $\frac{dV}{V}=3\frac{dR}{R}$
Error in the determination of the volume
$=3 \times 2 \%$
$=6 \%$