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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Aicharya · Accepted Answer

Percentage error in radius is given as 2% i.e. $\frac{\Delta r}{r} \times 100=2 \%$
Volume of sphere $V=\frac{4 \pi_2}{3}$
Percentage error in volume $\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=3 \times \frac{\Delta \mathrm{r}}{\mathrm{r}} \times 100=3 \times 2=6 \%$

If the error in the measurement of radius of a sphere is $2 \%$, then the error in the determination of…

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