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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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$6 \%$
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 \(\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=3 \times \frac{\Delta \mathrm{r}}{\mathrm{r}} \times 100=3 \times 2=6 \%\)
Volume of sphere \(V=\frac{4 \pi_2}{3}\)
Percentage error in volume \(\frac{\Delta \mathrm{V}}{\mathrm{V}} \times 100=3 \times \frac{\Delta \mathrm{r}}{\mathrm{r}} \times 100=3 \times 2=6 \%\)
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