Question

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

A)8% b)2% c)4% d)6%

Answer 1

Explanation:

$volume \: of \: sphere = \frac{4}{3} \pi {r}^{3} \\ \%error \: in \: radius = 2\% \\ \%error \: in \: volume \: = 3 \times \%error \: in \: radius \\ = 3 \times 2 \\ = 6\%$