Question

if the radius of a sphere has a percentage error of 2 , what is the % error in its volume ?

Answer 1

The % error in radius say "r" be 2.
i.e the error is:
$2 \times {10}^{ - 2}$

Volume of a sphere =
$\frac{4}{3} \pi {r}^{3}$

Therefore the mistaken volume will be,
$\frac{4}{3} \pi{(2 \times {10}^{ - 2} r)}^{3} \\ = \frac{4}{3} \pi {r}^{3} \times 8 \times {10}^{ - 6}$

So there'll be an error of
$8 \times {10}^{ - 6}$

Changing it into %

$8 \times {10}^{ - 4}$


Hence there'll be an error of 8×10^-4 %.

Answer 2

$\huge\textbf{Answer}$


V = 4/3 πr³ [Volume of sphere]

∆V/V = ± 3 (∆r/r)

∆V/V = ± (3 × 2)

∆V/V = ± 6%