Question

If there is an error of $\pm 0.04$ cm in the measurement of the diameter of a sphere then the approximate percentage error in its volume, when the radius is $10cm$, is

• A. $\pm 1.2$
• B. $\pm 0.06$
• C. $\pm 0.006$
• D. $\pm 0.6$

Answer 1

Answer: (A)

$\pm 1.2$

$\frac{V+dV-V}{V}=\frac{\frac{4\pi }{3}(r+dr{)}^3-\frac{4\pi }{3}{r}^3}{\frac{4\pi }{3}{r}^3}$
$=\frac{(r+dr{)}^3-r^3}{r^3}$
$=\frac{(r+dr{)}^3}{r^3}-\frac{r^3}{r^3}$
$=(\frac{r+dr}{r}{)}^3-1$
$=(1+\frac{dr}{r}{)}^3-1$
Since $\frac{dr}{r}<<1$
Hence
$(1+\frac{dr}{r}{)}^3-1$
$=1+\frac{3dr}{r}-1$
$=\frac{3dr}{r}$
$=\frac{0.12}{10}$
$=0.012$
Hence
% Error$=0.012\times 100$
$=1.2$
Thus error in volume can be $\pm 1.2$