Question:
The radius of a sphere is measured to be $(7.50 \pm 0.85) \mathrm{cm}$. Suppose the percentage error in its volume is $x$. The value of $x$, to the nearest $x$, is__________.
Solution:
$\because \mathrm{v}=\frac{4}{3} \pi \mathrm{r}^{3}$
taking $\log \&$ then differentiate
$\frac{\mathrm{dV}}{\mathrm{V}}=3 \frac{\mathrm{dr}}{\mathrm{r}}$
$=\frac{3 \times 0.85}{7.5} \times 100 \%=34 \%$