The circumference of a circle is measured as 56 cm with an error 0.02cm. the percentage error in its area is

$\begin{array}{l}\mathrm{s}=2\mathrm{\pi r}\\ \mathrm{r}=\frac{\mathrm{s}}{2\mathrm{\pi }}\\ \mathrm{\Delta s}=0.02\\ \mathrm{A}={\mathrm{\pi r}}^2\\ \mathrm{A}=\mathrm{\pi }{\left(\frac{\mathrm{S}}{2\mathrm{\pi }}\right)}^2=\frac{{\mathrm{S}}^2}{4\mathrm{\pi }}\\ \mathrm{\delta A}=\frac{2\mathrm{S}}{4\mathrm{\pi }}\mathrm{\delta S}\\ =\frac{\mathrm{s}.\mathrm{\delta S}}{2\mathrm{\pi }}\\ \text{ Percentage error in area }\mathrm{A}=\frac{\mathrm{\Delta A}}{\mathrm{A}}\times 100\\ \begin{array}{l}=\frac{\mathrm{S}.\mathrm{\delta S}}{2\mathrm{\pi }}\times \frac{4\mathrm{\pi }}{{\mathrm{S}}^2}\times 100\\ =\frac{2\mathrm{\delta S}}{\mathrm{S}}\times 100\end{array}\end{array}$
$\begin{array}{l}=\frac{2\times 0.02}{56}\times 100\\ =\frac{1}{14}\end{array}$