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

Let $r$ be the radius, $C$ be the circumference and $A$

be the area of the circle. Then,

$C=2\pi r\text{ and }A=\pi r^2$

$\therefore \mathrm{\Delta }C=\frac{dC}{dr}\mathrm{\Delta }r$ and $\mathrm{\Delta }A=\frac{dA}{dr}\mathrm{\Delta }r$

$\Rightarrow \mathrm{\Delta }C=2\pi \mathrm{\Delta }r\text{ and }\mathrm{\Delta }A=2\pi r\mathrm{\Delta }r$

We have, $C=56\text{ and }\mathrm{\Delta }C=0.02$

$\begin{array}{l}\therefore \frac{\mathrm{\Delta }C}{C}=\frac{0.02}{56}=\frac{1}{2800}\\ \Rightarrow \frac{2\pi \mathrm{\Delta }r}{2\pi r}=\frac{1}{2800}\Rightarrow \frac{\mathrm{\Delta }r}{r}=\frac{1}{2800}\\ \therefore \frac{\mathrm{\Delta }A}{A}\times 100=\frac{2\pi r\mathrm{\Delta }r}{\pi r^2}\times 100=2\left(\frac{\mathrm{\Delta }r}{r}\times 100\right)=2\times \frac{1}{28}=\frac{1}{14}\end{array}$