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

$\frac{1}{14}$

Let r be the radius, C be the circumference and A be the area of the circle. Then,

$C=2 \pi r$ and $A=\pi r^2$

∴  $\Delta C=\frac{d C}{d r} \Delta r$ and $\Delta A=\frac{d A}{d r} \Delta r$

$\Rightarrow \Delta C=2 \pi \Delta r$ and $\Delta A=2 \pi r \Delta r$

We have, $C=56$ and $\Delta C=0.02$

∴  $\frac{\Delta C}{C}=\frac{0.02}{56}=\frac{1}{2800}$

$\Rightarrow \frac{2 \pi \Delta r}{2 \pi r}=\frac{1}{2800} \Rightarrow \frac{\Delta r}{r}=\frac{1}{2800}$

∴  $\frac{\Delta A}{A} \times 100=\frac{2 \pi r \Delta r}{\pi r^2} \times 100=2\left(\frac{\Delta r}{r} \times 100\right)=2 \times \frac{1}{28}=\frac{1}{14}$