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

$s=2\pi r$

$r=\frac{s}{2\pi }$

$\Delta s=0.02$

$A=\pi r^2$

$A=\pi {\left(\frac{S}{2\pi }\right)}^2=\frac{S^2}{4\pi }$

$\delta A=\frac{2S}{4\pi }\delta S$

$=\frac{s.\delta s}{2\pi }$

Percentage error in area $A=\frac{△A}{A}\times 100$

$=\frac{S.\delta S}{2\pi }\times \frac{4\pi }{S^2}\times 100$

$=\frac{2\delta S}{S}\times 100$

$=\frac{2\times 0.02}{56}\times 100$

$=\frac{1}{14}$