In measuring the circumference $100$ $cm$ of a circle, there is an error of $0.05$ $cm$ then percentage error in its area is?
$
  A.\dfrac{1}{{10}} \\
  B.\dfrac{1}{5} \\
  C.\dfrac{1}{{100}} \\
  D.1 \\
$

Hint: In the given question, we have to find out the value of error percentage in Area. Since circumference is given, find out the value of the radius of the circle and differentiate the ratio, $C = 2\pi r$ (where $C$ is circumference) and work out the value of $\delta r$. Find out the area and again differentiate the relation $'A = \pi {r^2}'$to find its error. Take out the error $\% $ of area as asked.

Complete step-by-step answer:
Consider, C= Circumferences of circle, A= Area, r= radius.
Since,
$C = 100 cm$
$\therefore 2\pi r = 100$ [$\because $Circumference of circle $ = 2\pi r$]
$ \Rightarrow r = \dfrac{{50}}{\pi }$
$C = 2\pi r$
Again, differentiate both sides –
$ \Rightarrow \delta c = 2\pi \delta r$
Therefore, $\delta r$ is the error in radius.
$\therefore \left[ {\delta r = \dfrac{{0.05}}{{2\pi }}} \right]$ $\left[ {\ since, \delta c = 2\pi \delta r = 0.05} \right]$
$A = \pi {r^2}$
Differentiate both sides –
$\delta A = 2\pi r\delta r$
Error $\% $
\[
   \Rightarrow {\left( {\dfrac{{\delta A \times 100}}{A}} \right)\% } = {\left( {\dfrac{{2\pi r\delta r \times 100}}{{\pi {r^2}}}} \right)\% } \\
   = {\left( {\dfrac{{2\delta r \times 100}}{r}} \right)\% } \\
   = {\left( {\dfrac{{2 \times \dfrac{{0.05}}{{2\pi }} \times 100}}{{\dfrac{{50}}{{\pi }}}}} \right)\% } \\
\]
Error \[\% \] in area \[ = \dfrac{1}{10} = 0.1\% \]

So, the correct answer is “Option A”.

Note: The form $\Delta y = \dfrac{{dy}}{{dx}}\Delta x$, is very useful in the calculation of small changes (or errors) in dependent variable corresponding to small changes (or errors) in the independent variable and is of great importance in the theory of errors in engineering physics, statistics and several $O$ the branches of the science.