The circumference of a circle is measured as $56 \mathrm{~cm}$ with an error $0.02 \mathrm{~cm}$. The…
The circumference of a circle is measured as $56 \mathrm{~cm}$ with an error $0.02 \mathrm{~cm}$. The percentage error in its area is
- $1 / 7$
- $1 / 28$
- $1 / 14$
- $1 / 56$
Solution
Given circumference of a circle $S=2 \pi r=56$
$\begin{array}{ll}\Rightarrow & r=\frac{28}{\pi} \\ \text { Error } & \delta S=2 \pi \delta r=0.02 \\ \Rightarrow & \delta r=\frac{0.02}{2 \pi}\end{array}$
Area of circle, $A=\pi r^2$
$\therefore$ Percentage error in $\begin{aligned} A & =\frac{\delta A}{A} \times 100 \\ & =2 \times \frac{\delta r}{r} \times 100 \\ & =2 \times \frac{0.02 \times \pi}{2 \pi \times 28} \times 100 \\ & =\frac{1}{14}\end{aligned}$
Asked in: AP EAMCET 2007
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