What is the area and its approximate uncertainty of a circle of the radius of$\mathbf{3}\mathbf{.}\mathbf{1}\times {\mathbf{10}}^{\mathbf{4}}\mathbf{cm}$?

The given radius of the circle is $3.1\times {10}^4\mathrm{cm}$.

The area of a circle represents the region covered by a circular object in the 2-D plane. In the standard international system, the area of a circle is measured in meter square.

The relation for the area of a circle is $A=\pi r^2$.

Substituting the variables by their values in the above relation, you get

$\begin{array}{c}A=\pi {\left(3.1\times {10}^4\mathrm{cm}\right)}^2\\ =3.01\times {10}^9{\mathrm{cm}}^2.\end{array}$

Uncertainty represents the range within which the actual value of measurement lies.

The relation for uncertainty in the area is

$\frac{\Delta A}{A}=\frac{2\left(\Delta r\right)}{r}.$

Here, $\Delta r$ is the smallest possible uncertainty of the radius with the value $0.1\times {10}^4\mathrm{cm}$.

Substitute the variables by their values in the above relation, you get

$\begin{array}{c}\frac{\Delta A}{A}=\left[\frac{2\left(0.1\times {10}^4\mathrm{cm}\right)}{\left(3.1\times {10}^4\mathrm{cm}\right)}\right]\\ \Rightarrow \frac{\Delta A}{A}=0.0645\times 100\%\\ \Rightarrow \frac{\Delta A}{A}=6.45\%.\end{array}$

Thus, the area of the circle and the approximation uncertainly are $A=3.01\times {10}^9{\mathrm{cm}}^2$and $\frac{\Delta A}{A}=6.45\%$, respectively.