91Ó°ÊÓ

The radius of a semicircular Antarctica is 2000km .

The average thickness of the ice cover is 3000m.

Unit conversion is the conversion between different units of measure for the same quantity, usually through multiplier conversion factors.Use the given conversion factors to convert the quantities from one unit to other.

The expression for the volume of semicircle is given as follows:

$V=\frac{\mathrm{π}{R}^2}{2}t$ … (i)

Here, r is the radius and t is the thickness.

Convert the radius and thickness into cm.

$$\begin{gathered} {\text{1km=10}}^3{\text{cmand1m=10}}^2\text{m} \\ \text{Radiusofthesemicircleis,} \\ R=2000km\left(\frac{{10}^{3}m}{1km}\right)\left(\frac{{10}^{2}cm}{1km}\right) \\ =2000×{10}^{5}cm \\ \text{Thicknessofsemicircleis,} \\ t=3000m\left(\frac{{10}^{2}cm}{1m}\right) \\ =3000×{10}^{2}cm \\ \text{Thus,theradiusis}2000×{10}^{5}cm\text{andthicknessis}3000×{10}^{2}cm\text{.} \end{gathered}$$

Using equation (i), the volume of the ice is calculated as follows:

$$\begin{gathered} V=\frac{\mathrm{π}{R}^2}{2}t \\ =\frac{\mathrm{π}{\left(2000×{10}^{5}cm\right)}^2}{2}×3000×{10}^{2}cm \\ =1.9×{10}^{22}c{m}^3 \end{gathered}$$

Thus, the volume of ice is$1.9×{10}^{22}c{m}^3$ .