A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m.

Diameter of the cylindrical well = 7 m

$\Rightarrow$ Radius of the cylindrical $(\mathrm{r})=\frac{\mathbf{7}}{\mathbf{2}} \mathrm{m}$

Depth of the well (h) = 20 m

$\therefore$ Volume $=\pi \mathrm{r}^{2} \mathrm{~h}=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20 \mathrm{~m}^{2}$

$=22 \times 7 \times 5 \mathrm{~m}^{3}$

$\Rightarrow$ Volume of the earth taken out $=22 \times 7 \times 5 \mathrm{~m}^{3}$

Now this earth is spread out to form a cuboidal platform having length = 22 m, breadth = 14m

Let h be the height of the platform.

$\therefore$ Volume of the platform $=22 \times 14 \times \mathrm{h} \mathrm{m}^{3}$

$\therefore 22 \times 14 \times \mathrm{h}=22 \times 7 \times 5$

$\Rightarrow \mathrm{h}=\frac{\mathbf{2 2} \times \mathbf{7} \times \mathbf{5}}{\mathbf{2 2} \times \mathbf{1 4}}=\frac{\mathbf{5}}{\mathbf{2}} \mathrm{m}=2.5 \mathrm{~m}$

Thus, the required height of the platform is 2.5 m