Given,
Depth of well = 20 m
Diameter of well = 7 m
Radius of well = \(\frac{7}{2}\) m
Dimension of rectangular field = 22 m × 14 m
so,
Amount of earth dug out from well = πr2h = \(\frac{22}{7}\) x \(\frac{7}{2}\) x \(\frac{7}{2}\) x 20 = 770 m3
when this earth is spread on rectangular field:
Then height of platform formed on rectangular field = \(\frac{volume\,of\, earth\,dug\,out}{area\,of\,rectangular\,out}\) = \(\frac{770}{22\times 14}\) = 2.5 m