Question:
In the middle of a rectangular field measuring 30 m × 20 m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.
Solution:
Radius of dug $=\frac{7}{2} \mathrm{~m}$
Depth of dug = 10 m
The volume of dug
$=\pi\left(\frac{7}{2}\right)^{2} \times 10$
The volume of dug = 385 m3
Let the height through which the level of the field is raised will be x.
But, volume of dug = volume of earth spread over field.
$385=30 \times 20 \times x$
$x=\frac{385}{30 \times 20}$
$x=68.6 \mathrm{~cm}$