Question

In the middle of a rectangular field measuring $30m\times 20m$, a well of 7m diameter and 10m depth is dug. The earth so removed is evenlyspread over the remaining part of the field. Find the height through which the level of the field is raised.

Answer 1

Diameter of wall = 7m
$\therefore$ Radius (r) = $\frac{7}{2}$ m
Depth (h) = 10 m.
$\therefore$ Volume of earth dug out = $\pi r^{2}h$
= $\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times 10m^3=385m^3$
Area of field = $l\times b=30\times 20=600m^2$
Area of well = $\pi r^2=\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}=\frac{77}{2}{m}^2$
= 38.5 $m^2$
$\therefore$ Reamining area of the field = 600 - 38.5
= 561.5 $m^2$
$\therefore$ Height of level raised = $\frac{\text{Volume of earth}}{\text{Area of field}}$
= $\frac{385}{561.5}=\frac{385\times 10}{5615}m$
= 0.6856 m = 68.6 cm