# A well of diameter 3 m is dug 14 m deep.

Question:

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution:

Diameter of cylindrical well (d) = 3 m

$\Rightarrow$ Radius of the cylindrical well $=\frac{\mathbf{3}}{\mathbf{2}} \mathrm{m}=1.5 \mathrm{~m}$

Depth of the well $(\mathrm{h})=14 \mathrm{~m}$

$\therefore$ Volume $=\pi \mathrm{r}^{2} \mathrm{~h}=\frac{\mathbf{2 2}}{\mathbf{7}} \times\left(\frac{\mathbf{1 5}}{\mathbf{1 0}}\right)^{2} \times 14 \mathrm{~m}^{3}$

$=\frac{\mathbf{2 2} \times \mathbf{1 5} \times \mathbf{1 5} \times \mathbf{1 4}}{\mathbf{7} \times \mathbf{1 0} \times \mathbf{1 0}} \mathrm{m}^{3}=99 \mathrm{~m}^{3}$

Let the height of the embankment = H metre.

Internal radius of the embankment (r) = 1.5 m .

External radius of the embankment R = (4 + 1.5)m

$=5.5 \mathrm{~m}$

$\therefore$ Volume of the embankment

$=\pi R^{2} H-\pi r^{2} H=\pi H\left[R^{2}-r^{2}\right]=\pi H(R+r)(R-r)$

$=\frac{\mathbf{2 2}}{\mathbf{7}} \times \mathrm{H}(5.5+1.5)(5.5-1.5)$

$=\frac{22}{7} \times \mathrm{H} \times 7 \times 4 \mathrm{~m}^{3}$

Since, Volume of the embankment = Volume of the cylindrical well

$\therefore\left[\frac{22}{7} \times \mathbf{H} \times \mathbf{7} \times \mathbf{4}\right]=99$

$\Rightarrow \mathrm{H}=99 \times \frac{\mathbf{7}}{\mathbf{2 2}} \times \frac{\mathbf{1}}{\mathbf{7}} \times \frac{\mathbf{1}}{\mathbf{4}} \mathrm{m}=\frac{\mathbf{9}}{\mathbf{8}} \mathrm{m}=1.125 \mathrm{~m}$

$=\frac{\mathbf{9}}{\mathbf{8}} \mathrm{m}=1.125 \mathrm{~m}$

Thus, the required height of the embankment

= 1.125 m