In a corner of a rectangular field with dimensions
Question:

In a corner of a rectangular field with dimensions $35 \mathrm{~m} \times 22 \mathrm{~m}$, a well with $14 \mathrm{~m}$ inside diameter is dug $8 \mathrm{~m}$ deep. The earth dug out is spread evenly over the remaining part of the field. Find the rise in the level of the field.

Solution:

We have,

Length of the fiel, $l=35 \mathrm{~m}$,

Width of the field, $b=22 \mathrm{~m}$,

Depth of the well, $H=8 \mathrm{~m}$ and

Radius of the well, $R=\frac{14}{2}=7 \mathrm{~m}$,

Let the rise in the level of the field be $h$.

Now,

Volume of the earth on remaining part of the field = Volume of earth dug out

$\Rightarrow$ Area of the remaining field $\times h=$ Volume of the well

$\Rightarrow$ (Area of the field – Area of base of the well) $\times h=\pi R^{2} H$

$\Rightarrow\left(l b-\pi R^{2}\right) \times h=\pi R^{2} H$

$\Rightarrow\left(35 \times 22-\frac{22}{7} \times 7 \times 7\right) \times h=\frac{22}{7} \times 7 \times 7 \times 8$

$\Rightarrow(770-154) \times h=1232$

$\Rightarrow 616 \times h=1232$

$\Rightarrow h=\frac{1232}{616}$

$\therefore h=2 \mathrm{~m}$

So, the rise in the level of the field is 2 m.