Question

An agriculture field is in the form of a rectangle of length $20m$ width $14m$. A $10m$ deep well of diameter $7m$ is dug in a corner of the field and the earth taken out of the well is spread evenly over the remaining part of the field. Find the rise in its bed.

Answer 1

We have
Radius of the well $=\frac{7}{2}m$
Depth of the well $=10m$
Volume of the earth dug $=\pi {\left(\frac{7}{2}\right)}^2\times 10m^3=\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}\times 10m^3=358m^3$
Also we have
Length of the field$=14m$
breadth of the field $=14m$
Area of the field $=20\times 14m^2=280m^2$

Area of the base of the well $=\pi {\left(\frac{7}{2}\right)}^2{m}^2=\frac{22}{7}\times \frac{7}{2}\times \frac{7}{2}{m}^2=\frac{77}{2}{m}^2$

Area of the remaining part of the field = Area of the field - Area of the base of the field

$=(280-\frac{77}{2})m^2=\frac{483}{2}{m}^3$

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

Volume of the raised field = Area of the base $\times$ height $=(\frac{483}{2}\times h)m^3$

But volume of the raised field = Volume of the earth dugout

$\frac{483}{2}\times h=385\Rightarrow h=\frac{2\times 385}{483}=1.594m$

Rise in the level of the field$=1.594m$