Question

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

Answer 1

Hint: We need to find the relation between the shape of the well and the earth dig out of it. We will get that it is a cylinder. We then find the amount of earth and the volume of the embankment area using the formula for volume as $\pi {{r}^{2}}h$. We assume the height to find the volume and find a linear relation. We solve the equation to find the solution of the problem.

Complete step-by-step answer:
The well is of diameter 3m and is dug 14m deep. The shape of the well is of a cylinder.
The earth taken out of it has been spread evenly all around it to a width of 4m to form an embankment.
The earth in that well acts as the volume of the well. So, the volume of the well is used to make an embankment.
Let us take the dimensions of the well.
The diameter is 3m. This means radius will be half of it. So, radius (r) = $\dfrac{3}{2}$ m.
The height of the well is $h=14$ m.

The formula of volume of a cylindrical shape is $\pi {{r}^{2}}h$.
Now we put the values to find the volume of the well which is the amount of earth.
So, volume being $V=\pi {{\left( \dfrac{3}{2} \right)}^{2}}\left( 14 \right)=\dfrac{9\times 14\times 22}{7\times 4}=99{{m}^{3}}$.
This volume of earth has been used to form an embankment of 4m.
The area of the embankment is made of the same earth.
We find the area of the embankment which makes a hollow ring. It will have two radius, inner and outer due to the width of the embankment. We need to find the height of the embankment. Let’s assume the height as h.
The inner radius is $r=\dfrac{3}{2}$ m. and the outer radius will be \[R=r+4=\dfrac{3}{2}+4=\dfrac{11}{2}\] m.
The formula for area will be $\pi \left( {{R}^{2}}-{{r}^{2}} \right)h$.

We put the values to get the amount as $\dfrac{22}{7}\times \left\{ {{\left( \dfrac{11}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}} \right\}h$.
The amount of earth being same we can equate them and get $\dfrac{22}{7}\times \left\{ {{\left( \dfrac{11}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}} \right\}h=99$.
We solve the linear equation of h to find
\[\begin{align}
  & \dfrac{22}{7}\times \left\{ {{\left( \dfrac{11}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}} \right\}h=99 \\
 & \Rightarrow h\left\{ \left( \dfrac{121}{4} \right)-\left( \dfrac{9}{4} \right) \right\}=\dfrac{99\times 7}{22} \\
 & \Rightarrow 28h=\dfrac{99\times 7}{22} \\
 & \therefore h=\dfrac{99\times 7}{22\times 28}=\dfrac{9}{8} \\
\end{align}\]
The height of the embankment is \[\dfrac{9}{8}m\].

Note: We need to remember that due to the width the ring-shaped form will be created in the embankment. Sometimes students make the mistake of considering the embankment to be in the shape of a rectangle. The amount of earth creates volume in both cases. So, the relation has to create a linear equation to solve the problem.