Question

A well with 14 m inside diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment?
(a) 53.3cm
(b) 3.3 cm
(c) 43.3 cm
(d) None of these

Answer 1

Hint: In the problem, the well with inside diameter is dug 8 m deep so the volume taken out with 8 m height is the volume of the cylinder having diameter 14 m and height 8 m. The volume of the cylinder is calculated using the formula which is $ \pi {{r}^{2}}h $ where “r” is the radius and “h” is the height. We have given the diameter so convert it into radius by dividing the diameter of 14 m by 2. Now, this volume of earth is spread all around the well with a width of 21 m with some height so let us assume that the height of the embankment is “h” so find the volume of this embankment by using this formula $ \pi \left( {{\left( {{r}_{o}} \right)}^{2}}-{{\left( {{r}_{i}} \right)}^{2}} \right)h $ where $ {{r}_{0}}\And {{r}_{i}} $ are the outer and inner radius respectively. Outer radius is calculated by adding 7 m and 21 m and inner radius is 7m. Equate this volume of embankment with the volume of the earth that we dug and hence, find the value of height “h”.

Complete step-by-step answer:
It is given that a well with diameter 14 m dug with a depth of 8 m. Dividing diameter by 2 we will get the radius of the well i.e.
 $ \text{Radius}=\dfrac{14}{2}=7m $
In the below diagram, we have shown the top view of the well with inner radius as AD and after digging the embankment width (DE) is 21 m.

In the above diagram, the inner radius of the well is AD which is equal to 7 m and outer radius is given by AE with a length of $ \left( 21+7 \right)m $ .
We are going to find the dug volume of the earth with radius 7 m and height of 8 m by using the formula of volume of cylinder.
Volume of cylinder $ =\pi {{r}^{2}}h $
In the above formula, “r” is the radius and “h” is the height of the cylinder.
Substituting the value of “r” as 7 m and “h” as 8 m in the formula we get,
Volume of the cylinder is equal to:
  $ \begin{align}
  & \dfrac{22}{7}{{\left( 7 \right)}^{2}}\left( 8 \right) \\
 & =22\left( 56 \right) \\
 & =1232{{m}^{2}} \\
\end{align} $
This volume of earth is applied on the well to make the embankment so the volume of embankment is equal to the volume of the dug earth.
Volume of embankment is calculated by subtracting inner volume from the outer volume.
Let us assume the height of the embankment as “h”.
Outer volume is equal to:
 $ \pi {{r}_{o}}^{2}h $
Inner volume is equal to:
 $ \pi {{r}_{i}}^{2}h $
Subtracting inner volume from outer volume we get,
 $ \begin{align}
  & \pi {{r}_{o}}^{2}h-\pi {{r}_{i}}^{2}h \\
 & =\pi h\left( r_{o}^{2}-r_{i}^{2} \right) \\
\end{align} $
In the above expression, $ {{r}_{o}} $ is the outer radius which is equal to 28 m and $ {{r}_{i}} $ is the inner radius which is equal to 7 m so substituting these values in the above expression we get,
 $ \pi h\left( {{\left( 28 \right)}^{2}}-{{\left( 7 \right)}^{2}} \right) $
 $ \begin{align}
  & =\dfrac{22}{7}h\left( 784-49 \right) \\
 & =\dfrac{22}{7}h\left( 735 \right) \\
\end{align} $
In the above expression, 735 is divisible by 7 by 105 times.
 $ \begin{align}
  & 22h\left( 105 \right) \\
 & =2310h \\
\end{align} $
Equating the above volume to volume of the earth dug from the well we get,
 $ \begin{align}
  & 1232=2310h \\
 & \Rightarrow h=\dfrac{1232}{2310} \\
 & \Rightarrow h=0.533m \\
\end{align} $
But the height given in option is in cm so multiplying the above height by 100 to get the value of height in cm.
 $ h=53.3cm $
From the above solution, we have got the height of embankment is 53.3 cm.
Hence, the correct option is (a).

Note: In the above solution, in finding the volumes you can skip the substitution of the value of $\pi $ as $\dfrac{22}{7}$ because when we equate the volume of the dug earth with the volume of embankment then $\pi $ will be cancelled out from both the sides and calculations will be lot more easier than substituting the value of $\pi $ as $\dfrac{22}{7}$.
Volume of dug earth is equal to:
$=\pi {{\left( 7 \right)}^{2}}\left( 8 \right)$
Volume of embankment is equal to:
$\pi h\left( {{\left( 28 \right)}^{2}}-{{\left( 7 \right)}^{2}} \right)$
Equating the above two volumes to get the value of “h” we get,
$\pi h\left( {{\left( 28 \right)}^{2}}-{{\left( 7 \right)}^{2}} \right)=\pi {{\left( 7 \right)}^{2}}\left( 8 \right)$
Now, $\pi $ will be cancelled out from both the sides and we are left with:
$h\left( {{\left( 28 \right)}^{2}}-{{\left( 7 \right)}^{2}} \right)={{\left( 7 \right)}^{2}}\left( 8 \right)$
We can take ${{\left( 7 \right)}^{2}}$ as common from the left hand side of the above equation we get,
$\begin{align}
  & h{{\left( 7 \right)}^{2}}\left( {{\left( 4 \right)}^{2}}-{{\left( 1 \right)}^{2}} \right)={{\left( 7 \right)}^{2}}\left( 8 \right) \\
 & \Rightarrow h\left( 15 \right)=8 \\
 & \Rightarrow h=\dfrac{8}{15}=0.533m \\
\end{align}$
As you can see that we are getting the same value of “h” as we have shown in the above solution but here the calculation is easy as compared to that in the solution part.