Answer
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Hint: In this question, we will first find volume of well and volume of embankment. Then equate both the volumes and solve the equation to find the value of height of embankment.
Complete step-by-step answer:
We know that formula for volume of cylinder, $V$, is given by,
$V=\pi {{r}^{2}}h\cdots \cdots \left( i \right)$.
Where, $r$ is the radius of the cylinder and $h$ is the height of the cylinder.
Now, for given well, which is of the form of cylinder, we have,
$r=\dfrac{\text{diameter}}{2}=\dfrac{2\text{m}}{2}=1\text{m}$ and $h=14\text{m}$.
Therefore, using equation$\left( i \right)$
The volume of earth taken out of well
=The volume of well
$=\pi {{r}^{2}}h$
Putting values of $r$ and $h$, we get,
$=\left( \dfrac{22}{7}\times {{1}^{2}}\times 14 \right){{\text{m}}^{\text{3}}}$ m
Diving 14 by 7, we get,
$=22\times 2{{\text{m}}^{\text{3}}}$
$=44{{\text{m}}^{\text{3}}}\cdots \cdots \left( ii \right)$
Now, embankment around the well is of the form of the cylinder out of which another cylinder of smaller radius, that is the radius of well and same height is removed. Also, for embankment, the radius of the bigger cylinder is 5m more than that of well as given in question. Let the radius of the bigger circle be $R$. Then, $R=\left( 5+r \right)\text{m}=\left( 5+1 \right)\text{m}=6\text{m}$
And let the height of the embankment be $x$m.
Therefore, using equation$\left( i \right)$, we have,
Volume of embankment
$=\pi {{R}^{2}}x-\pi {{r}^{2}}x$
Taking $\pi x$ common, we can write,
$=\pi x\left( {{R}^{2}}-{{r}^{2}} \right)$
Putting values of $R$ and $r$, we have,
=$\begin{align}
& =\dfrac{22}{7}x\left( {{6}^{2}}-{{1}^{2}} \right){{\text{m}}^{\text{3}}} \\
& =\dfrac{22}{7}x\left( 36-1 \right){{\text{m}}^{\text{3}}} \\
& =\dfrac{22}{7}x\times 35\,{{\text{m}}^{\text{3}}} \\
\end{align}$
Dividing 35 by 7, we get,
$\begin{align}
& =22\times 5x\,{{\text{m}}^{\text{3}}} \\
& =110x\,{{\text{m}}^{\text{3}}}\cdots \cdots \left( iii \right) \\
\end{align}$
Also, according to the question, total earth taken out of the well is used to form an embankment around it.
Therefore,
Volume of well = Volume of embankment.
Using equation $\left( ii \right)$ and $\left( iii \right)$ here, we get,
$\begin{align}
& 44{{\text{m}}^{\text{3}}}=110x\,{{\text{m}}^{3}} \\
& \Rightarrow 44=110x \\
\end{align}$
Diving both side of the equation by 110, we get,
$x=\dfrac{44}{110}=\dfrac{2}{5}=0.4$
Hence, the height of the embankment is 0.4 m.
Therefore, the correct answer is option (c).
Note: While finding volume of the embankment, keep in mind to subtract the cylinder of smaller volume. And, also that, not whole well will be subtracted but only the part of well of height $x$.
Complete step-by-step answer:
We know that formula for volume of cylinder, $V$, is given by,
$V=\pi {{r}^{2}}h\cdots \cdots \left( i \right)$.
Where, $r$ is the radius of the cylinder and $h$ is the height of the cylinder.
Now, for given well, which is of the form of cylinder, we have,
$r=\dfrac{\text{diameter}}{2}=\dfrac{2\text{m}}{2}=1\text{m}$ and $h=14\text{m}$.
Therefore, using equation$\left( i \right)$
The volume of earth taken out of well
=The volume of well
$=\pi {{r}^{2}}h$
Putting values of $r$ and $h$, we get,
$=\left( \dfrac{22}{7}\times {{1}^{2}}\times 14 \right){{\text{m}}^{\text{3}}}$ m
Diving 14 by 7, we get,
$=22\times 2{{\text{m}}^{\text{3}}}$
$=44{{\text{m}}^{\text{3}}}\cdots \cdots \left( ii \right)$
Now, embankment around the well is of the form of the cylinder out of which another cylinder of smaller radius, that is the radius of well and same height is removed. Also, for embankment, the radius of the bigger cylinder is 5m more than that of well as given in question. Let the radius of the bigger circle be $R$. Then, $R=\left( 5+r \right)\text{m}=\left( 5+1 \right)\text{m}=6\text{m}$
And let the height of the embankment be $x$m.
Therefore, using equation$\left( i \right)$, we have,
Volume of embankment
$=\pi {{R}^{2}}x-\pi {{r}^{2}}x$
Taking $\pi x$ common, we can write,
$=\pi x\left( {{R}^{2}}-{{r}^{2}} \right)$
Putting values of $R$ and $r$, we have,
=$\begin{align}
& =\dfrac{22}{7}x\left( {{6}^{2}}-{{1}^{2}} \right){{\text{m}}^{\text{3}}} \\
& =\dfrac{22}{7}x\left( 36-1 \right){{\text{m}}^{\text{3}}} \\
& =\dfrac{22}{7}x\times 35\,{{\text{m}}^{\text{3}}} \\
\end{align}$
Dividing 35 by 7, we get,
$\begin{align}
& =22\times 5x\,{{\text{m}}^{\text{3}}} \\
& =110x\,{{\text{m}}^{\text{3}}}\cdots \cdots \left( iii \right) \\
\end{align}$
Also, according to the question, total earth taken out of the well is used to form an embankment around it.
Therefore,
Volume of well = Volume of embankment.
Using equation $\left( ii \right)$ and $\left( iii \right)$ here, we get,
$\begin{align}
& 44{{\text{m}}^{\text{3}}}=110x\,{{\text{m}}^{3}} \\
& \Rightarrow 44=110x \\
\end{align}$
Diving both side of the equation by 110, we get,
$x=\dfrac{44}{110}=\dfrac{2}{5}=0.4$
Hence, the height of the embankment is 0.4 m.
Therefore, the correct answer is option (c).
Note: While finding volume of the embankment, keep in mind to subtract the cylinder of smaller volume. And, also that, not whole well will be subtracted but only the part of well of height $x$.
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