Answer

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**Hint:**The base of a cylinder is circular, so its circumference will be equal to $2\pi R$. Equating this with the given value of the circumference, we will get the radius of the cylinder. For finding the volume of the cylinder, we have to use the formula for the volume of a cylinder, which is given by $V=\pi {{R}^{2}}h$. Substituting the values of the radius obtained and the given height, we will get the required volume of the cylinder.

**Complete step by step solution:**

Let the radius of the cylinder be $R$. So the given cylinder can be drawn as

We know that a cylinder has a circular shaped base. So the circumference of the base of the cylinder will be equal to the circumference of a circle of radius $R$, which is given by

$\Rightarrow C=2\pi R$

According to the question, the circumference of the cylinder is equal to $6m$. Therefore on substituting $C=6m$ in the above equation, we get

$\begin{align}

& \Rightarrow 6=2\pi R \\

& \Rightarrow R=\dfrac{6}{2\pi }m \\

& \Rightarrow R=\dfrac{3}{\pi }m \\

\end{align}$

Now, we know that the volume of a cylinder is given by

$\Rightarrow V=\pi {{R}^{2}}h$

Substituting the radius of the cylinder from (i) we get

\[\begin{align}

& \Rightarrow V=\pi {{\left( \dfrac{3}{\pi } \right)}^{2}}h \\

& \Rightarrow V=\pi \left( \dfrac{9}{{{\pi }^{2}}} \right)h \\

& \Rightarrow V=\dfrac{9h}{\pi } \\

\end{align}\]

According to the question, the height of the cylinder is equal to $44m$. Therefore, we substitute $h=44m$ in the above equation to get

$\Rightarrow V=\dfrac{9\times 44}{\pi }$

Putting $\pi =\dfrac{22}{7}$ we get

$\begin{align}

& \Rightarrow V=\dfrac{7\times 9\times 44}{22} \\

& \Rightarrow V=126{{m}^{3}} \\

\end{align}$

**Hence, the required volume of the cylinder is equal to $126{{m}^{3}}$.**

**Note:**The actual value of pi is not equal to $\dfrac{22}{7}$, but it is equal to approximately $3.14$. But we took it to $\dfrac{22}{7}$ so as to ease out the calculation. Otherwise $3.14$ in the denominator would have made the calculation for the volume very much complicated.

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