A solid iron rectangular block of dimension $\left( 2.2m\times 1.2m\times 1m \right)$ is cost into a hollow cylindrical pipe of internal radius 35 cm and thickness 5 cm. Find the length of the pipe.
a. 20.5m
b. 24.5m
c. 22.4m
d. 18.4m
Answer
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Hint: In order to find the solution of this question we need to remember that whenever a solid is transformed into another solid, the volume remains the same. So, we can write that the volume of iron rectangular block = volume of hollow cylindrical pipe, where volume of rectangular block is $l\times b\times h$ and volume of hollow cylindrical is $\pi h\left( r_{1}^{2}-r_{2}^{2} \right)$. By using this concept we can find the answer to this question.
Complete step-by-step answer:
In this question, we have been asked to find the length of the hollow cylinder pipe whose inner radius is 35 cm and thickness is 5 cm which is transformed from a solid rectangular block of dimensions, $\left( 2.2m\times 1.2m\times 1m \right)$.
Now, we know that when a solid is transformed from one shape to another, then the volume remains constant. So, we can say that,
Volume of rectangular block = volume of hollow cylindrical pipe ……… (i)
Now, we know that the volume of the rectangular block is calculated as $l\times b\times h$. We have been given that the length = 2.2m, breadth = 1.2m and height = 1m. So, we can say,
Volume of rectangular block = $2.2m\times 1.2m\times 1m=2.64{{m}^{3}}.........\left( ii \right)$
We also know that the volume of a hollow cylindrical pipe is calculated as, $\pi h\left( r_{1}^{2}-r_{2}^{2} \right)$, where ${{r}_{1}}$ represents the radius of the outer cylinder and ${{r}_{2}}$ represents the radius of the inner cylinder.
Let us consider the l meter as the height of the cylinder. So, we can say that, h = l. We have been given that the internal radius of the cylindrical pipe is 35cmand the thickness is 5cm. So, we can say that the outer radius of the cylinder is 35cm + 5cm = 40cm. Now, we know that 1m = 100cm, so we get $1cm=\dfrac{1}{100}m$. Therefore, we can say that, ${{r}_{1}}=\dfrac{40}{100}=0.4m$ and ${{r}_{2}}=\dfrac{35}{100}=0.35m$.
Hence, we can say that the volume of the cylindrical pipe is,
Volume of hollow cylindrical pipe = $\pi l\left( {{\left( 0.4 \right)}^{2}}-{{\left( 0.35 \right)}^{2}} \right)$
$\begin{align}
& \Rightarrow \pi l\left( 0.16-01225 \right) \\
& \Rightarrow \pi l\left( 0.0375 \right){{m}^{3}}.........\left( iii \right) \\
\end{align}$
Now, from equation (ii) and equation (iii), we will put the values in equation (i). So, we get,
$\begin{align}
& 2.64=\pi l\left( 0.0375 \right) \\
& l=\dfrac{2.64}{\pi \times 0.0375} \\
\end{align}$
Now, we know that $\pi =\dfrac{22}{7}$. So, we get,
$\begin{align}
& l=\dfrac{2.64}{\dfrac{22}{7}\times 0.0375} \\
& l=\dfrac{2.64\times 7}{22\times 0.0375} \\
& l=22.4m \\
\end{align}$
Hence, we can say that the length of the cylindrical pipe is 22.4m.
Therefore, option (c) is the correct answer.
Note: While solving this question, the possible mistake we can make is by not converting the dimension of the cylinder from cm to m, which will give us the wrong answer. We can also think of converting the dimensions of the rectangular block from m to cm, but in the end we want the answer in m. So, it is better to solve the question with the dimensions in m. We can also find the volume of outer and inner cylindrical pipes separately and then subtract them if we don’t remember the formula for the volume of the hollow cylinder.
Complete step-by-step answer:
In this question, we have been asked to find the length of the hollow cylinder pipe whose inner radius is 35 cm and thickness is 5 cm which is transformed from a solid rectangular block of dimensions, $\left( 2.2m\times 1.2m\times 1m \right)$.
Now, we know that when a solid is transformed from one shape to another, then the volume remains constant. So, we can say that,
Volume of rectangular block = volume of hollow cylindrical pipe ……… (i)
Now, we know that the volume of the rectangular block is calculated as $l\times b\times h$. We have been given that the length = 2.2m, breadth = 1.2m and height = 1m. So, we can say,
Volume of rectangular block = $2.2m\times 1.2m\times 1m=2.64{{m}^{3}}.........\left( ii \right)$
We also know that the volume of a hollow cylindrical pipe is calculated as, $\pi h\left( r_{1}^{2}-r_{2}^{2} \right)$, where ${{r}_{1}}$ represents the radius of the outer cylinder and ${{r}_{2}}$ represents the radius of the inner cylinder.
Let us consider the l meter as the height of the cylinder. So, we can say that, h = l. We have been given that the internal radius of the cylindrical pipe is 35cmand the thickness is 5cm. So, we can say that the outer radius of the cylinder is 35cm + 5cm = 40cm. Now, we know that 1m = 100cm, so we get $1cm=\dfrac{1}{100}m$. Therefore, we can say that, ${{r}_{1}}=\dfrac{40}{100}=0.4m$ and ${{r}_{2}}=\dfrac{35}{100}=0.35m$.
Hence, we can say that the volume of the cylindrical pipe is,
Volume of hollow cylindrical pipe = $\pi l\left( {{\left( 0.4 \right)}^{2}}-{{\left( 0.35 \right)}^{2}} \right)$
$\begin{align}
& \Rightarrow \pi l\left( 0.16-01225 \right) \\
& \Rightarrow \pi l\left( 0.0375 \right){{m}^{3}}.........\left( iii \right) \\
\end{align}$
Now, from equation (ii) and equation (iii), we will put the values in equation (i). So, we get,
$\begin{align}
& 2.64=\pi l\left( 0.0375 \right) \\
& l=\dfrac{2.64}{\pi \times 0.0375} \\
\end{align}$
Now, we know that $\pi =\dfrac{22}{7}$. So, we get,
$\begin{align}
& l=\dfrac{2.64}{\dfrac{22}{7}\times 0.0375} \\
& l=\dfrac{2.64\times 7}{22\times 0.0375} \\
& l=22.4m \\
\end{align}$
Hence, we can say that the length of the cylindrical pipe is 22.4m.
Therefore, option (c) is the correct answer.
Note: While solving this question, the possible mistake we can make is by not converting the dimension of the cylinder from cm to m, which will give us the wrong answer. We can also think of converting the dimensions of the rectangular block from m to cm, but in the end we want the answer in m. So, it is better to solve the question with the dimensions in m. We can also find the volume of outer and inner cylindrical pipes separately and then subtract them if we don’t remember the formula for the volume of the hollow cylinder.
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