# Ex.13.6 Q2 Surface Areas and Volumes Solution - NCERT Maths Class 9

## Question

The inner diameter of a cylindrical wooden pipe is \(24 \;cm\) and its outer diameter is \(28 \;cm\). The length of the pipe is \(35\ cm\). Find the mass of the pipe, if \(\begin{align}1\,\,c{m^3} \end{align}\) of wood has a mass of \(0.6\; g\).

## Text Solution

**Reasoning:**

Since the cylindrical wooden pipe is made up of two concentric circles at the top and bottom, we have to find the volume of both the cylinders.

**Diagram**

**What is known?**

Diameter of the inner and outer cylinder. Length of the cylinder.

**What is unknown?**

Mass of the pipe if \(1 \mathrm{cm}^{3}\) of pipe has mass \(6\; g\).

**Steps:**

Volume of the outer cylinder \(V_{1}=\pi R^{2} h\)

Outer diameter *\((2R) = 28\; cm \)*

Outer radius \(\begin{align}(R)=\frac{28}{2}=14 \mathrm{cm}\end{align}\)

Height \((h) = 35\ cm\)

Volume of outer cylinder

\(\begin{align}V_{1}&=\frac{22}{7} \times 14 \times 14 \times 35\\&=21560 \;\rm{cm}^{3}\end{align}\)

Volume of inner cylinder \(V_{2}=\pi r^{2} h\)

Inner diameter \(\begin{align} \left( {2r} \right) = 24\,cm. \end{align}\)

Inner radius \(\begin{align} (r) = \frac{{24}}{2} = 12\,cm \end{align}\)

Height \((h) = 35\ cm\)

Volume of outer cylinder

\(\begin{align} {V_2} &= \frac{{22}}{7} \times 12 \times 12 \times 35 \\&= 15840\,c{m^3} \end{align}\)

Volume of the wood used \(=\) Outer volume \(-\) Inner volume

\(\begin{align} &= 21560 - 15840 \\ & = 5720\;{\rm{cm}}^{\rm{3}} \\ \end{align}\)

Mass of \(\mathrm{1cm}^{3} \text { pipe }=0.6 \mathrm{g}\)

Therefore,

Mass of \(5720 \,{cm}^{3}\) pipe

\(\begin{align} &={ 0}{\rm{.6 \times 5720}}\\ &= 3432\,\,g\\&\quad\rm OR\\ &= 3.432\,\,kg \end{align}\)

**Answer:** Mass of the pipe \(= 3.432\ kg\).