# Ex.13.6 Q7 Surface Areas and Volumes Solution - NCERT Maths Class 9

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## Question

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is $$7 \; \rm mm$$ and the diameter of the graphite is $$1\; \rm mm$$. If the length of the pencil is $$14 \; \rm cm$$, find the volume of the wood and that of the graphite.

Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-6 | Question 7

## Text Solution

Reasoning:

Volume of cylinder $$\pi {r^2}h$$

What is known?

Diameter of the pencil, diameter of graphite and length of the pencil.

What is unknown?

Volume of the wood and that if graphite.

Steps:

For cylinder graphite.

Diameter $$(2r) = 1 \; \rm mm$$

Radius \begin{align}(r) = \frac{1}{2}\,\, \rm mm \end{align}

Length of the pencil $$(h) = 14 \; \rm cm = 140 \;\rm mm$$

\begin{align}\text{Capacity}&= \text{Volume} \\ &=\,\pi {r^2}h\\ &= \frac{{22}}{7} \times \frac{1}{2} \times \frac{1}{2} \times 140\\ &= 110\,\, \rm mm^3 \end{align}

\begin{align} \rm{In} \,\,c{m^3} &= \frac{{110}}{{10 \times 10 \times 10}} = 0.11\,\, \rm cm^3 \end{align}

For cylinder of wood:

To find the volume of the wood:

Total volume of the pencil $$-$$ Volume of graphite

\begin{align}\pi {R^2}h - \pi {r^2}h \pi h({R^2} - {r^2})\end{align}

Diameter of pencil $$(2R) = 7 \; \rm mm$$

Radius $$(r) = \frac{7}{2} \rm mm$$

Length of the pencil $$(h) = 14 \; \rm cm = 140 \; \rm mm$$

Volume of wood \begin{align}=\,\pi h({R^2} - {r^2}) \end{align}

\begin{align} &= \frac{{22}}{7} \times 140 \times [{(\frac{7}{2})^2} - {(\frac{1}{2})^2}]\\ &= \frac{{22}}{7} \times 140 \times [\frac{{49}}{4} - \frac{1}{2}]\\ &= 5280\,\, \rm mm^3\\ &= \frac{{5280}}{{10 \times 10 \times 10}}\, \rm cm^3 \\ &= 5.28\,\, \rm cm^3 \end{align}

Volume of the wood $$= 5.28\,\, \rm cm^3$$
Volume of graphite $$= 0.11\,\,\, \rm cm^3$$