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 \mathrm{~mm} $ and the diameter of the graphite is $ 1 \mathrm{~mm} $. If the length of the pencil is $ 14 \mathrm{~cm} $, find the volume of the wood and that of the graphite.
Given:
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 \( 7 \mathrm{~mm} \) and the diameter of the graphite is \( 1 \mathrm{~mm} \).
The length of the pencil is \( 14 \mathrm{~cm} \).
To do:
We have to find the volume of the wood and that of the graphite.
Solution:
Diameter of the graphite cylinder $=1 \mathrm{~mm}$
$=\frac{1}{10} \mathrm{~cm}$
This implies,
Radius of the graphite $r=\frac{\frac{1}{10}}{2} \mathrm{~cm}$
$r=\frac{1}{20} \mathrm{~cm}$
Length of the graphite $h=14 \mathrm{~cm}$
Therefore,
Volume of the graphite cylinder $=\pi r^{2} h$
$=\frac{22}{7} \times (\frac{1}{20})^2 \times 14 \mathrm{cm}^{3}$
$=0.11 \mathrm{~cm}^{3}$
Diameter of the pencil $=7 \mathrm{~mm}$
$=\frac{7}{10} \mathrm{~cm}$
This implies,
Radius of the pencil $r=\frac{\frac{7}{10}}{2} \mathrm{~cm}$
$r=\frac{7}{20} \mathrm{~cm}$
Length of the pencil $h=14 \mathrm{~cm}$
Therefore,
Volume of the pencil $=\pi r^{2} h$
$=\frac{22}{7} \times (\frac{7}{20})^2 \times 14 \mathrm{cm}^{3}$
$=5.39 \mathrm{~cm}^{3}$
This implies,
Volume of the wood $=$ Volume of the pencil $-$ Volume of the graphite
$=(5.39-0.11) \mathrm{cm}^{3}$
$=5.28 \mathrm{~cm}^{3}$
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