# 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.

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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

$=(5.39-0.11) \mathrm{cm}^{3}$

$=5.28 \mathrm{~cm}^{3}$

Updated on 10-Oct-2022 13:46:38