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 mm and the diameter of the graphite is 1 mm.If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
$[Assume\pi =\frac{22}{7}]$

• A. Volume of wood = $=5.28c{m}^3$ Volume of the graphite$=0.11c{m}^3$
• B. Volume of wood = $=9.28c{m}^3$ Volume of the graphite$=0.41c{m}^3$
• C. Volume of wood = $=3.28c{m}^3$ Volume of the graphite$=0.11c{m}^3$
• D. Volume of wood = $=5.28c{m}^3$ Volume of the graphite$=0.19c{m}^3$

Answer 1

The correct option is A

Volume of wood = $=5.28c{m}^3$

Volume of the graphite$=0.11c{m}^3$

Radius $\left(r_1\right)$ of the pencil $=\left(\frac{7}{2}\right)mm=(\frac{0.7}{2}cm=0.35cm)$
Radius $\left(r_2\right)$ of the graphite $=\left(\frac{1}{2}\right)mm=\left(\frac{0.1}{2}\right)cm=0.05cm$
Height (h) of the pencil = 14 cm
Volume of wood in the pencil $=\pi (r_1^2-r_2^2)h$
$=\left[\frac{22}{7}\left\{\right(0.35{)}^2-(0.05{)}^2\times 14\}\right]c{m}^3$
$=\left[\frac{22}{7}\right(0.1225-0.0025)\times 14]c{m}^3$
$=(44\times 0.12)c{m}^3$
$=5.28c{m}^3$
Volume of the graphite $=\pi r_2^{2}h=[\frac{22}{7}\times (0.05{)}^2\times 14]c{m}^3$
$=(44\times 0.0025)c{m}^3$
$=0.11c{m}^3$