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# A lead pencil consists of a cylinder of wood with solid cylinder of graphite filled in the interior. The diameter of the pencil is $7mm$and the diameter of the graphite is $1mm$. If the length of the pencil is $14cm$, find the volume of the wood and that of the graphite. (Assume$\pi =\frac{22}{7}$)

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Solution

## Step 1: Write the given dimensionsHeight of wood (and graphite) cylinder$h=14cm=140mm$Radius of pencil $R=\frac{7}{2}mm$Radius of graphite $r=\frac{1}{2}mm$Step 2: Find the volume of the graphiteVolume of Graphite $=\left(\pi {r}^{2}h\right)$ $=\frac{22}{7}×\frac{1}{2}×\frac{1}{2}×140m{m}^{3}\phantom{\rule{0ex}{0ex}}=110m{m}^{3}\phantom{\rule{0ex}{0ex}}=0.11c{m}^{3}$Step 3: Find the volume of the graphiteVolume of wood $=$ Volume of pencil – Volume of graphite $=\left(\pi {R}^{2}h\right)–\left(\pi {r}^{2}h\right)\phantom{\rule{0ex}{0ex}}=\pi \left({R}^{2}–{r}^{2}\right)h\phantom{\rule{0ex}{0ex}}=\frac{22}{7}\left[{\left(\frac{7}{2}\right)}^{2}-{\left(\frac{1}{2}\right)}^{2}\right]x140\phantom{\rule{0ex}{0ex}}=22×20×\left(49/4-1/4\right)\phantom{\rule{0ex}{0ex}}=22×20×12\phantom{\rule{0ex}{0ex}}=5280mm3\phantom{\rule{0ex}{0ex}}=52.80cm3$Hence, the volume of the wood and that of the graphite are $52.80c{m}^{3}$ and $0.11c{m}^{3}$ respectively.  Suggest Corrections  1      Explore more