Question

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled into it. The diameter of the pencil is $7mm$, the diameter of the graphite is $1mm$ and the length of the pencil is $10cm$. Calculate the weight of the whole pencil, if the specific gravity of the wood is $0.7gm/{cm}^3$ and that of the graphite is $2.1gm/{cm}^3$.

Answer 1

We have
Diameter of the graphite cylinder $=1mm=\frac{1}{10}cm$

Radius $=\frac{1}{20}cm$

Length of the graphite cylinder $=10cm$

Volume of the graphite cylinder $=(\frac{22}{7}\times \frac{1}{20}\times \frac{1}{20}\times 10){cm}^3$

Weight of graphite = Volume $\times$ Specific gravity

Weight of graphite $=(\frac{22}{7}\times \frac{1}{20}\times \frac{1}{20}\times 10\times 2.1)gm$

$=(\frac{22}{7}\times \frac{1}{20}\times \frac{1}{20}\times 10\times \frac{21}{10})gm=0.165gm$

Diameter of pencil $=7mm=\frac{7}{10}cm$

Radius of the pencil $=\frac{7}{20}cm$ and length of pencil $=10cm$

Volume of pencil $=\frac{22}{7}\times \frac{7}{20}\times \frac{7}{20}\times 10{cm}^3$

Volume of wood $=(\frac{22}{7}\times \frac{7}{20}\times \frac{7}{20}\times 10-\frac{22}{7}\times \frac{1}{20}\times \frac{1}{20}\times 10){cm}^3$

Volume of wood $=\frac{22}{7}\times \frac{1}{20}\times \frac{1}{20}\times 10(7\times 7-1){cm}^3=\frac{11}{7}\times \frac{1}{20}\times 48{cm}^3$

Weight of wood $=(\frac{11}{7}\times \frac{1}{20}\times 48\times 0.7)gm$

$=(\frac{11}{7}\times \frac{1}{20}\times 48\times \frac{7}{10})gm=2.64gm$

Total weight $=(2.64+0.165)gm=2.805gm$