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.

Volume of wood = π(R² - r²)h, where R = Radius of pencil and r = Radius of graphite.

Radius of pencil = $\dfrac{\text{Diameter of pencil}}{2}$

= $\dfrac{7}{2}$ = 3.5 mm

Radius of graphite = $\dfrac{\text{Diameter of graphite}}{2}$

= $\dfrac{1}{2}$ = 0.5 mm

Given height = 14 cm = 140 mm.

Putting values in formula,

$\Rightarrow \text{ Volume of wood} = \dfrac{22}{7} \times ((3.5)^2 - (0.5)^2) \times 140 \\[1em] = \dfrac{22}{7} \times (12.25 - 0.25) \times 140 \\[1em] = \dfrac{22}{7} \times 12 \times 140 \\[1em] = 22 \times 12 \times 20 \\[1em] = 5280 \text{ mm}^3 \\[1em] = 5280 \times \Big(\dfrac{1}{10}\Big)^3 \text{ cm}^3 \\[1em] = 5.28 \text{ cm}^3.$

Volume of graphite = πr²h.

Putting values we get,

$\text{Volume of graphite } = \dfrac{22}{7}\times (0.5)^2 \times 140 \\[1em] = \dfrac{22 \times 0.25 \times 140}{7} \\[1em] = \dfrac{770}{7} \\[1em] = 110 \text{ mm}^3 \\[1em] = 110 \times \Big(\dfrac{1}{10}\Big)^3 \\[1em] = 0.11 \text{ cm}^3$

Hence, the volume of wood = 5.28 cm³ and volume of graphite = 0.11 cm³.