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, and the volume of the wood and that of the graphite.

Given :
Radius (R) of pencil = \(\frac{7}{2} mm = 3.5 mm = 0.35 cm\)
Radius (r) of graphite = \(\frac{1}{2} mm = 0.5 mm = 0.05 cm\)
Height (h) of pencil = 14 cm

We know that,

Volume of wood in pencil
= \({\pi} ({R}^2 - {r}^2) h\)
= \(\frac{22}{7} × ({0.35}^2 - {0.05}^2) × 14\) \({cm}^3\)
= \(\frac{22}{7} × (0.1225 - 0.0025) × 14\) \({cm}^3\)
= 44 × 0.12 \({cm}^3\)
= 5.28 \({cm}^3\)

Similarly, Volume of graphite in pencil
= \({\pi} {r}^2 h\)
= \(\frac{22}{7} × {0.05}^2 × 14\) \({cm}^3\)
= 44 × 0.0025 \({cm}^3\)
= 0.11 \({cm}^3\)