Question

A lead pencil consists of a cylinder of wood with a solid cylinder of graphite fitted into it. The diameter of the pencil is 7 mm, the diameter of the graphite is 1 mm and the length of the pencil is 10 cm. Calculate the weight of the whose pencil, if the specific gravity of the wood is 0.7 g/cm² and that of the graphite is 2.1 g/cm².

Answer 1

Here, r = $\frac{7}{2}\mathrm{mm}=0.35\mathrm{cm}$; h = 10 cm, where r and h are the radius and height of the pencil, respectively.

$$\begin{gathered} \mathrm{Volume}\mathrm{of}\mathrm{graphite}=\mathrm{\pi }{r}^{\mathit{2}}h \\ =\frac{22}{7}\times 0.{05}^2\times 10=\frac{0.55}{7}{\mathrm{cm}}^3 \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{Weight}\mathrm{of}\mathrm{graphite}=\mathrm{volume}\times \mathrm{specfic}\mathrm{gravity}\mathrm{of}\mathrm{graphite} \\ =\frac{0.55}{7}\times 2.1=0.165\mathrm{g} \end{gathered}$$

Similarly, weight of the wood = volume of the wood $\times$specific gravity of wood
$$\begin{gathered} \mathrm{Now},\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{wood}=\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{pencil}-\mathrm{volume}\mathrm{of}\mathrm{graphite} \\ =\frac{22}{7}\times \left[{\left(0.35\right)}^2-{\left(0.05\right)}^2\right]\times 10\times 0.7 \\ =\frac{22}{7}\times \left(0.1225-0.0025\right)\times 7 \\ =22\times 0.12=2.64\mathrm{g} \end{gathered}$$

∴ Total weight of the pencil = weight of the wood + weight of graphite
= (0.165 + 2.64) g = 2.805 g