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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 $1.4 cm$ and the diameter of the graphite is $0.2 cm$ . If the length of the pencil is $28 cm$ , find the volume of the wood and that of the graphite.

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44.28 {cm}^{2} and 0.24 {cm}^{2}

40.28 {cm}^{2} and 0.34 {cm}^{2}

42.28 {cm}^{2} and 0.44 {cm}^{2}

40.8 {cm}^{2} and 0.54 {cm}^{2}

answer is C.

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Detailed Solution

Given that, diameter of pencil is 1.4cm, diameter of graphite is 0.2cm, length of pencil is 28cm.
Volume of cylinder

= π r^{2} h

The radius of the pencil is

\frac{1.4}{2} = 0.7 cm

The radius of the graphite is

\frac{0.2}{2} = 0.1 cm

The height of the pencil

= 28 cm

The volume of graphite

= π r_{g}^{2} h

where,

r_{g}

is the radius of graphite.
The volume of pencil

= π r_{p}^{2} h

where,

r_{p}

is the radius of pencil.
The volume of the wood in the pencil

= π (r_{p}^{2} - r_{g}^{2}) h

\Rightarrow \frac{22}{7} ({0.7}^{2} - {0.1}^{2}) 28

\Rightarrow \frac{22}{7} \times 0.48 \times 28

\Rightarrow 42.28 {cm}^{3}

The volume of graphite

= π r_{gh}^{2}

\Rightarrow \frac{22}{7} \times {(0.1)}^{2} \times 14

\Rightarrow 0.44 {cm}^{2}

Thus, the volume of the wood in the pencil is

42.28 {cm}^{2}

and the volume of the graphite is

0.44 {cm}^{2}

.
Therefore, option (3) is correct.

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