Question

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 .​

Answer 1

VOLUME:

The space occupied by an object solid body is called the volume of the particular object solid body volume is always measured in cubic unit.

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Solution:

Given:

Diameter of the pencil =7mm

Radius of the pencil (R) = 7/2mm

Diameter of the graphite cylinder =1mm

Radius of the graphite(r) = 1/2mm

Height(h)=14cm= 140mm

[1cm= 10 mm]

Volume of a cylinder= πr²h

Volume of graphite cylinder = πr2h = 22/7 ×1/2 × 1/2 ×140

Volume of graphite = 110 mm³

Volume of pencil = πR²h

= 22/7 ×7/2 × 7/2 ×140

= 490×11= 5390 mm²

Volume of pencil=5390mm³

Volume of wood = Volume of pencil - Volume of graphite

Volume of wood= 5390- 110= 5280 mm³

= 5280/1000= 5.28 cm³

[1mm³= 1/1000cm³]

Volume of wood = 5.28 cm³

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Hope this will help you...

Answer 2

Given :

• Radius of pencil $\tt{ = \frac{7}{2} \: mm = \frac{0.7}{2} \: cm arrow0.35 \: cm}$

• Radius of graphite $\tt{ = \frac{1}{2} \: mm = \frac{0.1}{2} \: cm arrow0.05 \: cm}$

• Height of pencil = 14 cm

___________

$\sf \gray{Volume \: of \: wood \: in \: pencil = \pi( {r}^{2} _{1} - {r}^{2} _{2} )h}$

$\sf{ = \frac{22}{7} \times ( {0.35)}^{2} - ( {0.05)}^{2} \times 14 \: cm {}^{3} }$

$\sf{ = \frac{22}{7} \times 0.1225 \times 0.0025 \times 14 \: cm {}^{3} }$

$\sf{ =44 \times 0.12 } \: cm {}^{3}$

$\sf \boxed {\sf \red{ = 5.28 \: cm {}^{3} }}$

___________

$\therefore \sf \gray{Volume \: of \: graphite = \pi {r}^{2} h}$

$\sf{ = \frac{22}{7} \times {(0.05)}^{2} \times 14 \: cm {}^{3} }$

$\sf{ = 44 \times 0.0025 \: cm {}^{3} }$

$\boxed{\sf \red{ = 0.11\: cm {}^{3}}}$