Ex.13.6 Q7 Surface Areas and Volumes Solution - NCERT Maths Class 9

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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 \(7 \; \rm mm\) and the diameter of the graphite is \(1\; \rm mm\). If the length of the pencil is \(14 \; \rm cm\), find the volume of the wood and that of the graphite.

 Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-6 | Question 7

Text Solution

Reasoning:

Volume of cylinder \( \pi {r^2}h \)

What is known?

Diameter of the pencil, diameter of graphite and length of the pencil.

What is unknown?

Volume of the wood and that if graphite.

Steps:

For cylinder graphite.

Diameter \((2r) = 1 \; \rm mm\)

Radius \(\begin{align}(r) = \frac{1}{2}\,\, \rm mm \end{align}\)

Length of the pencil \((h) = 14 \; \rm cm = 140 \;\rm mm\)

\(\begin{align}\text{Capacity}&= \text{Volume} \\ &=\,\pi {r^2}h\\ &= \frac{{22}}{7} \times \frac{1}{2} \times \frac{1}{2} \times 140\\ &= 110\,\, \rm mm^3 \end{align}\)

\(\begin{align}  \rm{In} \,\,c{m^3} &= \frac{{110}}{{10 \times 10 \times 10}} = 0.11\,\, \rm cm^3  \end{align}\)

For cylinder of wood:

To find the volume of the wood:

Total volume of the pencil \(-\) Volume of graphite

\(\begin{align}\pi {R^2}h - \pi {r^2}h \pi h({R^2} - {r^2})\end{align}\)

Diameter of pencil \((2R) = 7 \; \rm mm\)

Radius \((r) = \frac{7}{2} \rm mm \)

Length of the pencil \((h) = 14 \; \rm cm = 140 \; \rm mm\)

Volume of wood \(\begin{align}=\,\pi h({R^2} - {r^2}) \end{align}\)

\[\begin{align} &= \frac{{22}}{7} \times 140 \times [{(\frac{7}{2})^2} - {(\frac{1}{2})^2}]\\ &= \frac{{22}}{7} \times 140 \times [\frac{{49}}{4} - \frac{1}{2}]\\ &= 5280\,\, \rm mm^3\\ &= \frac{{5280}}{{10 \times 10 \times 10}}\, \rm cm^3 \\ &= 5.28\,\, \rm cm^3 \end{align}\]

Answer:

Volume of the wood \( = 5.28\,\, \rm cm^3 \)

Volume of graphite \( = 0.11\,\,\, \rm cm^3 \)

  
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