The difference between outside and inside surface areas of cylindrical metallic pipe $ 14 \mathrm{~cm} $ long is $ 44 \mathrm{~m}^{2} $. If the pipe is made of $ 99 \mathrm{~cm}^{3} $ of metal, find the outer and inner radii of the pipe.

Given:

The difference between outside and inside surface areas of cylindrical metallic pipe \( 14 \mathrm{~cm} \) long is \( 44 \mathrm{~m}^{2} \).

The pipe is made of \( 99 \mathrm{~cm}^{3} \) of metal.

To do:

We have to find the outer and inner radii of the pipe.

Solution:

Length of the cylindrical metallic pipe $= 14\ cm$
Difference between outside and inside surface areas $= 44\ m^2$

Volume of the pipe $= 99\ cm^3$
Let $R$ and $r$ be the outer and inner radii of the pipe respectively.

This implies,

Outer surface area $-$ inner surface area $= 44\ cm^2$

Therefore,

$2 \pi \mathrm{R} h-2 \pi r h=44$

$\Rightarrow 2 \pi h(\mathrm{R}-r)=44$

$\Rightarrow 2 \times \frac{22}{7} \times 14(\mathrm{R}-r)=44$

$\Rightarrow \mathrm{R}-r=\frac{44 \times 7}{22 \times 14 \times 2}$

$\Rightarrow \mathrm{R}-r=\frac{1}{2}$

$\Rightarrow \mathrm{R}-r=0.5 \mathrm{~cm}$.........(i)

Volume of the metal used $=99 \mathrm{~cm}^{3}$

$\Rightarrow \pi \mathrm{R}^{2} h-\pi r^{2} h=99$

$\Rightarrow \pi h(\mathrm{R}^{2}-r^{2})=99$

$\Rightarrow \frac{22}{7} \times 14(\mathrm{R}+r)(\mathrm{R}-r)=99$

$\Rightarrow 44(\mathrm{R}+r) \times \frac{1}{2}=99$        [From (i)]

$\Rightarrow \mathrm{R}+r=\frac{99 \times 2}{44}$

$\Rightarrow \mathrm{R}+r=\frac{9}{2}$

$\Rightarrow \mathrm{R}+r=4.5$........(ii)

Adding equations (i) and (ii), we get,

$\mathrm{R}+r+\mathrm{R}-r=4.5+0.5$

$2 \mathrm{R}=5$

$\Rightarrow \mathrm{R}=\frac{5}{2}$

$\Rightarrow \mathrm{R}=2.5 \mathrm{~cm}$

Substituting $\mathrm{R}$in (i), we get,

$2.5-r=0.5 \mathrm{~cm}$

$r=2.5-0.5$

$\Rightarrow r=2 \mathrm{~cm}$

The outer and inner radii of the pipe are $2.5\ cm$ and $2\ cm$ respectively.

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Updated on: 10-Oct-2022

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