A solid cuboid of iron with dimensions $33\ cm \times 40\ cm \times 15\ cm$ is melted and recast into a cylindrical pipe. The outer and inner diameters of pipe are $8\ cm$ and $7\ cm$ respectively. Find the length of pipe.

Given:

A solid cuboid of iron with dimensions $33\ cm \times 40\ cm \times 15\ cm$ is melted and recast into a cylindrical pipe. The outer and inner diameters of pipe are $8\ cm$ and $7\ cm$ respectively.

To do:

We have to find the length of pipe.

Solution:

Volume of the cuboid $=lbh$

$= 33 \times 40 \times 15\ cm^3$
$= 19800\ cm^3$

Volume of cylindrical pipe $=$ Volume of the cuboid

$=19800\ cm^3$

Inner diameter of the pipe $= 7\ cm$

Outer diameter of the pipe $= 8\ cm$

Therefore,

Outer radius of the pipe $R=\frac{8}{2}$

$=4\ cm$

Inner radius of the pipe $r=\frac{7}{2} \mathrm{~cm}$

Let $h$ be the length of the pipe.

$\Rightarrow \pi \mathrm{R}^{2} h-\pi r^{2} h=19800$

$\Rightarrow \frac{22}{7} h(4^{2}-\frac{7}{2})^{2}=19800$

$\Rightarrow \frac{22}{7} h(16-\frac{49}{4})=19800$

$\Rightarrow \frac{22}{7} h \times \frac{64-49}{4}=19800$

$\Rightarrow \frac{22}{7} \times \frac{15}{4} h=19800$

$\Rightarrow h=19800 \times \frac{7}{22} \times \frac{4}{15}$

$\Rightarrow h=\frac{554400}{330} \mathrm{~cm}$

$\Rightarrow h=1680 \mathrm{~cm}$

The length of pipe is $1680\ cm$.

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Updated on: 10-Oct-2022

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