A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is $10.4\ cm$ and its length is $25\ cm$. The thickness of the metal is $8\ mm$ everywhere. Calculate the volume of the metal.

 Given:

A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is $10.4\ cm$ and its length is $25\ cm$. The thickness of the metal is $8\ mm$ everywhere. 

To do:

We have to find the volume of the metal.

Solution:

Length of the metallic tube $= 25\ cm$

Inner diameter $= 10.4\ cm$

This implies,

Radius $(r) =\frac{10.4}{2}$

$= 5.2\ cm$

Thickness of the metal $= 8\ mm$

This implies,

Outer radius $(R) = 5.2 + 0.8$

$= 6.0\ cm$

Volume of the metal used $= \pi (R^2 - r^2) \times h$

$=\frac{22}{7}(6^{2}-5.2^{2}) \times 25$

$=\frac{22}{7}(36-27.04) \times 25$

$=\frac{22}{7} \times 8.96 \times 25$

$=704 \mathrm{~cm}^{3}$

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Updated on: 10-Oct-2022

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