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A copper wire, $ 3 \mathrm{~mm} $ in diameter, is wound about a cylinder whose length is $ 12 \mathrm{~cm} $, and diameter $ 10 \mathrm{~cm} $, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be $ 8.88 \mathrm{~g} \mathrm{per} \mathrm{cm}^{3} $.
Given:
A copper wire, \( 3 \mathrm{~mm} \) in diameter, is wound about a cylinder whose length is \( 12 \mathrm{~cm} \), and diameter \( 10 \mathrm{~cm} \), so as to cover the curved surface of the cylinder.
The density of copper is \( 8.88 \mathrm{~g} \mathrm{per} \mathrm{cm}^{3} \).
To do:
We have to find the length and mass of the wire.
Solution:
Diameter of the cylinder $(d)= 10\ cm$
This implies,
Radius of the cylinder $(r) = \frac{10}{2}\ cm$
$= 5\ cm$
Length of the wire one complete round $= 2 \pi r$
$= 2\times3.14\times5\ cm$
$= 31.4\ cm$
The diameter of the wire $= 3\ mm$
$= \frac{3}{10}\ cm$
This implies,
Radius of the wire $= \frac{3}{10}\times\frac{1}{2}$
$= 0.15\ cm$
The thickness(height) of the cylinder covered in one complete round $= \frac{3}{10}\ cm$
Therefore,
The number of rounds of the wire required to cover the height of the cylinder $=\frac{12}{\frac{3}{10}}$
$=4\times10$
$=40$
The length of the wire required to cover the surface of the cylinder $=$ The length of the wire required to complete 40 rounds
$=40 \times 31.4\ cm$
$= 1256\ cm$
Volume of the wire required $=$ Area of the wire $\times$ Length of the wire
$= \pi (0.15)^2\times1256$
$=3.14\times0.0225\times1256$
$=88.7364\ cm^3$
We know that,
Mass $=$ Volume $\times$ Density
$= 88.7364\times8.88$
$= 787.98\ g$
The length of the wire is $1256\ cm$ and the mass of the wire is $787.98$ grams.