# 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 $8.88 \mathrm{~g} \mathrm{per} \mathrm{cm}^{3}$.

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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.

Updated on 10-Oct-2022 13:47:38

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