
A cubic cm of gold is drawn into a wire $0.1mm$ in diameter, find the length of the wire.
Answer
594.9k+ views
Hint: Cubic cm of gold is $1c{m^3}$ then, it is drawn into wire. It means the gold object is changing its shape. While changing the shape or drawing into the wire volume remains unchanged. Using this logic we will solve the given question.
Complete step by step answer:
First of all we are given gold whose volume is $1c{m^3}$.
Now it is saying that the $1c{m^3}$gold is drawn into a wire where diameter is given which is equal to $0.1mm$
We have to find the length of the wire.
So as we know that full volume of gold gets converted into wire. It means the volume of gold is equal to the volume of wire.
Now you are wondering if we know the volume of gold is $1c{m^3}$ but what about wire.
So let me clear, wire is a cylindrical shape. And we know that the volume of cylinder is $\pi {R^2}H$
Where $R$ is the radius of the wire and $H$ is the length of wire.
Now according to question,
Diameter of wire is given and that is equal to $0.1mm$
And we know the relationship between diameter and radius.
radius$ = $diameter$ \div 2$
so radius of wire is
$ = \dfrac{{0.1mm}}{2} = 0.05mm$
And that we know that $1cm = 10mm$
So we can say that $1mm = \dfrac{1}{{10}}cm$
$
\Rightarrow 0.05mm = \dfrac{{0.05}}{2}cm \\
= 0.005cm \\
$
Now lets equate the volume.
Volume of gold$ = $volume of wire
$
1c{m^3} = \pi {R^2}H \\
\Rightarrow 1c{m^3} = \dfrac{{22}}{7} \times {\left( {0.005} \right)^2}c{m^2} \times H \\
\Rightarrow H = \dfrac{7}{{22}} \times \dfrac{1}{{{{\left( {0.005} \right)}^2}}}cm \\
\text{On simplifying the above equation,}
\Rightarrow H = \dfrac{7}{{22}} \times \dfrac{1}{{25 \times {{10}^{ - 6}}}}cm \\
\Rightarrow H = \dfrac{7}{{22}} \times \dfrac{{{{10}^6}}}{{25}}cm \\
$
And we know
$1m = 100cm$
So, $1cm = {10^{ - 2}}m$
$
H = \dfrac{7}{{22}} \times \dfrac{{{{10}^6}}}{{25}} \times {10^{ - 2}}m \\
H = \dfrac{7}{{22}} \times \dfrac{{{{10}^4}}}{{25}}m \\
\text{On simplifying the above equation,} \\
= \dfrac{7}{{550}} \times 10000m \\
H = 127.27m $
$\therefore$ The length of wire is $127.27m$.
Note:Whenever we draw the volume into a wire, then both must have the same volume. Because we are not changing the volume but only the shape of the object. While substituting values and doing calculations, we need to take proper care.
Complete step by step answer:
First of all we are given gold whose volume is $1c{m^3}$.
Now it is saying that the $1c{m^3}$gold is drawn into a wire where diameter is given which is equal to $0.1mm$
We have to find the length of the wire.
So as we know that full volume of gold gets converted into wire. It means the volume of gold is equal to the volume of wire.
Now you are wondering if we know the volume of gold is $1c{m^3}$ but what about wire.
So let me clear, wire is a cylindrical shape. And we know that the volume of cylinder is $\pi {R^2}H$
Where $R$ is the radius of the wire and $H$ is the length of wire.
Now according to question,
Diameter of wire is given and that is equal to $0.1mm$
And we know the relationship between diameter and radius.
radius$ = $diameter$ \div 2$
so radius of wire is
$ = \dfrac{{0.1mm}}{2} = 0.05mm$
And that we know that $1cm = 10mm$
So we can say that $1mm = \dfrac{1}{{10}}cm$
$
\Rightarrow 0.05mm = \dfrac{{0.05}}{2}cm \\
= 0.005cm \\
$
Now lets equate the volume.
Volume of gold$ = $volume of wire
$
1c{m^3} = \pi {R^2}H \\
\Rightarrow 1c{m^3} = \dfrac{{22}}{7} \times {\left( {0.005} \right)^2}c{m^2} \times H \\
\Rightarrow H = \dfrac{7}{{22}} \times \dfrac{1}{{{{\left( {0.005} \right)}^2}}}cm \\
\text{On simplifying the above equation,}
\Rightarrow H = \dfrac{7}{{22}} \times \dfrac{1}{{25 \times {{10}^{ - 6}}}}cm \\
\Rightarrow H = \dfrac{7}{{22}} \times \dfrac{{{{10}^6}}}{{25}}cm \\
$
And we know
$1m = 100cm$
So, $1cm = {10^{ - 2}}m$
$
H = \dfrac{7}{{22}} \times \dfrac{{{{10}^6}}}{{25}} \times {10^{ - 2}}m \\
H = \dfrac{7}{{22}} \times \dfrac{{{{10}^4}}}{{25}}m \\
\text{On simplifying the above equation,} \\
= \dfrac{7}{{550}} \times 10000m \\
H = 127.27m $
$\therefore$ The length of wire is $127.27m$.
Note:Whenever we draw the volume into a wire, then both must have the same volume. Because we are not changing the volume but only the shape of the object. While substituting values and doing calculations, we need to take proper care.
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