Question

The diameter of the metallic sphere is equal to 9 cm. It is melted and drawn into a long wire of diameter of 2 mm having uniform cross – section. Find the length of the wire.

Answer 1

Hint: We will calculate the radius of the sphere and then the volume of the sphere as the volume of sphere will be equal to the volume of the long wire as the sphere is re – casted into the wire. The volume of the sphere will be calculated by the formula: $\dfrac{4}{3}\pi {r^3}$ where r is the radius of the sphere and the volume of the wire will be $\pi {{\text{R}}^2}{\text{L}}$ where R is the radius of the wire and L is the length of the wire. We will equate both the equations and then simplify them to find the length of the wire.

Complete step-by-step answer:
We are told that a metallic sphere is melted and drawn into a long wire.

The diameter of the metallic sphere is 9 cm. Therefore, its radius is $r = \dfrac{{{\text{diameter}}}}{2} = \dfrac{9}{2} = 4.5{\text{cm}}$
Now, the diameter of the wire is 2 mm.
Therefore, the radius R of the wire is $R = \dfrac{{{\text{diameter}}}}{2} = \dfrac{2}{2}{\text{ = 1mm}}$
Converting it into cm using the relation: 1mm = $\dfrac{1}{{10}}{\text{cm}}$ , we get the radius of the wire as:
$ \Rightarrow R = 0.1cm$
Now, we know that the volume of the sphere will be equal to the volume of the wire.
The volume of the sphere is given by $\dfrac{4}{3}\pi {r^3}$and the volume of the wire is given by$\pi {{\text{R}}^2}{\text{L}}$
Therefore, we can write
$ \Rightarrow \dfrac{4}{3}\pi {r^3} = \pi {{\text{R}}^2}{\text{L}}$
Substituting the values of r, R and L, we get
$
   \Rightarrow \dfrac{4}{3}\pi {\left( {4.5} \right)^3} = \pi {\left( {0.1} \right)^2}L \\
   \Rightarrow \dfrac{4}{3}{\left( {4.5} \right)^3} = {\left( {0.1} \right)^2}L \\
   \Rightarrow L = \dfrac{{4{{\left( {4.5} \right)}^3}}}{{3{{\left( {0.1} \right)}^2}}} \\
   \Rightarrow L = \dfrac{{364.5}}{{0.03}} = 12,150{\text{cm}} \\
 $
Therefore, the length of the wire is found to be 12,150 cm.

Note: As the sphere is re-casted into wire so the volume of the sphere is equal to the volume of the wire. You may go wrong with the units of the length for e.g., while using the radius of the wire as it is mm but we are finding the length in cm.