The diameter of a metallic sphere is equal to $ 9 \mathrm{~cm} $. It is melted and drawn into a long wire of diameter $ 2 \mathrm{~mm} $ having uniform cross-section. Find the length of the wire.

Given:

The diameter of a metallic sphere is equal to \( 9 \mathrm{~cm} \). It is melted and drawn into a long wire of diameter \( 2 \mathrm{~mm} \) having uniform cross-section.

To do:

We have to find the length of the wire.

Solution:

Diameter of the metallic sphere $=9 \mathrm{~cm}$

This implies,

Radius of the sphere $r=\frac{9}{2} \mathrm{~cm}$

Volume of the sphere $=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi(\frac{9}{2})^{3}$

$=\frac{243}{2} \pi \mathrm{cm}^{3}$

Diameter of the wire $=2 \mathrm{~mm}$

This implies,

Radius of the wire $=\frac{2}{2} \mathrm{~mm}$

$=1 \mathrm{~mm}$

$=\frac{1}{10} \mathrm{~cm}$ Let $h$ be the length of the wire, this implies,

$\pi r^{2} h=\frac{243}{2} \pi$

$\Rightarrow r^{2} h=\frac{243}{2}$

$\Rightarrow (\frac{1}{10})^{2} h=\frac{243}{2}$

$\Rightarrow \frac{1}{100} h=\frac{243}{2}$

$\Rightarrow h=\frac{243}{2} \times 100$

$\Rightarrow h=12150 \mathrm{~cm}$

The length of the wire is $12150\ cm$.

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