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The diameters of the internal and external surfaces of a hollow spherical shell are $ 6 \mathrm{~cm} $ and $ 10 \mathrm{~cm} $ respectively. If it is melted and recast into a solid cylinder of diameter $ 14 \mathrm{~cm} $, find the height of the cylinder.
Given:
The diameters of the internal and external surfaces of a hollow spherical shell are \( 6 \mathrm{~cm} \) and \( 10 \mathrm{~cm} \) respectively.
It is melted and recast into a solid cylinder of diameter \( 14 \mathrm{~cm} \).
To do:
We have to find the height of the cylinder.
Solution:
Outer diameter of the hollow spherical shell $= 10\ cm$
Inner diameter of the hollow spherical shell $= 6\ cm$
This implies,
Outer radius $R =\frac{10}{2}$
$= 5\ cm$
Inner radius $r =\frac{6}{2}$
$= 3\ cm$
Volume of the metal used $=\frac{4}{3} \pi(\mathrm{R}^{3}-r^{3})$
$=\frac{4}{3} \pi[5^{3}-3^{3}]$
$=\frac{4}{3} \pi[125-27]$
$=\frac{4}{3} \pi \times 98 \mathrm{~cm}^{3}$
Volume of the solid cylinder $=$ Volume of the metal used
$=\frac{4}{3} \pi \times 98 \mathrm{~cm}^{3}$
Diameter of the solid cylinder $=14 \mathrm{~cm}$
Radius of the solid cylinder $r_{1}=\frac{14}{2}$
$=7 \mathrm{~cm}$
Let $h$ be the height of the cylinder.
Therefore,
$\pi r_{1}^{2} h=\frac{4}{3} \pi \times 98$
$\Rightarrow \pi(7)^{2} h=\frac{4}{3} \pi \times 98$
$\Rightarrow 49 \pi h=\frac{98 \times 4}{3} \pi$
$\Rightarrow h=\frac{98 \times 4 \pi}{3 \times 49 \times \pi}$
$\Rightarrow h=\frac{8}{3}$
The height of the solid cylinder is $\frac{8}{3} \mathrm{~cm}$.