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