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 $=7 \times 2$

$=14 \mathrm{~cm}$

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Updated on: 10-Oct-2022

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