The diameters of internal and external surfaces of a hollow spherical shell are $ 10 \mathrm{~cm} $ and $ 6 \mathrm{~cm} $ respectively. If it is melted and recast into a solid cylinder of length of $ 2 \frac{2}{3} \mathrm{~cm} $, find the diameter of the cylinder.

Given:

The diameters of internal and external surfaces of a hollow spherical shell are \( 10 \mathrm{~cm} \) and \( 6 \mathrm{~cm} \) respectively.

It is melted and recast into a solid cylinder of length of \( 2 \frac{2}{3} \mathrm{~cm} \).

To do:

We have to find the diameter of the cylinder.

Solution:

Diameter of the external surface of the sphere $=10 \mathrm{~cm}$

Internal diameter of the sphere $=6 \mathrm{~cm}$

Therefore,

External radius $\mathrm{R}=\frac{10}{2}$

$=5 \mathrm{~cm}$

Internal radius $r=\frac{6}{2}$

$=3 \mathrm{~cm}$

Volume of the metal used $=\frac{4}{3} \pi[\mathrm{R}^{3}-r^{3}]$

$=\frac{4}{3} \pi[5^{3}-3^{3}]$

$=\frac{4}{3} \pi(125-27)$

$=\frac{4}{3} \pi \times 98$

$=\frac{392 \pi}{3} \mathrm{~cm}^{3}$

Therefore,

Volume of the solid cylinder so formed $=\frac{392}{3} \pi \mathrm{cm}^{3}$

Length of the cylinder $h=2 \frac{2}{3} \mathrm{~cm}$

$=\frac{8}{3} \mathrm{~cm}$
Let $r$ be the radius.

$\Rightarrow \pi r^{2} h=\frac{392}{3} \pi$

$\Rightarrow \pi r^{2} \times \frac{8}{3}=\frac{392}{3} \pi$

$\Rightarrow r^{2}=\frac{392 \pi}{3} \times \frac{3}{8 \pi}$

$\Rightarrow r^{2}=49$

$\Rightarrow r=7\ cm$

$\Rightarrow$ Diameter $=2 r$

$=2 \times 7$

$=14 \mathrm{~cm}$

The diameter of the cylinder is $14\ cm$.

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Updated on: 10-Oct-2022

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