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 $\frac{8}{3} \mathrm{~cm}$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

29 Views

Kickstart Your Career

Get certified by completing the course

Get Started