A hollow sphere of internal and external radius $2\ cm$ and $4\ cm$ respectively is melted into a cone of base radius $4\ cm$. Find the height and slant height of the cone.

 Given:

A hollow sphere of internal and external radius $2\ cm$ and $4\ cm$ respectively is melted into a cone of base radius $4\ cm$. 

To do:

We have to find the height and slant height of the cone.

Solution:

Internal radius of the hollow sphere $(r) = 2\ cm$

External radius of the hollow sphere $(R) = 4\ cm$

Therefore,

Volume of the metal used $=\frac{4}{3} \pi(R^{3}-r^{3})$

$=\frac{4}{3} \pi[4^{3}-2^{3}]$

$=\frac{4}{3} \pi[64-8]$

$=\frac{224}{3} \pi \mathrm{cm}^{3}$

Therefore,

Volume of the cone $=\frac{224}{3} \pi \mathrm{cm}^{3}$

Radius of the cone $=4 \mathrm{~cm}$

This implies,

Height of the cone $(h)=\frac{\text { Volume } \times 3}{\pi r^{2}}$

$=\frac{224 \pi \times 3}{3 \times \pi \times 4 \times 4} \mathrm{~cm}$

$=14\ cm$

Slant height of the cone $=\sqrt{r^{2}+h^{2}}$

$=\sqrt{(4)^{2}+(14)^{2}}$

$=\sqrt{16+196}$

$=\sqrt{212} \mathrm{~cm}$

$=14.56 \mathrm{~cm}$

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Updated on: 10-Oct-2022

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