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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 $=\sqrt{r^{2}+h^{2}}$
$=\sqrt{(4)^{2}+(14)^{2}}$
$=\sqrt{16+196}$
$=\sqrt{212} \mathrm{~cm}$
$=14.56 \mathrm{~cm}$