A hollow sphere of external and internal diameters 8 cm and 4 cm, respectively is melted into a solid cone of base diameter 8 cm.

Solution:

We have,

External radius of the hollow sphere, $R_{1}=\frac{8}{2}=4 \mathrm{~cm}$,

Internal radius of the hollow sphere, $R_{2}=\frac{4}{2}=2 \mathrm{~cm}$ and

Base radius of the cone, $r=\frac{8}{2}=4 \mathrm{~cm}$

Let the height of the cone be $h$.

Now,

Volume of the cone $=$ Volume of the hollow sphere

$\Rightarrow \frac{1}{3} \pi r^{2} h=\frac{4}{3} \pi R_{1}^{3}-\frac{4}{3} \pi R_{2}{ }^{3}$

$\Rightarrow \frac{1}{3} \pi r^{2} h=\frac{4}{3} \pi\left(R_{1}{ }^{3}-R_{2}{ }^{3}\right)$

$\Rightarrow h=\frac{4}{r^{2}}\left(R_{1}{ }^{3}-R_{2}{ }^{3}\right)$

$\Rightarrow h=\frac{4}{4 \times 4}\left(4^{3}-2^{3}\right)$

$\Rightarrow h=\frac{1}{4}(64-8)$

$\Rightarrow h=\frac{1}{4} \times 56$

$\therefore h=14 \mathrm{~cm}$

So, the height of the cone is 14 cm.