A cylinder with base radius 8 cm and height 2 cm is melted to form a cone of height 6 cm.

Solution:

We have,

Base radius of the cylinder, $r=8 \mathrm{~cm}$,

Height of the cylinder, $h=2 \mathrm{~cm}$ and

Height of the cone, $H=6 \mathrm{~cm}$

Let the base radius of the cone be $R$.

Now,

Volume of the cone = Volume of the cylinder

$\Rightarrow \frac{1}{3} \pi R^{2} H=\pi r^{2} h$

$\Rightarrow R^{2}=\frac{3 r^{2} h}{H}$

$\Rightarrow R^{2}=\frac{3 \times 8 \times 8 \times 2}{6}$

$\Rightarrow R^{2}=64$

$\Rightarrow R=\sqrt{64}$

$\therefore R=8 \mathrm{~cm}$

So, the radius of the base of the cone is 8 cm.