A cone of height 20 cm and radius of base 5 cm is made up of modelling clay.

Solution:

We have,

the base radius of the cone, $r=5 \mathrm{~cm}$ and

the height of the cone, $h=20 \mathrm{~cm}$

Let the radius of the sphere be $R$.

As,

Volume of sphere $=$ Volume of cone

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

$\Rightarrow R^{3}=\frac{\pi r^{2} h \times 3}{3 \times 4 \pi}$

$\Rightarrow R^{3}=\frac{r^{2} h}{4}$

$\Rightarrow R^{3}=\frac{5 \times 5 \times 20}{4}$

$\Rightarrow R^{3}=125$

$\Rightarrow R=\sqrt[3]{125}$

$\Rightarrow R=5 \mathrm{~cm}$

$\Rightarrow$ Diameter of the sphere $=2 R=2 \times 5=10 \mathrm{~cm}$

So, the diameter of the sphere is 10 cm.