A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm.

We have,

Radius of the hemisphere, $R=6 \mathrm{~cm}$ and

Height of the cone, $h=75 \mathrm{~cm}$

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

Now,

Volume of the cone = Volume of the hemisphere

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

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

$\Rightarrow r^{2}=\frac{2 \times 6 \times 6 \times 6}{75}$

$\Rightarrow r^{2}=5.76$

$\Rightarrow r=\sqrt{5.76}$

$\therefore r=2.4 \mathrm{~cm}$

So, the radius of the base of the cone is 2.4 cm.