A solid metallic hemisphere of radius 8 cm

Let height of the cone be h.

Given, radius of the base of the cone = 6 cm

$\therefore \quad$ Volume of circular cone $=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi(6)^{2} h=\frac{36 \pi h}{3}=12 \pi h \mathrm{~cm}^{3}$

Also, given radius of the hemisphere $=8 \mathrm{~cm}$

$\therefore \quad$ Volume of the hemisphere $=\frac{2}{3} \pi r^{3}=\frac{2}{3} \pi(8)^{3}=\frac{512 \times 2 \pi}{3} \mathrm{~cm}^{3}$

According to the question,

Volume of the cone $=$ Volume of the hemisphere

$\Rightarrow \quad 12 \pi h=\frac{512 \times 2 \pi}{3}$

$\therefore$ $h=\frac{512 \times 2 \pi}{12 \times 3 \pi}$

$=\frac{256}{9}=28.44 \mathrm{~cm}$