A copper sphere of radius 3 cm is melted and recast into a right

The radius of the copper sphere is 3cm. Therefore, the volume of the copper sphere is

$V=\frac{4}{3} \pi \times(3)^{3} \mathrm{~cm}^{3}$

The copper sphere is melted to produce a right circular cone. The height of the right circular cone is 3cm. Let the base-radius of the right circular cone is r. Then, the volume of the right circular cone is

$V_{1}=\frac{1}{3} \pi \times(r)^{2} \times 3$

$=\pi \times r^{2} \mathrm{~cm}^{3}$

Since, the sphere is melted to recast the cone; the volumes of the sphere and the cone are equal. Hence, we have

$V=V_{1}$

$\Rightarrow \frac{4}{3} \pi \times(3)^{3}=\pi \times r^{2}$

$\Rightarrow \quad 4 \times(3)^{2}=r^{2}$

$\Rightarrow \quad r=2 \times 3$

$\Rightarrow \quad r=6$

Hence, the base-radius of the right circular cone is 6 cm.