A hollow sphere of internal and external radii 2 cm and 4 cm

The internal and external radii of the hollow sphere are 2cm and 4cm respectively. Therefore, the volume of the hollow sphere is

$V=\frac{4}{3} \pi\left\{(4)^{3}-(2)^{3}\right\}$

$=\frac{4}{3} \times \frac{22}{7} \times 56$

$=\frac{32 \times 22}{3}$

The hollow sphere is melted to produce a right circular cone of base-radius 4cm. Let, the height and slant height of the cone be h cm and l cm respectively. Then, we have

$l^{2}=(4)^{2}+h^{2}$

$\Rightarrow l^{2}=16+h^{2}$

The volume of the cone is

$V_{1}=\frac{1}{3} \pi r_{1}^{2} h_{1}$

$=\frac{1}{3} \times \frac{22}{7} \times(4)^{2} \times h$

Since, the volume of the cone and hollow sphere are same, we have

$V_{1}=V$

$\Rightarrow \frac{1}{3} \times \frac{22}{7} \times(4)^{2} \times h=\frac{32 \times 22}{3}$

$\Rightarrow \quad \frac{1}{7} \times(4)^{2} \times h=32$

$\Rightarrow \quad h=\frac{32 \times 7}{16}$

$\Rightarrow \quad=14$

Then, we have

$I^{2}=16+(14)^{2}$

$\Rightarrow=212$

$\Rightarrow I=14.56$

Therefore, the height and the slant height of the cone are 14 cm and 14.56 cm respectively.