A hollow sphere of internal and external diameters 4 cm and 8 cm

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

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

The hollow spherical shell is melted to recast a cone of base- radius 4cm. Let, the height of the cone is h. Therefore, the volume of the cone is

$V_{1}=\frac{1}{3} \pi \times(4)^{2} \times h \mathrm{~cm}^{3}$

Since, the volume of the cone is same as the volume of the hollow sphere, we have

$V_{1}=V$

$\Rightarrow \frac{1}{3} \pi \times(4)^{2} \times h=\frac{4}{3} \pi \times\left\{(4)^{3}-(2)^{3}\right\}$

$\Rightarrow \quad 16 \times h=4 \times 56$

$\Rightarrow \quad h=\frac{4 \times 56}{16}$

$\Rightarrow \quad=14$