A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm.

Solution:

Given that

Hollow sphere external radii = r2 = 4 cm

Internal radii = r1 = 2 cm

Cone base radius (R) = 4 cm

Height = h

Volume of cone = Volume of sphere

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

$4^{2} h=4\left(4^{3}-2^{3}\right)$

$\mathrm{h}=\frac{4 \times 56}{16}$

h = 14 cm

Slantheight $(\mathrm{l})=\sqrt{\mathrm{r}^{2}+\mathrm{h}^{2}}$

Slantheight $(\mathrm{l})=\sqrt{\mathrm{r}_{2}^{2}+\mathrm{h}^{2}}$

$1=\sqrt{4^{2}+\left(14^{2}\right)}$

$1=\sqrt{16+196}$

$1=\sqrt{212}$

 

$1=14.56 \mathrm{~cm}$