Question

A hollow sphere of external and internal diameters 8 cm and 4 cm respectively is melted into a solid cone of base diameter 8 cm. Find the height of the cone.

Answer 1

We have,

External radius of the hollow sphere, $R_1=\frac{8}{2}$ = 4 cm,

Internal radius of the hollow sphere, $R_2=\frac{4}{2}$= 2 cm and

Base radius of the cone, $r=\frac{8}{2}$ = 4 cm

Let the height of the cone be h.

Now,

Volume of the cone = Volume of the hollow sphere

⇒ $\frac{1}{3}$ $\pi r^{2}h$ = $\frac{4}{3}$$\pi (R_1{)}^3$ - $\frac{4}{3}$$\pi (R_2{)}^3$

⇒ $\frac{1}{3}$ $\pi r^{2}h$ = $\frac{4}{3}$ $\pi \left[\right(R_1{)}^3-\left(R_2{)}^3\right]$

⇒ $h=\frac{4}{r^2}\left[\right(R_1{)}^3-\left(R_2{)}^3\right]$

⇒ $h=\frac{4}{4\times 4}[4^3-2^3]$

⇒ $h=\frac{4}{4\times 4}[64-8]$

⇒$h=\frac{1}{4}\left[56\right]$

Therefore, h = 14 cm

So, the height of the cone is 14 cm.