Question

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

Answer 1

Given : Internal diameter of hollow sphere(d)= 4 cm.
Internal radius of hollow sphere (r) = $\frac{4}{2}$= 2 cm
external diameter of hollow sphere (D) = 8 cm.
external radius of hollow sphere( R )= $\frac{8}{2}$ = 4 cm.

Volume of the Hollow sphere

$=\frac{4}{3}\pi (R^3-r^3)$

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

$=\frac{4}{3}\pi (64-8)$

$=\frac{4}{3}\pi \left(56\right)c{m}^3$

Diameter of the cone$\left(d_1\right)=8cm$
radius of the cone$\left(r_1\right)=\frac{8}{2}=4cm$

Let the height of the cone be h cm.
Volume of the cone$=\frac{1}{3}\pi r_1^{2}h=\frac{1}{3}\pi \times 4^{2}h=\frac{16}{3}\pi h$

Volume of the cone = Volume of the hollow sphere

$\frac{16}{3}\pi h=\frac{4}{3}\pi \times 56$

$16h=4\times 56$

$h=\frac{4\times 56}{16}=\frac{56}{4}=14cm$

Hence, the height of the cone is 14 cm.