A hollow sphere of external diameter 10 cm and internal diameter 6 cm is melted and made into a solid right circular cone of height 8 cm. Find the radius of the cone so formed.
(Use $\pi = \dfrac{22}{7}$)

Given,

External radius of hollow sphere (R) = 5 cm, internal radius of hollow sphere (r) = 3 cm, height of cone (h) = 8 cm.

By formula,

Volume of metal in the hollow sphere V = $\dfrac{4}{3} \pi (R^3 − r^3)$

Substituting values we get :

$\Rightarrow V${\text{sphere}}=\dfrac{4}{3} \times \pi \times \Big(5^{3}-3^{3}\Big) \\[1em] \Rightarrow V{\text{sphere}} =\dfrac{4}{3} \times \pi \times (125-27) \\[1em] \Rightarrow V{\text{sphere}} =\dfrac{4}{3} \times \pi \times 98 \\[1em] \Rightarrow V{\text{sphere}} =\dfrac{392}{3} \pi

By formula,

Volume of cone = $\dfrac{1}{3} \pi \times \text{radius}^2h$

Let radius of cone formed be m cm.

$V_{\text{cone}}=\dfrac{1}{3}\pi m^{2}\times 8=\dfrac{8}{3}\pi m^{2}$

Since, the sphere is melted to form, the cone the volume of both objects will be equal.

$\Rightarrow \dfrac{392}{3}\pi=\dfrac{8}{3}\pi m^{2} \\[1em] \Rightarrow 392 = 8m^{2} \\[1em] \Rightarrow m^2 = \dfrac{392}{8} \\[1em] \Rightarrow m^{2} = 49 \\[1em] \Rightarrow m = 7 \text{ cm}.$

Hence, the radius of the cone formed = 7 cm.