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 $\pi = \frac{22}{7}$)
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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 = $\frac{4}{3} \pi (R^3 − r^3)$
Substituting values we get :
$$\begin{aligned}\Rightarrow V_{\text{sphere}}=\frac{4}{3} \times \pi \times \Big(5^{3}-3^{3}\Big) \\[1em] \Rightarrow V_{\text{sphere}} =\frac{4}{3} \times \pi \times (125-27) \\[1em] \Rightarrow V_{\text{sphere}} =\frac{4}{3} \times \pi \times 98 \\[1em] \Rightarrow V_{\text{sphere}} =\frac{392}{3} \pi\end{aligned}$$
By formula,
Volume of cone = $\frac{1}{3} \pi \times \text{radius}^2h$
Let radius of cone formed be m cm.
$V_{\text{cone}}=\frac{1}{3}\pi m^{2}\times 8=\frac{8}{3}\pi m^{2}$
Since, the sphere is melted to form, the cone the volume of both objects will be equal.
$$\begin{aligned}\Rightarrow \frac{392}{3}\pi=\frac{8}{3}\pi m^{2} \\[1em] \Rightarrow 392 = 8m^{2} \\[1em] \Rightarrow m^2 = \frac{392}{8} \\[1em] \Rightarrow m^{2} = 49 \\[1em] \Rightarrow m = 7 \text{ cm}.\end{aligned}$$
Hence, the radius of the cone formed = 7 cm.