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. [CBSE 2013C]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{External}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{hollow}\mathrm{sphere},R_{\mathit{1}}=\frac{8}{2}=4\mathrm{cm}, \\ \mathrm{Internal}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{hollow}\mathrm{sphere},R_2=\frac{4}{2}=2\mathrm{cm}\mathrm{and} \\ \mathrm{Base}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{cone},r=\frac{8}{2}=4\mathrm{cm} \\ \mathrm{Let}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{cone}\mathrm{be}h. \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cone}=\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{hollow}\mathrm{sphere} \\ \Rightarrow \frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{4}{3}\mathrm{\pi }{R_1}^3-\frac{4}{3}\mathrm{\pi }{R_2}^3 \\ \Rightarrow \frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{4}{3}\mathrm{\pi }\left({R_1}^3-{R_2}^3\right) \\ \Rightarrow h=\frac{4}{r^2}\left({R_1}^3-{R_2}^3\right) \\ \Rightarrow h=\frac{4}{4\times 4}\left(4^3-2^3\right) \\ \Rightarrow h=\frac{1}{4}\left(64-8\right) \\ \Rightarrow h=\frac{1}{4}\times 56 \\ \therefore h=14\mathrm{cm} \end{gathered}$$

So, the height of the cone is 14 cm.