Question

The radii of internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm, respectively. It is melted and recast into a solid cylinder of diameter 14 cm. Find the height of the cylinder. [CBSE 2012]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{the}\mathrm{internal}\mathrm{base}\mathrm{radius}\mathrm{of}\mathrm{spherical}\mathrm{shell},r_1=3\mathrm{cm}, \\ \mathrm{the}\mathrm{external}\mathrm{base}\mathrm{radius}\mathrm{of}\mathrm{spherical}\mathrm{shell},r_2=5\mathrm{cm}\mathrm{and} \\ \mathrm{the}\mathrm{base}\mathrm{radius}\mathrm{of}\mathrm{solid}\mathrm{cylinder},r=\frac{14}{2}=7\mathrm{cm} \\ \mathrm{Let}\mathrm{the}\mathrm{height}\mathrm{of}\mathrm{the}\mathrm{cylinder}\mathrm{be}h. \\ \mathrm{As}, \\ \mathrm{Volume}\mathrm{of}\mathrm{solid}\mathrm{cylinder}=\mathrm{Volume}\mathrm{of}\mathrm{spherical}\mathrm{shell} \\ \Rightarrow \mathrm{\pi }{r}^{2}h=\frac{4}{3}\mathrm{\pi }{r_2}^3-\frac{4}{3}\mathrm{\pi }{r_1}^3 \\ \Rightarrow \mathrm{\pi }{r}^{2}h=\frac{4}{3}\mathrm{\pi }\left({r_2}^3-{r_1}^3\right) \\ \Rightarrow r^{2}h=\frac{4}{3}\left({r_2}^3-{r_1}^3\right) \\ \Rightarrow 7\times 7\times h=\frac{4}{3}\left(5^3-3^3\right) \\ \Rightarrow 49\times h=\frac{4}{3}\left(125-27\right) \\ \Rightarrow h=\frac{4}{3}\times \frac{98}{49} \\ \therefore h=\frac{8}{3}\mathrm{cm} \end{gathered}$$

So, the height of the cylinder is $\frac{8}{3}$ cm.