Question

A solid metallic sphere of diameter 8 cm is melted and drawn into a cylindrical wire of uniform width. If the length of the wire is 12 m, then find its width. [CBSE 2013]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{metallic}\mathrm{sphere},R=\frac{8}{2}=4\mathrm{cm}\mathrm{and} \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{wire},h=12\mathrm{m}=1200\mathrm{cm} \\ \mathrm{Let}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{base}\mathrm{be}r. \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{wire}=\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{metallic}\mathrm{sphere} \\ \Rightarrow \mathrm{\pi }{r}^{2}h=\frac{4}{3}\mathrm{\pi }{R}^3 \\ \Rightarrow r^2=\frac{4R^3}{3h} \\ \Rightarrow r^2=\frac{4\times 4\times 4\times 4}{3\times 1200} \\ \Rightarrow r^2=\frac{16}{225} \\ \Rightarrow r=\sqrt{\frac{16}{225}} \\ \Rightarrow r=\frac{4}{15}\mathrm{cm} \\ \therefore \mathrm{The}\mathrm{width}\mathrm{of}\mathrm{the}\mathrm{wire}=2r \\ =2\times \frac{4}{15} \\ =\frac{8}{15}\mathrm{cm} \end{gathered}$$

So, the width of the wire is $\frac{8}{15}$ cm.