Question

A solid metallic right circular cone 20 cm high and whose vertical angle is 60°, is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter $\frac{1}{12}$ cm, then find the length of the wire.
[HOTS] [CBSE 2014]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{solid}\mathrm{metallic}\mathrm{cone},H=20\mathrm{cm}, \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{frustum},h=\frac{20}{2}=10\mathrm{cm}\mathrm{and} \\ \mathrm{Radius}\mathrm{of}\mathrm{the}\mathrm{wire}=\frac{1}{24}\mathrm{cm} \\ \mathrm{Let}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{wire}\mathrm{be}l,\mathrm{EG}=r\mathrm{and}\mathrm{BD}=R\mathit{.} \\ \mathrm{In}∆\mathrm{AEG}, \\ \tan 30°=\frac{\mathrm{EG}}{\mathrm{AG}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{r}{H-h} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{r}{20-10} \\ \Rightarrow r=\frac{10}{\sqrt{3}}\mathrm{cm} \\ \mathrm{Also},\mathrm{in}∆\mathrm{ABD}, \\ \tan 30°=\frac{\mathrm{BD}}{\mathrm{AD}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{R}{H} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{R}{20} \\ \Rightarrow R=\frac{20}{\sqrt{3}}\mathrm{cm} \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{wire}=\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{frustum} \\ \Rightarrow \mathrm{\pi }{\left(\frac{1}{24}\right)}^{2}l=\frac{1}{3}\pi h\left(R^2+r^2+Rr\right) \\ \Rightarrow \frac{l}{576}=\frac{1}{3}\times 10\times \left[{\left(\frac{20}{\sqrt{3}}\right)}^2+{\left(\frac{10}{\sqrt{3}}\right)}^2+\left(\frac{20}{\sqrt{3}}\right)\left(\frac{10}{\sqrt{3}}\right)\right] \\ \Rightarrow l=\frac{576}{3}\times 10\times \left[\frac{400}{3}+\frac{100}{3}+\frac{200}{3}\right] \\ \Rightarrow l=\frac{576}{3}\times 10\times \frac{700}{3} \\ \Rightarrow l=448000\mathrm{cm} \\ \therefore l=4480\mathrm{m} \end{gathered}$$

So, the length of the wire is 4480 m.