Question

A metallic right circular cone is 20 m high and has a vertical angle of 60°. It is cut into two parts at the middle of its height by a plane parallel to the base. If the frustum so obtained is drawn into a wire of diameter $\frac{1}{16}$ cm, find the length of the wire.

Answer 1

Consider the frustum created by cutting the original cone.
Consider the vertical axis.
Vertical angle of the cone = 60^o
Angle of the central axis with slant height = 30^o

Height of the cone = 20 cm
Height of the smaller cone = Height of the frustum = 10 cm

In the smaller cone,
$$\begin{gathered} \frac{\mathrm{Radius}}{\mathrm{Height}}=\tan 30° \\ \frac{\mathrm{Radius}}{10}=\frac{1}{\sqrt{3}} \\ \mathrm{Radius}=\frac{10}{\sqrt{3}}\mathrm{cm} \end{gathered}$$

Height of the original cone = 20 m

$$\begin{gathered} \frac{\mathrm{Radius}}{\mathrm{Height}}=\tan 30° \\ \frac{\mathrm{Radius}}{20}=\frac{1}{\sqrt{3}} \\ R=\frac{20}{\sqrt{3}}\mathrm{cm} \end{gathered}$$

Volume of the frustum
$$\begin{gathered} =\frac{1}{3}\mathrm{\pi }h\left(R^2+r^2+Rr\right) \\ =\frac{1}{3}\mathrm{\pi }\times 10\left({\left(\frac{20}{\sqrt{3}}\right)}^2+{\left(\frac{10}{\sqrt{3}}\right)}^2+\frac{20\times 10}{\sqrt{3}\times \sqrt{3}}\right) \\ =\frac{10\mathrm{\pi }}{3}\left(\frac{400}{3}+\frac{100}{3}+\frac{200}{3}\right) \\ =\frac{10\mathrm{\pi }}{3}\times \frac{700}{3}{\mathrm{cm}}^3 \\ =\frac{7000\mathrm{\pi }}{9}{\mathrm{cm}}^3 \end{gathered}$$

Let the length of the wire be x cm.
Diameter$=\frac{1}{16}cm$
Radius$=\frac{1}{32}cm$

Volume of the wire $=\mathrm{\pi }\times {\left(\frac{1}{32}\right)}^2\times \mathrm{x}{\mathrm{cm}}^3$

Volume of the wire = Volume of the frustum
Therefore,
$$\begin{gathered} \frac{7000\mathrm{\pi }}{9}=\mathrm{\pi }\times {\left(\frac{1}{32}\right)}^2\times x \\ \Rightarrow x=\frac{7000\times 32\times 32}{9} \\ \Rightarrow x=796444\mathrm{cm}=7964.44\mathrm{m} \end{gathered}$$

So, the length of the wire drawn is 7964.44 m.