A solid metallic right circular cone $20cm$ height whose vertical angle is ${60}^o$, is cut into two parts at the middle of its height by a plane parallel to its base. if the frustum so obtained to be drawn in to a wire of diameter $\frac{1}{12}cm$, find the length of the wire?

Let, $OCD$ be the metallic cone and $ABDC$ be the required frustum.

Since, frustum is drawn into wire of volume of frustum $ABCD=$ Volume of cylindrical wire.

$\begin{array}{c}OQ=QP=10cm\\ \angle QOB={30}^{\circ }\end{array}$

Height of frustum $=h=QP=10cm$

Let $PD=r_1$ and $QB=r_2$

$\begin{array}{c}\tan {30}^o=\frac{PD}{OP}andtan{30}^o=\frac{QB}{OQ}\\ \tan {30}^{\circ }=\frac{1}{\sqrt{3}}=\frac{r_1}{20}andtan{30}^{\circ }=\frac{1}{\sqrt{3}}=\frac{r_2}{10}\\ r_1=\frac{20}{\sqrt{3}}cm{r}_2=\frac{10}{\sqrt{3}}cm\\ \therefore Volumeoffrustum=\frac{\pi h}{3}\left(r_1^2+r_2^2+r_1{r}_2\right)\\ =\frac{\pi \times 10}{3}\left({\left(\frac{20}{\sqrt{3}}\right)}^2+{\left(\frac{10}{\sqrt{3}}\right)}^2+\frac{20}{\sqrt{3}}\times \frac{10}{\sqrt{3}}\right)\\ =\frac{10\pi }{3}\left(\frac{400}{3}+\frac{100}{3}+\frac{200}{3}\right)\\ =\frac{10\pi }{3}\left(\frac{700}{3}\right)\\ =\frac{7000\pi }{9}c{m}^3\end{array}$

Given diameter $=\frac{1}{12}cm$

Volume of wire $=$ volume of frustum

Let $h$ be the length of the wire

$\begin{array}{c}\pi r^{2}h=\frac{7000\pi }{9}\\ h=\frac{7000}{9}\times {\left(\frac{1}{\frac{1}{24}}\right)}^2=4480m\\ \therefore h=4480m\end{array}$