Question

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

Answer 1

Let VAB be the solid metallic right circular cone of height $20cm$. Suppose this cone is cut by a plane parallel to its base at a point $O^{\prime }$ such that $V{O}^{\prime }=O^{\prime }O$ i.e O' is the mid-point of $VO$. Let $r_1$ and $r_2$ be the radii of circular ends of the frustum $AB{B}^{\prime }{A}^{\prime }$.

In triangles $VOA$ and $V{O}^{\prime }{A}^{\prime }$, we have

$\tan {30}^o=\frac{OA}{AO};\tan {30}^o=\frac{O^{\prime }{A}^{\prime }}{V{O}^{\prime }}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{r_1}{20};\frac{1}{\sqrt{3}}=\frac{r_2}{10}$

$\Rightarrow r_1=\frac{20}{\sqrt{3}}cm;r_2=\frac{10}{\sqrt{3}}cm$

Volume of the frustum $=\frac{1}{3}\pi (r_1^2+r_2^2+r_1{r}_2)h$

Volume of the frustum $=\frac{1}{3}\pi (\frac{400}{3}+\frac{100}{3}+\frac{200}{3})\times 10{cm}^2=\frac{7000}{9\pi }{cm}^2$

Let the length of the wire of $\frac{1}{16}cm$ diameter be $l$ cm. Then
volume of the metal used in wire $=\pi \times {\left(\frac{1}{32}\right)}^2\times l{cm}^2$

volume of the metal used in wire $=\frac{\pi l}{1024}{cm}^2$
Since the frustum is recast into a wire of length $l$ cm and diameter $\frac{1}{16}cm$

Volume of the metal used in wire = volume of the frustum

$\frac{\pi l}{1024}=\frac{7000}{9\pi }$

$\Rightarrow l=\frac{7000\pi }{9}\times \frac{1024}{\pi }cm=7964.4m$