Question

A metallic right circular cone $20cm$ high and whose vertical angle is ${60}^{\circ }$ 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 wire of diameter $\frac{1}{16}cm$, find the length of wire.

Answer 1

Let $ABC$ is the metallic cone and $DECB$ is the required frustum.

Let the radii of frustum are $r_1$ and $r_2$

i.e. $DP=r_1$ and $BO=r_2$

Now from $△ADP$ and $△ABO$,

$r_2=h_1\times \tan 30$

$\Rightarrow r_2=10\times \frac{1}{\sqrt{3}}$

$\Rightarrow r_2=\frac{10}{\sqrt{3}}$

$r_1=(h_1+h_2)\tan 30$

$\Rightarrow r_1=20\times \frac{1}{\sqrt{3}}$

$\Rightarrow r_1=\frac{20}{\sqrt{3}}$

Now volume of the frustum$DECB$

$=(\pi \times \frac{h_2}{3})\times (r_1^2+r_1{r}_2+r_2^2)$

$=\pi \times \frac{10}{3}\times \frac{700}{3}$

$=\frac{7000\pi }{9}$ on simplification

Now let $l$ is the length of the wire.

Given diameter of the wire $d=\frac{1}{16}$

So radius of the wire $R=\frac{d}{2}=\frac{1}{16}\times \frac{1}{2}=\frac{1}{32}$

Now volume of the frustum$=$ volume of the wire drawn from it

$\Rightarrow \frac{7000\pi }{9}=\pi R^2\times l$

$\Rightarrow l=\frac{7000\pi }{\pi R^2}$

$\Rightarrow l=\frac{7000\pi }{\pi {\left(\frac{1}{32}\right)}^2}$

$\Rightarrow l=\frac{796444.444}{100}$m since $100$cm$=1$m

$\therefore l=7964.444$m