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# A 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 $\frac{1}{16}\ cm$, find the length of the wire.

**Given: **

A metallic right circular cone 20 cm high and whose vertical angle is $60^{o}$, is cut into two parts at the middle of its height by a plane parallel to its base.

The frustum so obtained is drawn into a wire of diameter $\frac{1}{16}$cm.

**To do:**

We have to find the length of the wire.

**Solution:**

**
**

Let ACB be the cone whose vertical angle $\angle ACB = 60^{o} $.

Let $R$ and $x$ be the radii of the lower and upper end of the frustum.

Here, height of the cone, $OC = 20 cm=H$

Height $CP = h = 10\ cm $

Let us consider P as the mid-Point of OC

Cutting the cone into two parts through P.

OP =$\frac{20}{2}= 10\ cm$

Also,$\angle ACO$ and $\angle OCB =\frac{1}{2} \times 60^{o} =30^{o} $

After cutting cone CQS from cone CBA, the remaining solid obtained is a frustum.

Now, in triangle CPQ

$tan30^{o}=\frac{x}{10}$

$\frac{1}{\sqrt{3}} =\frac{x}{10}$

$\Rightarrow x=\frac{10}{\sqrt{3}}\ cm$

In triangle COB

$tan30^{o}=\frac{R}{20}$

$\Rightarrow \frac{1}{\sqrt{3}} =\frac{R}{20}$

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

Volume of the frustum, $V=\frac{1}{3} \pi \left( R^{2} H-x^{2} h\right)$

$\Rightarrow V=\frac{1}{3} \pi \left(\left(\frac{20}{\sqrt{3}}\right)^{2} .20-\left(\frac{10}{\sqrt{3}}\right)^{2} .10\right)$

$\Rightarrow V=\frac{1}{3} \pi \left(\frac{400\times 20}{3} -\frac{100\times 10}{3}\right)$

$\Rightarrow V=\frac{1}{3} \pi \left(\frac{8000-1000}{3}\right)$

$\Rightarrow V=\frac{7000}{9} \pi \ cm^{3}$

Let us assume the length of the wire $l$.

Given the diameter of the wire obtained from the frustum$=\frac{1}{16}\ cm$

Radius of the wire, $r=\frac{1}{2} \times \frac{1}{16} =\frac{1}{32}\ cm$

The volume of the wire$=\pi r^{2} l$

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

$=\frac{\pi l}{1024} cm^{3}$

The volumes of the frustum and the wire formed are equal,

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

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

$\Rightarrow l=\frac{7000\times 1024}{9}$

$\Rightarrow l=796444.44\ cm$

$\Rightarrow l=7964.44\ m$

Therefore, the length of the wire is 7964.4 m.

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