A solid metallic right circular cone 20 cm high and whose vertica…

Question

A solid metallic right circular cone $20 \ cm$ 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 a wire of diameter $\frac{1}{12} \ cm$, find the length of the wire.

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Answer

Let $\text{ACB}$ be the cone whose vertical angle $\angle A C B=60^{\circ}$.
Let $R$ and $x$ be the radii of the lower and upper end of the frustum.
Here, height of the cone, $O C=20 \ cm=H$ and $C P=h=10 \ cm$
Let us consider $P$ as the mid $-$ Point of $O C$.
After cutting the cone into two parts through $P$.
$O P=\frac{20}{2}=10 \ cm$
Also, $\angle A C O$ and $\angle O C B=\frac{1}{2} \times 60^{\circ}=30^{\circ}$
After cutting cone $\text{CQS}$ from cone $\text{CBA}$, the remaining solid obtained is a frustum.
Now, in triangle $\text{CPQ}$ :
$\tan 30^{\circ}=\frac{x}{10}$
$\frac{1}{\sqrt{3}}=\frac{x}{10}$
$x=\frac{10}{\sqrt{3}} \ cm$
In triangle $C O B$ :
$\tan 30^{\circ}=\frac{R}{C O}$
$\frac{1}{\sqrt{3}}=\frac{R}{20}$
$R=\frac{20}{\sqrt{3}} \ cm$
Volume of the frustum
$V=\frac{1}{3} \pi\left(R^2 H-x^2 h\right)$
$V=\frac{1}{3} \pi\left(\left(\frac{20}{\sqrt{3}}\right)^2 \cdot 20-\left(\frac{10}{\sqrt{3}}\right)^2 \cdot 10\right)$
$V=\frac{1}{3} \pi\left(\frac{8000}{3}-\frac{1000}{3}\right)$
$V=\frac{1}{3} \pi\left(\frac{7000}{3}\right)$
$V=\frac{1}{9} \pi \times 7000$
$V=\frac{7000}{9} \pi$
Thev olumes of the frustum and the wire formed are equal
$\pi \times\left(\frac{1}{24}\right)^2 \times l=\frac{7000}{9} \pi\left(\text { Volume of wire }=\pi r^2 h\right)$
$l=\frac{7000}{9} \times 24 \times 24$
$l=448000 \ cm$
$l=4480\ m$
Hence the length of the wire is $4480\ m$