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.

Answer 1

Let ACB be the cone whose vertical angle $\angle ACB={60}^{\circ }$. 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.
After cutting the cone into two parts through P.
$OP=\frac{20}{2}=10cm$
Also, $\angle ACO$ and $\angle OCB=\frac{1}{2}\times {60}^{\circ }={30}^{\circ }$

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

Now, in triangle CPQ:

tan 30° = $\frac{x}{10}$

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

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

In triangle COB:

Tan 30° = $\frac{R}{CO}$

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

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

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

V=$\frac{1}{3}\pi \left(\right(\frac{20}{\sqrt{3}}{)}^2.20-(\frac{10}{\sqrt{3}}{)}^2.10$)

=$\frac{1}{3}\pi (\frac{8000}{3}-\frac{1000}{3}$)

= $\frac{1}{3}\pi (\frac{7000}{3}$)

= $\frac{1}{9}\pi \times 7000$

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

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

π×$(\frac{1}{24}{)}^2\times l=\frac{7000}{9}\pi [Volumeofwire=\pi r^{2}h$]

⇒l = $\frac{7000}{9}\times 24\times 24$

⇒ l = 448000 cm = 4480 m

Hence, the length of the wire is 4480 m.