Question

Question 5
A 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 $\frac{1}{16}$ cm, find the length of the wire.

Answer 1

Volume of frustum will be equal to the volume of wire and by using this relation we can calculate the length of the wire.


In the given figure; AO = 20 cm and hence height of frustum PO = 10 cm
In triangle AOC we have angle CAO = ${30}^{\circ }$ (halft of vertical angle of cone BAC)
Therefore;
$tan{30}^{\circ }=\frac{OC}{AO}$
$Or,\frac{1}{\sqrt{3}}=\frac{OC}{20}$
$Or,OC=\frac{20}{\sqrt{3}}$
Using similarity criteria in triangles AOC and ADE it can be shown that $DE=\frac{10}{\sqrt{3}}$ (because DE bisects the cone through its height)
Similarly, PO = 10 cm
Volume of frustum can be calculated as follows:
$V=\frac{1}{3}\pi h(r_1^2+r_2^2+r_1{r}_2)$
$=\frac{1}{3}\pi \times \pi 10[{\left(\frac{20}{\sqrt{3}}\right)}^2+{\left(\frac{10}{\sqrt{3}}\right)}^2+\frac{20}{\sqrt{3}}\times \frac{10}{\sqrt{3}}]$
$=\frac{1}{3}\pi \times 10(\frac{400}{3}+\frac{100}{3}+\frac{200}{3})$
$=\frac{7000}{9}\pi c{m}^3$
Volume of cylinder = $\pi r^{2}h$
$Or,\pi \times {\left(\frac{1}{32}\right)}^2\times h$
$=\frac{7000}{9}\pi$
$Or,h=\frac{7000}{9}\times 1024$
$=796444.44cm$