# 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 is drawn into a wire of diameter 1/16 cm, find the length of the wire.

Solution

From ΔAEG

$$\frac{EG}{AG}=\tan 30$$ $$\frac{10}{\sqrt{3}}=EG\Rightarrow EG=\frac{10\sqrt{3}}{3}$$

From ΔAEG

$$\frac{BD}{AD}=\tan 30$$ $$BD=\frac{20}{\sqrt{3}}=\frac{20\sqrt{3}}{3}$$

Radius (r1) of upper end of frustum = (10√3)/3 cm

Radius (r2) of lower end of container = (20√3)/3 cm

Height (r3) of container = 10 cm

Now,

Volume of the frustum = (⅓)×π×h(r12+r22+r1r2)

$$\frac{1}{3}\times\pi\times10={\frac{(10\sqrt{3}}{3})^{2}+\frac{(20\sqrt{3}}{3})^{2}+\frac{(10\sqrt{3}\times20\sqrt{3})}{3\times 3}}$$

= 7000/0 cm3 ………………………..(1)

Volume of Cylinder = 1/3 π(radius)2H

1/3π(1/1622H …………………(2)

We know that (1) = (2) , then

7000π / 9 = 1/3 π(1/(16×2))2H

H = 70003232/9

H = 796444.443 cm

H =7964.44 m

Hence, the length of the wire = 7964.44 m