A solid 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 1/12 cm, find the length of the wire.

#### Solution

Let ACB be the cone whose vertical angle ∠ACB = 60°. 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 = 20/2=10 cm

Also, ∠ACO and ∠OCB = (1/2)×60°=30°

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

Now, in triangle CPQ:

`tan30^@=x/10`

`⇒1/sqrt3=x/10`

`⇒x=10/sqrt3cm`

In triangle COB:

`tan30^@=R/"CO"`

`⇒1/sqrt3=R/20`

`⇒R=20/sqrt3 cm`

Volume of the frustum, V=1/3π(R^{2}H − x^{2}h)

`⇒V=1/3π((20/sqrt3)^2⋅20−(10/sqrt3)^2⋅10)`

`=1/3π(8000/3−1000/3)`

`= 1/3π(7000/3)`

`=1/9π×7000`

`=7000/9π`

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

`πxx(1/24)^2xxl=7000/9π [Volume of wire =πr2h]`

`⇒l=7000/9xx24xx24`

⇒l=448000 cm=4480 m

Hence, the length of the wire is 4480 m.