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.
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:
In triangle COB:
Volume of the frustum, V=1/3π(R2H − x2h)
The volumes of the frustum and the wire formed are equal.
`πxx(1/24)^2xxl=7000/9π [Volume of wire =πr2h]`
⇒l=448000 cm=4480 m
Hence, the length of the wire is 4480 m.