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

FRUSTUM:
Let r1 and r2 be the radii of the top and bottom surface, respectively of the frustum.
In ΔAOF, tan30° = FO/OA
1/√3 = FO/10 ⇒ FO = 10/√3 = 10√3/3 cm
So, r1 = 10√3/3 cm
In ΔABD, BD/AD = tan30° ⇒ BD = 20√3/3 cm.
So, r2 = 20√3/3 cm
And height of frustum (h) = 10 cm.
So, Volume of frustum = (1/3)πh(r1² + r2² + r1r2) = (1/3)π × 10 [(10√3/3)² + (20√3/3)² + (10√3/3)(20√3/3)]
= (10/3)π[100/3 + 400/3 + 200/3] = (10/3) × (22/7) × (700/3) = 22000/9 cm³
WIRE:
Let the radius of the wire be r and l be the length of the wire.
So, r = 1/32 cm
Volume of wire = Volume of cylinder
Volume of wire = πr²l
22000/9 = (22/7) × (1/32)² × l
l = 22000 × 32 × 32 × 7 / 9 × 22
l = 7,96,444.44 cm
l = 7964.44 m.
So, length of the wire is 7964.44 cm.