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.
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
Volume of the frustum = (1/3) × π × h(r12+r22+r1r2)
= \(\frac{1}{3} \times \pi \times 10 \big[(\frac{10 \sqrt 3}{3})^2 + (\frac{20 \sqrt 3}{3})^2 + \frac{(10 \sqrt 3)(20 \sqrt 3)}{3 \times 3} \big]\)
Solving this we get,
Volume of the frustum = 22000/9 cm3
The radius (r) of wire = (1/16) × (1/2) = 1/32 cm
Let the length of wire be “l”.
Volume of wire = Area of cross-section x Length
= (πr2) x l
= π(1/32)2 x l
Now, Volume of frustum = Volume of wire
22000/9 = (22/7) x (1/32)2 x l
Solving this we get,
l = 7964.44 m