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 cm, find the length of the wire.

Given: Height of solid metallic right circular cone, h = 20cm

Diameter of wire = 1/12 cm

Radius of wire = (1/12)/2 = 1/24 cm

A solid cone AFE has been cut by BC || FE.

AD ⊥ FE.

B is the midpoint of the AD.

After cutting into two parts through B,

BD = 20/2 = 10cm

Given ∠FAE = 60°

So, ∠FAD = ∠EAD = 30°

After cutting cone AGC from cone AFE, the remaining solid is a frustum.

Consider ΔABC,

⇒ tan30° = r/10

⇒ 1/√3 = r/10

∴ r = 10/√3 cm

Consider ΔADE,

⇒ tan30° = R/AD

⇒ 1/√3 = R/20

∴ R = 20/√3 cm

We know that volume of frustum of cone = [πh(r₁² + r₂² + r₁r₂)]/3

Substituting the obtained values in the above formula,

⇒ Volume of frustum GCEF = [10π((10/√3)² + (20/√3)² + (10/√3)(20/√3)]/3

= [10π(700/3)]/3

= 7000π/9 cm³

Volume of wire formed = Volume of frustum of cone

We know that volume of cylinder = πr²l

So, π(1/24)²(l) = 7000π/9

⇒ l = 7000(24)²/9

∴ l = 4, 48, 000 cm

Ans. The length of the wire required is 4, 48, 000 cm.