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Question:- A metallic right circular cone 20 cm high and whose vertical angle is 60 degree 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 1/16cm,find the length of the wire
Answer: 7964.44 m
Step-by-step explanation:
Let us suppose that there is right circular cone ABC which is cut from the middle by the plane EOD.
AO = 20/2 = 10cm
Since this is right circular cone, AP is bisector of ∠BAC
∠EAO = ∠BAP = 60°/2 = 30°
Now,
In ΔEAO
tan30° = EO/AO
1/√3 = EO/10
EO = 10/√3 cm
Similarly,
In ΔABP
tan30° = BP/AP
1/√3 = BP/20
BP = 20√3 cm
Now, In frustum BEDC,
lower radius (r₁) = EO = 10/√3 cm
Upper radius (r₂) = EP = 20/√3 cm
Height (h) = 10 cm
Volume of frustum = πh/3(r₁² + r₂² + r₁.r₂)
=10π/3{(10/√3)² + (20/√3)² + (10/√3 × 20/√3)}
=10π/3{100/3 + 400/3 + 200/3)
=10π/3(700/3)
= 7000π/9 cm²
Diameter of the wire = 1/16cm
Radius of wirte = 1/32 cm
If the length of the wire l,
Volume of wire = πr²l
=π × 1/32 × 1/32 × l
=πl/1024 cm²
Since, Frustum is formed in a wire,
Volume of wire = Volume of Frustum
πl/1024 = 7000π/9
l/1024 = 7000/9
l = 796444.44 cm
l = 7964.44 m
