Question

A metallic right circular cone is 20 cm high and has a vertical angle of 60° this is cut into two parts at the middle of it's height by a plane parallel to the base if the frustom so obtained is drawn into a wire of diameter 1/16 cm find the length of the wire​

Answer 1

Solution :

Let OAB be the cone in which angle AOB= 60°.

Clearly angle DOE = 30°, OE = 10 cm, OF = 20 cm.

Let ED = r and FB = R.

ED/OE = tan 30°

ED/10cm = 1/√3

ED = 10 x 1/√3 cm

r = 10/√3 cm

and FB/OF = tan30°

FB/20cm = 1/√3

FB = 20 x 1/√3 cm

R = 20/√3cm

Also, EF = 10 cm.

Thus, ABCD is the frustum of a cone in which

R = 20/√3 cm,

r = 10/√3cm

h = 10 cm.

Volume of this frustum= ⅓πh(R² +r² +Rr)

⇛⅓ x π x 10 {400/3 + 100/3 + 200/3} cm³

⇛π x 10/3 x 700/3 cm³

⇛700π/9 cm³

Let the length of the wire be l.

Radius of the wire, r_1 = 32 Cm.

Volume of the wire πr_1²l = π x (1/32)² x l

⇛ 7000π/9 = πl/32

⇛ l=(7000 x 32 x 32/9) cm

⇛ l=(7000 x 32 x 32/9 x 100)m

⇛ 71680/9 m

⇛ 7964.44m