A metallic right circular cone 20 cm high and whose vertical angle is 60º 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 wire of diameter \( \frac{1}{16} \) cm, find the length of the wire.

Let ABC be the metallic cone, DECB is the required frustum.



Let the two radii of the frustum be \( DO' \ = \ r_2 \) and \( BO \ = \ r_1 \)

From the ∆ ADO' and ABO,

\( r_2 \ = \ h_1 \) tan 30º \( = \ 10 \ × \ \frac{1}{ \sqrt{3}} \)

\( r_1 \ = \ (h_1 \ + \ h_2 \) tan30º \( = \ 20 \ × \ \frac{1}{ \sqrt{3}} \)

Volume of the frustum DBCE \(= \ \frac{1}{3} \pi h_2 \ × [ (r_1)^2 \ + \ (r_1)(r_2) \ + \ (r_2)^2 ] \)

\(= \ \frac{1}{3} \pi h_2 \ × \ [ \frac{400}{3} \ + \ \frac{200}{3} \ + \ \frac{100}{3}] \ \)
\( = \frac{ \pi \ × \ 10}{3} \ × \ \frac{700}{3} \)

Volume of the wire of length l and diameter D

\(= \ \pi ( \frac{D}{2})^2 \ × \ l \ \)
\( = \ \frac{ \pi D^2}{4} \ × \ l \)
[ \( V \ = \ \pi r2h \) ]

\(\therefore \) Volume of the frustum = Volume of the wire drawn from it

\(\Rightarrow \ \frac{ \pi \ × \ 10}{3} \ × \ \frac{700}{3} \ \)
\( = \ \frac{ \pi D^2}{4} \ × \ l \)
[ \(\because \) D=116]

\(l \ = \ \frac{10 \ × \ 700 \ × \ 4}{3 \ × \ 3 \ × \ 16 \ × \ 16} \ \)
\( = \ 7964.44 \) m