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# 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