Question

A solid metallic right circular cone 20cm high and whose vertical angle is 60 degree is cut into two equal 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 1/12 cm find the length of the wire.

Answer 1

Answer:

Step-by-step explanation:

Length of the wire is 4480m

Answer 2

Answer:

 977383.463636 = cm

Step-by-step explanation:

Let ABC is the metallic cone  

A metallic right circular cone 20cm high and whose Vertical angle is 60 degree is cut at the middle of its height

So, AO = OF =$\frac{AF}{2} =\frac{20}{2}=10 cm$

And $\angle DAO =\angle OAE = 30^{\circ}$

In ΔAFC

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan \theta 30^{\circ}= \frac{FC}{AF}$

$\frac{1}{\sqrt{3}}= \frac{FC}{20}$

$\frac{20}{\sqrt{3}}= FC$

$11.547= FC$

So, R =11.547

In ΔDAO  

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan \theta 30^{\circ}= \frac{OD}{OA}$

$\frac{1}{\sqrt{3}}= \frac{OD}{10}$

$\frac{10}{\sqrt{3}}= OD$

$5.7735= OD$

So, r =5.7735

Volume of frustum =$\frac{1}{3}\pi h(R^2+r^2+R \times r)$

                              = $\frac{1}{3}\times \frac{22}{7} \times 10 (11.547^2+5.7735^2+11.547 \times 5.7735 )$

                                = $7679.4415 cm^3$

Diameter of wire =$\frac{1}{12} cm$

Radius = $\frac{\frac{1}{12}}{2}=\frac{1}{24}$

Wire is in the shape of cylinder

So, volume of wire =$\pi r^2 h$

                              =$\frac{22}{7} \times(\frac{1}{24})^2 h$

Since the frustum so obtained is drawn into a wire So, volume remain same

So, $7679.4415=\frac{22}{7} \times(\frac{1}{20})^2 h$

$7679.4415 \times \frac{7}{22} \times 400=h$

$977383.463636 =h$

Hence The length of the wire is   977383.463636 cm.