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A metallic right circular cone $20 cm$ high and whose vertical angle is $6 0^{\circ}$ is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained is drawn into a wire of diameter $\frac{1}{16}$ cm, find the length of the wire. Use $π = \frac{22}{7}$

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796444.44 cm

790044.44 cm

796400.44 cm

796004.44 cm

answer is A.

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Detailed Solution

Given,
Height of cone AG=20 cm. Thus, length of half cone,
AF=FG= 10 cm
Suppose
Upper end radius of frustrum,

FB = r_{1}

Lower end radius of frustrum,

DG = r_{2}

Now in

△ AFB

\Rightarrow \tan 3 0^{\circ} = \frac{FB}{AF}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{FB}{10}

\Rightarrow r_{1} = \frac{10}{\sqrt{3}} cm

Similarly, in

△ AGD

\Rightarrow \tan 3 0^{\circ} = \frac{GD}{AG}

= \frac{1}{\sqrt{3}} = \frac{GD}{20}

\Rightarrow GD = \frac{20}{\sqrt{3}}

\Rightarrow r_{2} = \frac{20}{\sqrt{3}} cm

Now, diameter of wire

= \frac{1}{16} cm

So, radius =

\frac{1}{16} \times \frac{1}{2} = \frac{1}{32} cm

Here, volume of wire,

= π {(radius of wire)}^{2} \times lengt h of wire

And now volume of the frustrum,
=

\frac{1}{3} π \times h ({r_{1}}^{2} + {r_{2}}^{2} + r_{1} \times r_{2})

We have,
Volume of frustum = volume of wire

\Rightarrow \frac{1}{3} π \times h ({r_{1}}^{2} + {r_{2}}^{2} + r_{1} \times r_{2}) = π r^{2} \times lengt h of wire (l)

Put all the values in above equation

\Rightarrow \frac{1}{3} \times \frac{22}{7} \times 10 \times [{(\frac{10}{\sqrt{3}})}^{2} + {(\frac{20}{\sqrt{3}})}^{2} + {(\frac{10}{\sqrt{3}} \times \frac{20}{\sqrt{3}})}^{2}] = \frac{22}{7} \times {(\frac{1}{32})}^{2} \times l

\Rightarrow \frac{1}{3} \times 10 \times [(\frac{100}{3}) + (\frac{400}{3}) + (\frac{200}{3})] = (\frac{1}{1024}) \times l

\Rightarrow \frac{10}{3} \times \frac{700}{3} = \frac{1}{1024} \times l \Rightarrow \frac{7000}{9} \times 1024 = l

So, length of the wire

l = 796444.44 cm

Option 1 is correct.

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