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

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The figure for the problem is shown below :

                                                   Surface area and volume ,     25864

Using geometry we can write :

                                        EG\ =\ \frac{10\sqrt{3}}{3}\ cm              and            BD\ =\ \frac{20\sqrt{3}}{3}\ cm

Thus the volume of the frustum is given by :

                                                        \\=\ \frac{1}{3}\pi h\left ( r^2_1\ +\ r^2_2\ +\ r_1r_2 \right )\\\\=\ \frac{1}{3}\pi \times 10\left ( \left ( \frac{10\sqrt{3}}{3} \right )^2\ +\ \left ( \frac{20\sqrt{3}}{3} \right )^2\ +\ \left ( \frac{10\sqrt{3}}{3} \right )\times \left ( \frac{20\sqrt{3}}{3} \right ) \right )\\\\=\ \frac{22000}{9}\ cm^3                                   

Now, the radius of the wire is : 

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

Thus the volume of wire is given by :     =\ \pi r^2\times l

                                                        =\ \pi \times \left ( \frac{1}{32} \right )^2\times l

Now equating volume of frustum and wire, we get :

                                          \frac{22000}{9}\ =\ \pi\times \left ( \frac{1}{32} \right )^2\times l

                                                   l\ =\ 796444.44\ cm

                                                 l\ =\ 7964.44\ m 

Thus the length of wire drawn is 7964.44 m.

Posted by

Devendra Khairwa

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