# 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 $\dpi{100} \frac{1}{16}\textup{cm}$, find the length of the wire.

The figure for the problem is shown below :

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.

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