A solid metallic right circular cone 20cm high and whose vertical…

Question

A solid metallic right circular cone 20cm 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 $\frac{1}{12}\text{cm},$ then find the length of the wire.

✓

Answer

We have,
Height of the solid metallic cone, H = 20cm,
Height of the frustom, $\text{h}=\frac{20}{2}=10\text{cm}$ and
Radius of the wire $=\frac{1}{24}\text{cm}$
Let the length of the wire be l, BG = r and BO = R.
In $\triangle\text{AEG},$
$\tan30^\circ=\frac{\text{EG}}{\text{AG}}$
$\Rightarrow\frac{1}{\sqrt3}=\frac{\text{r}}{\text{H}-\text{h}}$
$\Rightarrow\frac{1}{\sqrt3}=\frac{\text{r}}{\text{20}-\text{10}}$
$\Rightarrow\text{r}=\frac{10}{\sqrt3}\text{cm}$
Also, in $\triangle\text{ABD},$
$\tan30^\circ=\frac{\text{BD}}{\text{AD}}$
$\Rightarrow\frac{1}{\sqrt3}=\frac{\text{R}}{\text{H}}$
$\Rightarrow\frac{1}{\sqrt3}=\frac{\text{R}}{\text{20}}$
$\Rightarrow\text{R}=\frac{20}{\sqrt3}\text{cm}$
Now,
Volume of the wire = Volume of the frustum
$\Rightarrow\pi\Big(\frac{1}{24}\Big)^2\text{l}=\frac{1}{3}\pi\text{h}\big(\text{R}^2+\text{r}^2+\text{Rr}\big)$
$\Rightarrow\frac{\text{l}}{576}=\frac{1}{3}\times10\times\bigg[\Big(\frac{20}{\sqrt3}\Big)^2+\Big(\frac{10}{\sqrt3}\Big)^2+\Big(\frac{20}{\sqrt3}\Big)\Big(\frac{10}{\sqrt3}\Big)\bigg]$
$\Rightarrow\text{l}=\frac{576}{3}\times10\times\Big[\frac{400}{3}+\frac{100}{3}+\frac{200}{3}\Big]$
$\Rightarrow\text{l}=\frac{576}{3}\times10\times\frac{700}{3}$
$\Rightarrow\text{l}=448000\text{cm}$
$\therefore\text{l}=4480\text{m}$
So, the length or the wire Is 4460m.