Question

A metallic right circular cone 20 cm high and whose vertical angle is ${60}^{\circ }$ 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 $\frac{1}{16}cm$, find the length of the wire. [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 3 Marks

Height of right circular cone $=h=20cm$

Right circular cone is cut into two parts at the middle.

We get right circular cone of height 10 cm and frustum of cone of height 10 cm.

Height of the frustum of the cone $=h_1=10cm$

Let radius of the upper end of the frustum of the cone $=r_{1}cm$

Let radius of the lower end of the frustum of the cone $=r_{2}cm$



We can find $r_1$ from $\Delta APQ$.

$\frac{r_1}{AQ}=tan{30}^{\circ }$

$\Rightarrow \frac{r_1}{10}=\frac{1}{\sqrt{3}}$

$\Rightarrow r_1=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3}cm$

Similarly, we can find $r_2$ from $\Delta ABC$

$\frac{r_2}{AC}=tan{30}^{\circ }$

$\Rightarrow \frac{r_2}{20}=\frac{1}{\sqrt{3}}$

$\Rightarrow r_2=\frac{20}{\sqrt{3}}=\frac{20\sqrt{3}}{3}cm$

Let length of wire = $l$ cm

Radius of wire $=r=\frac{1}{16\times 2}=\frac{1}{32}cm$

According to given condition we have volume of the frustum of the cone = volume of wire

$\Rightarrow \frac{1}{3}.\pi .h_1\left(\right(r_1{)}^2+(r_2{)}^2+(r_1\left)\right(r_2\left)\right)=\pi .r^2.l$

$\Rightarrow \frac{1}{3}\times 10(\frac{100}{3}+\frac{400}{3}+\frac{200}{3})=\frac{1}{32}\times \frac{1}{32}\times l$

$\Rightarrow l=\frac{1}{3}\times 10\left(\frac{700}{3}\right)\times 32\times 32$

$=796444.44cm=7964.44m$ (100 cm= 1m)