**Question:**

A metallic right circular cone 20 cm high and whose vertical angle is 90° is cut into two parts at the middle point of its axis by a plane parallel to the base. If the frustum so obtained be drawn into a wire of diameter (1/16) cm, find the length of the wire.

**Solution:**

We have the following situation

Let ABC be the cone. The height of the metallic cone is AO=20cm. The cone is cut into two parts at the middle point of its axis. Hence, the height of the

frustum cone is AD=10cm. Since, the angle A is right angled, so each of the angles B and C are 45 degrees. Also, the angles E and F each are equal to 45

degrees. Let the radii of the top and bottom circles of the frustum cone are *r**1* cm and *r**2* cm respectively.

From the triangle ADE, we have

$\frac{\mathrm{DE}}{\mathrm{AD}}=\cot 45^{\circ}$

$\Rightarrow \frac{r_{1}}{10}=1$

$\Rightarrow r_{1}=10 \mathrm{~cm}$

From the triangle AOB, we have

$\frac{\mathrm{OB}}{\mathrm{OA}}=\cot 45^{\circ}$

$\Rightarrow \frac{r_{2}}{20}=1$

$\Rightarrow r_{2}=20 \mathrm{~cm}$

The volume of the frustum cone is

$V=\frac{1}{3} \pi\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right) \times h$

$=\frac{1}{3} \pi\left(10^{2}+10 \times 20+20^{2}\right) \times 10$

$=\frac{1}{3} \times \frac{22}{7} \times 700 \times 10$

$=\frac{22000}{3} \mathrm{~cm}^{3}$

The radius of the wire is $\frac{1}{32} \mathrm{~cm}$. Let the length of the wire be $/ \mathrm{cm}$. Then, the volume of the wire is

$V_{1}=\pi\left(\frac{1}{32}\right)^{2} \times l \mathrm{~cm}^{3}$

Since, the frustum is drawn in the wire, their volumes must be equal. Hence, we have

$V_{1}=V$

$\Rightarrow \pi\left(\frac{1}{32}\right)^{2} \times l=\frac{22000}{3}$

$\Rightarrow \quad l=\frac{22000 \times(32)^{2} \times 7}{3 \times 22}$

$\Rightarrow \quad l=\frac{1000 \times(32)^{2} \times 7}{3}$

$\Rightarrow \quad l=2389333.33 \mathrm{~cm}$

$\Rightarrow \quad l=23893.33 \mathrm{~m}$

Hence, the length of the wire is 23893.33 m.

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.