**Question:**

A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.

**Solution:**

Given data is as follows:

Internal diameter = 10.4 cm

Thickness of the metal = 8 mm

Length of the pipe = 25 cm

We have to find the volume of the metal used in the pipe.

We know that,

Volume of the hollow pipe $=\pi\left(R^{2}-r^{2}\right) h$

Given is the internal diameter which is equal to 10.4 cm .Therefore,

r = 10.4/2

r = 5.2 cm

Also, thickness is given as 8 mm. Let us convert it to centimeters.

Thickness = 0.8 cm

Now that we know the internal radius and the thickness of the pipe, we can easily find external radius 'R'.

R = 5.2 + 0.8

R = 6 cm

Therefore, Volume of metal in the pipe $=22 / 7 \times\left(6^{2}-5.2^{2}\right) \times 25$

$=704 \mathrm{~cm}^{3}$

Therefore, the volume of metal present in the hollow pipe is $704 \mathrm{~cm}^{3}$.