A cylindrical tube, open at both ends, is made of 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 $704 \mathrm{~cm}^{3}$.