Question

An alloy has a density of `rho=8.900xx10^3*kg*m^-3`...can you address the following question?

What would be the mass of a `1.5*m` length of a pipe made from this alloy if its OUTER radius was `1.75xx10^-2*m`, and the BORE of the pipe, its inner radius, was `1.25xx10^-2*m`?

Answer 1

Approx......`2*kg`.........

Explanation:

We work out (i) the mass of a solid pipe......

i.e. `pixx(1.75xx10^-2*m)^2xx8.900xx10^3*kg*m^-3xx1.5*m`

`=8.56*kg`

And (ii) we work the mass of an inner pipe, i.e. the mass that an internal pipe would have......

`pixx(1.25xx10^-2*m)^2xx8.900xx10^3*kg*m^-3xx1.5*m`

`=6.55*kg`

And the mass of the given pipe is simply `(i)-(ii)=2.01*kg`....

Note that it might help you to draw a diagram showing the cross-section of the pipe, and the areas of the two cross-sections we need. Good luck.

Note that in plumbing, when we talk of say `10*mm` or `1/2*"inch"` pipe, we always speak of the internal bore of the pipe. An engineer would speak of tube diameters, i.e. `10*mm` outer diameter.......pipe versus tube is always an issue in the lab.