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#?

1 Answer
Sep 3, 2017

Answer:

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.