An alloy has a density of ρ=8.900×103⋅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 1.75×10−2⋅m, and the BORE of the pipe, its inner radius, was 1.25×10−2⋅m?

1 Answer
anor277
Sep 3, 2017

Approx......2⋅kg.........

Explanation:

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

i.e. π×(1.75×10−2⋅m)2×8.900×103⋅kg⋅m−3×1.5⋅m

=8.56⋅kg

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

π×(1.25×10−2⋅m)2×8.900×103⋅kg⋅m−3×1.5⋅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 12⋅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.

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