# 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 $1.5 \cdot m$ length of a pipe made from this alloy if its OUTER radius was $1.75 \times {10}^{-} 2 \cdot m$, and the BORE of the pipe, its inner radius, was $1.25 \times {10}^{-} 2 \cdot m$?

Sep 3, 2017

Approx......$2 \cdot k g$.........

#### Explanation:

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

i.e. $\pi \times {\left(1.75 \times {10}^{-} 2 \cdot m\right)}^{2} \times 8.900 \times {10}^{3} \cdot k g \cdot {m}^{-} 3 \times 1.5 \cdot m$

$= 8.56 \cdot k g$

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

$\pi \times {\left(1.25 \times {10}^{-} 2 \cdot m\right)}^{2} \times 8.900 \times {10}^{3} \cdot k g \cdot {m}^{-} 3 \times 1.5 \cdot m$

$= 6.55 \cdot k g$

And the mass of the given pipe is simply $\left(i\right) - \left(i i\right) = 2.01 \cdot k g$....

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 $10 \cdot m m$ or $\frac{1}{2} \cdot \text{inch}$ pipe, we always speak of the internal bore of the pipe. An engineer would speak of tube diameters, i.e. $10 \cdot m m$ outer diameter.......pipe versus tube is always an issue in the lab.