Math Help - Weight of pipe

1. Weight of pipe

An iron pipe is 21 m long & its exterior diameter is 8 cm. If the thickness of the pipe is 1 cm & iron weight 8 g/cu.cm, find the weight of the pipe.

How do I do this? I have found the volume of the pipe by first finding the external volume, then the internal volume & then finding the difference but what do I do after that to find the weight?

Thanks,

Ron

2. Just multiply by the given density of iron. Note that you're asked to find the weight, not the mass. Thus, you'll need also to multiply by g.

3. I tried what you suggested but am not getting the correct answer. This is what I did:

External Volume=22*4*4*2100/7=105600 cu.cm
Internal Volume=22*3*3*2100/7=59400 cu.cm
Volume=105600-59400=46200 cu.cm

Weight (or mass...whatever it is)=46200*8=369600 gm=369.6 kg

But the answer is 92.4 kg.

Thanks,

Ron

4. I get essentially what you get. I think the figure of 92.4 kg is incorrect.

5. Assuming it's a cylinder then, as for all prisms the volume is cross sectional area multiplied by length or $V = \pi r^2 h = \dfrac{\pi}{4}d^2h$. You may also have a unit mismatch - your length is in metres but the diameter is cm!

$V_1 = \dfrac{\pi}{4} \cdot 8^2 \cdot 2100 = 33600\pi \text{ cm}^3$ -- this is the whole cylinder

Edited this lineBy the same token: $V_2 = 18900\pi \text{ cm}^3$ (you should see if you can figure out where this figure came from) -- this is the volume of empty air on the insde

Edited this lineHence the difference in volumes $\Delta V = V_1 - V_2 = 33600\pi - 18900\pi = 14700\pi \text{ cm}^3$ -- this is the volume taken up by the tube itself

$m = \rho V = 8 \cdot 10^{-3} \cdot 14700\pi = 117.6\pi \approx 369.45 \text{kg , (3sf)}$

The weight (your question erroneously uses kg to describe weight so they might mean find the mass) is given by $W = mg = 117.6g \pi \text{ N}$

I don't get your answer but I can't see where I went wrong.

Edited - thanks to Ackbeet pointing out where I went wrong

6. e^(i pi):

The thickness of the pipe is 1 cm. You used 7 as the inner diameter, but the thickness has to be subtracted twice from the outer diameter in order to get the inner diameter.