The reaction for complete combustion of acetylene is
$$\rm 2C_2H_2 + 5O_2 \rightarrow 4 CO_2 + 2H_2O$$
...you'll get complete combustion when acetylene and oxygen are mixed in a 2:5 mole (or volume) ratio. If you have a higher ratio than that, combustion will be incomplete. The incomplete combustion products will appear as soot and carbon monoxide.
Air is 1/5th O$_2$, so you need 5 times as much air as you'd need pure O$_2$. The optimal acetylene/air ratio should be about 2:25.
Look at the acetylene/air ratios for each tube. The gases are both at the same temperature and pressure, so the volume ratio is equal to the mole ratio. Any ratio that's greater than 2/25 will probably produce soot, and the higher the ratio, the more soot you'd expect. Any tube that contains an appreciable amount of oxygen mixed in with the acetylene will burn quickly.
The full tube burned slow because oxygen needed to mix with acetylene at the mouth of the tube. The little bit of acetylene that burned there produced soot, because the acetylene/air ratio was probably much higher than 2/25.
The tubes that were 3/4 and 1/2 full had 3:1 and 1:1 acetylene/air ratios. Both ratios are fuel-rich, so soot was produced (you'd expect more soot with the 3/4 full tube than the 1/2 full tube). Both tubes burned quickly because there was a fair amount of O$_2$ mixed with the acetylene.
The tube that was 1/10 full had a 1:9 acetylene/air ratio. 1/9 is just a bit higher than 2/25, so this tube was fuel-rich, too, but much less so than any of the other tubes. Very little soot would be expected; maybe none, if more air was drawn into the tube as the acetylene burned.
