Date | November 2019 | Marks available | 3 | Reference code | 19N.3.HL.TZ0.9 |
Level | Higher level | Paper | Paper 3 | Time zone | 0 - no time zone |
Command term | Determine | Question number | 9 | Adapted from | N/A |
Question
A Pitot tube shown in the diagram is used to determine the speed of air flowing steadily in a horizontal wind tunnel. The narrow tube between points A and B is filled with a liquid. At point B the speed of the air is zero.
Explain why the levels of the liquid are at different heights.
The density of the liquid in the tube is 8.7 × 102 kg m–3 and the density of air is 1.2 kg m–3. The difference in the level of the liquid is 6.0 cm. Determine the speed of air at A.
Markscheme
air speed at A greater than at B/speed at B is zero
OR
total/stagnation pressure «PB» – static pressure «PA» = dynamic pressure ✔
so PA is less than at PB (or vice versa) «by Bernoulli effect» ✔
height of the liquid column is related to «dynamic» pressure difference «hence lower height in arm B» ✔
«»
difference in pressure «Pa» ✔
correct substitution into the Bernoulli equation, eg: ✔
«ms–1» ✔