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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.

[3]
a.

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.

[3]
b.

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» ✔

a.

«ρliquid gh=0.5×ρair v2»

difference in pressure PB-PA=8.7×102×9.8×0.06=510 «Pa» ✔

correct substitution into the Bernoulli equation, eg: 12×1.2v2=510 ✔

v=29 «ms–1» ✔

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Option B: Engineering physics » Option B: Engineering physics (Additional higher level option topics) » B.3 – Fluids and fluid dynamics (HL only)
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Option B: Engineering physics » Option B: Engineering physics (Additional higher level option topics)
Option B: Engineering physics

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