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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]

The density of the liquid in the tube is 8.7 × 10²kg 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]

Markscheme

air speed at A greater than at B/speed at B is zero
OR
total/stagnation pressure «P_B» – static pressure «P_A» = dynamic pressure ✔

so P_A is less than at P_B (or vice versa) «by Bernoulli effect» ✔

height of the liquid column is related to «dynamic» pressure difference «hence lower height in arm B» ✔

« $ρ_{liquid} g h = 0.5 \times ρ_{air} v^{2}$ »

difference in pressure $P_{B} - P_{A} = 8.7 \times 10^{2} \times 9.8 \times 0.06 = 510$ «Pa» ✔

correct substitution into the Bernoulli equation, eg: $\frac{1}{2} \times 1.2 v^{2} = 510$ ✔

$v = 29$ «ms^–1» ✔

Examiners report

[N/A]

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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