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Question 19N.3.HL.TZ0.9

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Date November 2019 Marks available [Maximum mark: 6] Reference code 19N.3.HL.TZ0.9
Level HL Paper 3 Time zone TZ0
Command term Determine, Explain Question number 9 Adapted from N/A
9.
[Maximum mark: 6]
19N.3.HL.TZ0.9

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.

(a)

Explain why the levels of the liquid are at different heights.

[3]

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» ✔

(b)

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]

Markscheme

«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» ✔