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Date	May 2017	Marks available	2	Reference code	17M.3.HL.TZ1.9
Level	Higher level	Paper	Paper 3	Time zone	1
Command term	Show that	Question number	9	Adapted from	N/A

Question

An air bubble has a radius of 0.25 mm and is travelling upwards at its terminal speed in a liquid of viscosity 1.0 × 10^–3 Pa s.

The density of air is 1.2 kg m^–3 and the density of the liquid is 1200 kg m^–3.

Explain the origin of the buoyancy force on the air bubble.

[2]

With reference to the ratio of weight to buoyancy force, show that the weight of the air bubble can be neglected in this situation.

[2]

Calculate the terminal speed.

[2]

Markscheme

ALTERNATIVE 1

pressure in a liquid increases with depth

so pressure at bottom of bubble greater than pressure at top

ALTERNATIVE 2

weight of liquid displaced

greater than weight of bubble

[2 marks]

$\frac{{{\text{weight}}}}{{{\text{bouyancy}}}}\left( { = \frac{{V{\rho _a}g}}{{V{\rho _l}g}} = \frac{{{\rho _a}}}{{{\rho _l}}} = \frac{{1.2}}{{1200}}} \right) = {10^{ - 3}}$ $\frac{{{\text{weight}}}}{{{\text{bouyancy}}}}\left( { = \frac{{V{\rho _a}g}}{{V{\rho _l}g}} = \frac{{{\rho _a}}}{{{\rho _l}}} = \frac{{1.2}}{{1200}}} \right) = {10^{ - 3}}$

since the ratio is very small, the weight can be neglected

Award [1 max] if only mass of the bubble is calculated and identified as negligible to mass of liquid displaced.

[2 marks]

evidence of equating the buoyancy and the viscous force « ${\rho _l}\frac{4}{3}\pi {r^3}g = 6\pi \eta r{v_t}$ ${\rho _l}\frac{4}{3}\pi {r^3}g = 6\pi \eta r{v_t}$ »

v_t = « $\frac{2}{9}\frac{{1200 \times 9.81}}{{1 \times {{10}^{ - 3}}}}{\left( {0.25 \times {{10}^{ - 3}}} \right)^2} =$ $\frac{2}{9}\frac{{1200 \times 9.81}}{{1 \times {{10}^{ - 3}}}}{\left( {0.25 \times {{10}^{ - 3}}} \right)^2} =$ » 0.16 «ms^–1»

[2 marks]

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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