| Date | May 2017 | Marks available | 2 | Reference code | 17M.3.HL.TZ1.9 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Explain | 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.
With reference to the ratio of weight to buoyancy force, show that the weight of the air bubble can be neglected in this situation.
Calculate the terminal speed.
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]
\(\frac{{{\text{weight}}}}{{{\text{bouyancy}}}}\left( { = \frac{{V{\rho _a}g}}{{V{\rho _l}g}} = \frac{{{\rho _a}}}{{{\rho _l}}} = \frac{{1.2}}{{1200}}} \right) = {10^{ - 3}}\)
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 «\({\rho _l}\frac{4}{3}\pi {r^3}g = 6\pi \eta r{v_t}\)»
vt = «\(\frac{2}{9}\frac{{1200 \times 9.81}}{{1 \times {{10}^{ - 3}}}}{\left( {0.25 \times {{10}^{ - 3}}} \right)^2} = \)» 0.16 «ms–1»
[2 marks]
