| Date | November 2017 | Marks available | 2 | Reference code | 17N.3.HL.TZ0.11 |
| Level | Higher level | Paper | Paper 3 | Time zone | Time zone 0 |
| Command term | Calculate | Question number | 11 | Adapted from | N/A |
Question
The diagram shows a simplified model of a Galilean thermometer. The thermometer consists of a sealed glass cylinder that contains ethanol, together with glass spheres. The spheres are filled with different volumes of coloured water. The mass of the glass can be neglected as well as any expansion of the glass through the temperature range experienced. Spheres have tags to identify the temperature. The mass of the tags can be neglected in all calculations.
Each sphere has a radius of 3.0 cm and the spheres, due to the different volumes of water in them, are of varying densities. As the temperature of the ethanol changes the individual spheres rise or fall, depending on their densities, compared with that of the ethanol.
The graph shows the variation with temperature of the density of ethanol.
Using the graph, determine the buoyancy force acting on a sphere when the ethanol is at a temperature of 25 °C.
When the ethanol is at a temperature of 25 °C, the 25 °C sphere is just at equilibrium. This sphere contains water of density 1080 kg m–3. Calculate the percentage of the sphere volume filled by water.
The room temperature slightly increases from 25 °C, causing the buoyancy force to decrease. For this change in temperature, the ethanol density decreases from 785.20 kg m–3 to 785.16 kg m–3. The average viscosity of ethanol over the temperature range covered by the thermometer is 0.0011 Pa s. Estimate the steady velocity at which the 25 °C sphere falls.
Markscheme
density = 785 «kgm−3»
«\(\frac{4}{3}\pi {\left( {0.03} \right)^3} \times 785 \times 9.8\) =» 0.87 «N»
Accept answer in the range 784 to 786
\(\frac{{0.87}}{{\frac{4}{3}\pi {{\left( {0.03} \right)}^3} \times 1080 \times 9.8}}\)
OR
\(\frac{{0.87}}{{1080 \times 1.13 \times {{10}^{ - 4}}}}\)
OR
\(\frac{{785}}{{1080}}\)
0.727 or 73%
Allow ECF from (a)(i)
use of drag force to obtain \({\frac{4}{3}\pi }\)r3 x 0.04 x g = 6 x \(\pi \) x 0.0011 x r x v
v = 0.071 «ms–1»
