| Date | May 2015 | Marks available | 4 | Reference code | 15M.2.HL.TZ1.1 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 1 |
| Command term | Calculate, Comment, Deduce, Draw, Explain, and State | Question number | 1 | Adapted from | N/A |
Question
This question is about the flow of liquids.
A student carries out an experiment to investigate how the rate of flow R of water through a narrow tube varies with the pressure difference across the tube. The pressure difference is proportional to the height h shown in the diagram. The student measures h in cm with a metre ruler. R is obtained by measuring the volume of water collected in a measuring cylinder in a time of 100s.
The equation of the trend line shown in (b) is given by
R = −0.0005h2 + 0.0843h − 1.5632.
(i) Calculate the value of R for h = 0.
(ii) State why this value of R is not physically possible.
(iii) State the number of significant figures that you have used for your value in (c)(i).
(iv) Comment, with reference to the experimental data, on the number of significant figures that you have used for your value in (c)(i).
The student estimates that the uncertainty in timing 100s is ±1s. Using the data on the graph, deduce the absolute uncertainty in the volume of water collected when R = 2.1 units.
Markscheme
(i) R0 = –1.6(–1.5632)cm3 s–1 ;
(ii) it is negative / this would mean water running uphill / gaining potential energy / OWTTE;
(iii) an answer consistent with candidate’s value;
(iv) should be 2 significant figures (in line with data values);
% uncertainty in t = 1%;
% uncertainty in V (= 5 + 1)=6% or % uncertainty in V (= 5 − 1) = 4%;
V (= 2.1 100) = 210 (units);
absolute uncertainty (= 210 × 6%) = 12.6 / 13 / 10 (units) or absolute uncertainty (= 210 × 4%) = 8.4 / 8 (units);
