 Use the instrument precision.
 Instrument precision is defined as the
‘smallest non-zero reading measured by the
instrument’,
and refers to the actual graduations on the scale of the
instrument. A metre ruler with mm graduations would
have a precision of + 1 mm.
For single readings, or multiple identical readings, taken
by this type of ruler the estimate of uncertainty would be
+ 1mm
 Calculate the mean value
 Estimate the uncertainty using:
Uncertainty = 0.5 X spread
e.g. 5 readings of length:
0.15mm, 0.12mm, 0.16mm, 0.13mm, 0.14mm
Mean = 0.14mm
Uncertainty = 0.5 x (0.16-0.12) = 0.02mm
 Length = 0.14 ± 0.02mm
Calculate the % uncertainty in the area of a metal wire
given that diameter is 0.16 ± 0.01 mm
% uncertainty in diameter = (0.01/0.16) × 100 = ± 6.3 %
area is proportional to diameter squared
 % uncertainty in area = 6.3 + 6.3 = ± 12.6 %
Calculate % uncertainty in the speed measurement
given the following uncertainties in distance and time
measurements
% uncertainty in distance = ± 0.5 %
% uncertainty in time = ± 2.5 %
% uncertainty in speed = 0.5 + 2.5 = ± 3.0 %
 Use the format:
variable symbol – solidus – unit
 You should not name the variable in full.
e.g. eg ‘V/mV’ is correct, ‘output pd of solar cell in
millivolts’ is not
 Unless directed otherwise, put all data in one
table, from raw data on the left, to processed
values for plotting on the right.
1 mark for each of the following points:
 Labelling of axes
 Suitable Scales
 Line of best Fit
1 mark for each of the following points:
 Plotting of points
 Line of best fit
 Reading data for calculation of gradient, and suitable
size ‘triangle’
 Gradient value, with appropriate significant figures
Marking the origin correctly on a graph,
eg PHAB3X Sec A Part 1 Q2(b)
100
100
80
80
60
60
0
0
0
1
2
3
Unacceptable:
the marking of the origin as above
produces a non-linear scale
which will always be penalised.
0
1
2
3
Solution:
use of the broken scale
convention resolves the problem
but watch out if a gradient
calculation is then required.
100
100
80
80
60
60
40
0
1
2
3
Unacceptable:
leaving an origin unmarked on
either axis will not be accepted;
the scale will still be treated as
non-linear since the origin is now
ambiguous.
0
1
2
3
Solution:
use of a false origin is acceptable
but candidates should be careful if
they are then asked to calculate the
gradient.
Finding an intercept which cannot be read directly,
eg PHAB3X Section B Q1(a)
white rectangle represents
edge of grid printed in answer
booklet
While the intercept on the horizontal
axis can be read directly, the vertical
intercept can not.
Candidates gain no credit for
extending the line off the grid into
the margin.
When answering Section B,
candidates should not be given the
opportunity to re-plot graphs.
The use of algebra is expected if
the intercept cannot be read directly.
0
0
Significant Figures in a Previous Paper:
foruse
usein
inanswering
answering part
for
part(a)
(a)
L/cm
R/W
R/W
6.6
6.6
10.6
10.6
13.8
13.8
17.8
17.8
21.4
21.4
2.9
2.9
7.6
7.6
13.0
13.0
21.6
21.6
30.4
30.4
0.067
0.0666
[0.067]
0.068
0.0676
[0.068]
0.068
0.0683
0.068
0.0682
0.066
0.0664
To test the theory that R = kL2 candidates were expected to evaluate
R/L2 for every row of the table.
Many candidates forfeited marks because they truncated their results to
2 significant figures; in at least 3 rows of the table, 3 sf was justified.
Tables
 Independent variable must be in left hand column of table
 Quote to max possible significant figures (e.g. 1.10 V if
instrument precision is 0.01V)
Graphs
 Points on a graph should cover at least half the grid
horizontally and vertically.
 All points must be plotted
 Marks are forfeited if plotted > 1mm from correct position
 Marks can be forfeited if > 2mm from trend line
 Use the convention variable symbol – solidus – unit for table
headings and graph axes; a bracket is essential if the log of a
variable is involved, eg ln(variable/unit)
 When compressing a graph scale use the broken scale
convention if marking the origin (0, 0) otherwise mark a false
origin
 AS and A2 candidates should be able to calculate the intercept
on a graph if this cannot be read directly
 The result of a calculation should be to the same number of
significant figures as the least accurate data used in the
calculation