Pupil Notes
Measurements and their errors
A Level Physics
3.1 Measurements and their errors
Pupil Notes
3.1.1 Use of SI units and their prefixes
Any measurement/physical quantity consists of 2 important parts:
1.
2.
Its magnitude (size/the number itself);
Its unit.
Without both of these parts, the physical quantity has no meaning.
For example, a person’s mass may quoted as ‘50 kilograms’ but without the ‘kilograms’ the situation in
unclear. Is their mass 50 ounces, pounds, stone, grams, kilograms, tonnes?
Remember to include appropriate units at all times!
In addition, units must be stated in the accepted SI format. For example speed/velocity is measured in
metres per second which must be written as m s-1 as opposed to m/s which you might be more familiar
with. Other examples include acceleration which is measured in metres per second per second or
metres per second squared. This must be written as m s-2 as opposed to m/s/s or m/s2. Note that there
is a space between each individual unit: ms -1 is per millisecond and not m s-1 or metres per second.
The International System (SI) of units contains 7 fundamental (sometimes called base) units which are
as follows:
Physical Quantity
Symbol
SI Unit
SI Symbol
mass
m
kilogram
kg
length
l
metre
m
time
t
second
s
electric current
I
ampere
A
mole
mol
kelvin
K
candela
cd
amount of substance
temperature
T
luminous intensity
Physical quantities, except those that contain fundamental units, have units that are combinations of the
7 fundamental units. These are called derived units.
For example:
speed
=
distance/time
units of speed
=
units of distance/units of time
units of speed
=
metres/seconds
units of speed
=
m/s
units of speed
=
m s-1
acceleration
=
change in velocity/time taken for change
units of acceleration =
units of distance/units of time
units of acceleration =
metres per second/seconds
units of acceleration =
m s-1/s
1
units of acceleration =
(ms-1)s-1
units of acceleration =
m s-2
force
=
mass x acceleration
units of force
=
units of mass x units of acceleration
units of force
=
kg x ms-2
units of force
=
kg m s-2
In some cases, particular combinations of fundamental units are used so often that they are given a
new derived name. For example, the unit of force is a derived unit (kg m s-2). This unit is given a new
name, the newton (N), so that 1 N = 1 kg m s-2.
energy
=
force x distance
units of energy
=
units of force x units of distance
units of energy
=
kg m s-2 x m
units of energy
=
(kg m s-2)m
units of energy
=
kg m 2 s-2
The same goes for energy. The unit of energy is a derived unit (kg m2 s-2). This unit is given a new
name, the joule (J), so that 1 J = 1 kg m2 s-2. Alternatively, 1 J = 1 Nm.
power
=
energy/time
units of power
=
units of energy/units of time
units of power
=
kg m2 s-2/s
units of power
=
(kg m2 s-2)s-1
units of power
=
kg m2 s-3
The same goes for power. The unit of power is a derived unit (kg m2 s-3). This unit is given a new
name, the watt (W), so that 1 W = 1 kg m2 s-3. Alternatively, 1 W = 1 Js-1.
The period of a pendulum, T, is related to its length, l, and the acceleration due to gravity, g, using the
following equation:
T = 2
l
g
The unit for the period of a pendulum, T, can be deduced by deriving it from the units of the physical
quantities contained within the equation of which there are 2 in this case.
period of a pendulum
=
units of the period of a pendulum
=
constant x
length
acceleration
m
m s −2
2
units of the period of a pendulum
=
1
s −2
units of the period of a pendulum
=
s2
units of the period of a pendulum
=
s
The table below lists SI derived units:
SI
Derived
Unit
SI
Derived
Unit
Symbol
Physical Quantity
Symbol
SI
Base
Unit
Alternative
SI Derived
Unit
force
F
kg m s-2
newton
N
pressure
P
kg m-1 s-2
pascal
Pa
frequency
f
s-1
hertz
Hz
energy
E
kg m2 s-2
joule
J
Nm
power
P
kg m2 s-3
watt
W
Js-1
electric charge
Q
As
coulomb
C
electric potential difference
V
kg m2 s-3 A-1
volt
V
W/A
resistance
R
kg m2 s-3 A-2
ohm

V/A
flux

kg m2 s-2 A-1
weber
Wb
Vs
magnetic field strength
B
kg s-2 A-1
tesla
T
Wbm-2
activity
A
s-1
becquerel
Bq
absorbed dose
D
m2 s-2
gray
Gy
Jkg-1
dose equivalent
H
m2 s-2
sievert
Sv
Jkg-1
Nm-2
Conversion between Different Units of Physical Quantities
How many seconds are there in 1 year?
years
days
1 year =
365 days
x 365
1 year =
hours
=
8760 h
x 24
minutes
=
525 600 s
x 60
seconds
=
31 535 000 s
x 60
31 535 000 s
How many years are there in 1 s?
seconds
minutes
1s
0.017 min
=
 60
1s
=
hours
=
 60
2.8 x 10-4 h
days
=
 24
1.2 x 10-5 days
years
=
3.1 x 10-8 years
 365
3.1 x 10-8 years
3
Conversion rules:
years to seconds
x 31 536 000
seconds to years
 31 536 000
How many joules of energy are there in 1 kWh?
As energy = power x time, 1 J = 1 Ws so kilowatts must be converted into watts and hours must be
converted into seconds.
kilowatt hours
(joules)
1 kWh
watt hours
=
1000 Wh
x 1000
1 kWh
=
=
watt minutes
60 000 Wmin
x 60
watt
=
seconds
3 600 000 Ws (J)
x 60
3 600 000 J
Scientific Notation and Prefixes
In Physics you will be working with very large and very small numbers.
In Astronomy you will be dealing with very large distances. For example, the distance from Earth to the
nearest star is 82 000 000 000 000 000 metres.
In atomic Physics you will be dealing with very small distances. For example, the spacing between
atoms in a solid is about 0.000 000 000 1 metres.
It is not convenient to work with numbers written out in full. For this reason, it is usual when dealing
with very large or very small numbers to use scientific notation.
Large Numbers
▪
5 000 000 can be written as 5 x 106.
5 x 106
-
5 multiplied by 10 six times
5 x 10 x 10 x 10 x 10 x 10 x 10
Another way of looking at this is as follows:
Firstly insert a decimal point after the first number.
5.0 x 106
▪
-
5000000.0
the decimal point moves to the right 6 places
82 000 000 000 000 000 can be written as 8.2 x 1016.
8.2 x 1016
-
-
8.2 multiplied by 10 sixteen times
8.2 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10
Another way of looking at this is as follows:
4
8.2 x 1016
-
82000000000000000.0
the decimal point moves to the right 16 places
82 000 000 000 000 000 can also be written as 0.82 x 10 17or 82 x 1015.
▪
0.82 x 1017
-
0.82 multiplied by 10 seventeen times
- 0.82 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10
Another way of looking at this is as follows:
▪
0.82 x 1017
-
082000000000000000.0
the decimal point moves to the right 17 places
82 x 1015
-
82 multiplied by 10 fifteen times
- 82 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10
Another way of looking at this is as follows:
Firstly insert a decimal point after the first number.
82.0 x 1015
-
82 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 . 0
the decimal point moves to the right 15 places
Small Numbers
▪
0.0000005 can be written as 5 x 10-7.
5 x 10-7
-
5 divided by 10 seven times
5 / 10 / 10 / 10 / 10 / 10 / 10 / 10
Another way of looking at this is as follows:
Firstly insert a decimal point after the first number.
5.0 x 10-7
▪
-
0 .0 0 0 0 0 0 5 0
the decimal point moves to the left 7 places
0.0000000001 can be written as 1 x 10-10.
1 x 10-10
-
1 divided by 10 ten times
1 / 10 / 10 / 10 / 10 / 10 / 10 / 10 / 10 / 10 / 10
Another way of looking at this is as follows:
Firstly insert a decimal point after the first number.
1.0 x 10-10
-
0 .0 0 0 0 0 0 0 0 0 1 0
the decimal point moves to the left 10 places
0.0000000001 can also be written as 0.1 x 10-9, 10 x 10-11, etc.
5
▪
0.1 x 10-9
-
0.1 divided by 10 nine times
0.1 / 10 / 10 / 10 / 10 / 10 / 10 / 10 / 10 / 10
Another way of looking at this is as follows:
▪
0.1 x 10-9
-
0 .0 0 0 0 0 0 0 0 0 1
the decimal point moves to the left 9 places
10 x 10-11
-
10 divided by 10 eleven times
10 / 10 / 10 / 10 / 10 / 10 / 10 / 10 / 10 / 10 / 10 / 10
Another way of looking at this is as follows:
Firstly insert a decimal point after the first number.
10.0 x 10-11
-
0 .0 0 0 0 0 0 0 0 0 1 0 0
the decimal point moves to the left 11 places
Questions
Problem Set 1
Convert the following numbers into scientific notation form.
1.
5500
5.5 x 10
2.
7900000
7.9 x 10
3.
960000
9.6 x 10
4.
120
1.2 x 10
5.
6100000000
6.1 x 10
6.
47000
4.7 x 10
7.
8400
8.4 x 10
8.
300000000
3.0 x 10
9.
56000
5.6 x 10
10.
28000000
2.8 x 10
Problem Set 2
Convert the following numbers from scientific notation form to decimal form.
1.
4.0 x 107
2.
3.8 x 102
3.
5.2 x 108
4.
1.5 x 103
5.
6.7 x 109
6
6.
7.9 x 104
7.
9.6 x 1010
8.
2.4 x 105
9.
0.1 x 1011
10.
8.3 x 106
Problem Set 3
Convert the following numbers into scientific notation form.
1.
0.1
1.0 x 10
2.
0.00000065
6.5 x 10
3.
0.0029
2.9 x 10
4.
0.000000074
7.4 x 10
5.
0.00038
3.8 x 10
6.
0.0000000083
8.3 x 10
7.
0.000047
4.7 x 10
8.
0.00000000092
9.2 x 10
9.
0.0000056
5.6 x 10
10.
0.00000000001
1.0 x 10
Problem Set 4
Convert the following numbers from scientific notation form to decimal form.
1.
1.9 x 10-2
2.
6.4 x 10-7
3.
2.8 x 10-3
4.
7.3 x 10-9
5.
3.7 x 10-4
6.
8.2 x 10-9
7.
4.6 x 10-5
8.
9.1 x 10-10
9.
5.5 x 10-6
10.
0.1 x 10-11
7
Prefixes
Instead of using powers of 10, we sometimes use prefixes.
Prefixes are used to denote multiples and sub-multiples of any unit used to measure a physical
quantity.
For example:
▪
Instead of saying 1000 metres, we can say 1 kilometre.
1000 m
=
1 x 103 m
(1 multiplied by 10 three times)
Another way of looking at this is as follows:
Firstly insert a decimal point after the first number.
1.0 x 103
-
1000.0
the decimal point moves to the right 3 places
1 kilometre can be written as:
1000 m
=
1 x 103 m
=
1 km
The ‘k’ stands for kilo. The k is replaced by x 103.
▪
Instead of saying 0.001 metres, we can say 1 millimetre.
0.001 m
=
1 x 10-3 m
(1 divided by 10 three times)
Another way of looking at this is as follows:
Firstly insert a decimal point after the first number.
1.0 x 10-3
-
0.0010
the decimal point moves to the left 3 places
1 millimetre can be written as:
0.001 m
=
1 x 10-3 m
=
1 mm
The ‘m’ stands for milli. The m is replaced by x 10 -3.
A list of all prefixes used is shown in the following tables.
Prefix
Symbol
Power of 10
tera
T
x 1012
x 10 12 times
multiple
giga
G
x 109
x 10 9 times
multiple
mega
M
x 106
x 10 6 times
multiple
kilo
k
x 103
x 10 3 times
multiple
centi
c
x 10-2
 10 2 times
sub-multiple
milli
m
x 10-3
 10 3 times
sub-multiple
8
micro
μ
x 10-6
 10 6 times
sub-multiple
nano
n
x 10-9
 10 9 times
sub-multiple
pico
p
x 10-12
 10 12 times
sub-multiple
femto
f
x 10-15
 10 15 times
sub-multiple
Prefix
Symbol
Standard
Notation
tera
T
x 1012
giga
G
x 109
mega
M
x 106
kilo
k
x 103
1 km
= 1 x 103 m = 1 000 m
centi
c
x 10-2
1 cm
= 1 x 10-2 m = 0.01 m
milli
m
x 10-3
1 mm
= 1 x 10-3 m = 0.001 m
micro

x 10-6
nano
n
x 10-9
pico
p
x 10-12
femto
f
x 10-15
Example
= 1 x 1012 m = 1 000 000 000 000 m
1 Tm
= 1 x 109 m = 1 000 000 000 m
1 Gm
= 1 x 106 m = 1 000 000 m
1 Mm
1 m
= 1 x 10-9 m = 0.000000001 m
1 nm
= 1 x 10-12 m = 0.000000000001 m
1 pm
1 fm
= 1 x 10-6 m = 0.000001 m
= 1 x 10-15 m = 0.000000000000001 m
Questions
Problem Set 5
Rewrite the quantities as shown in the examples below.
Example 1
56.7 mm
56.7 x 10-3 m
0.0567 m
Example 2
34 km
34.0 x 103 m
34 000 m
1.
94 cm
x 10
m
m
2.
2.5 km
x 10
m
m
3.
1.6 Gm
x 10
m
m
4.
21.3 nm
x 10
m
m
5.
60 mm
x 10
m
m
6.
0.4 cm
x 10
m
m
7.
18.8 Mm
x 10
m
m
8.
9.2 m
x 10
m
m
9.
76 km
x 10
m
m
10.
3.7 nm
x 10
m
m
9
Problem Set 6
Rewrite the quantities as shown in the examples below.
Example 1
1200 m
1.2 x 103 m
1.2 km
Example 2
340 000 m
0.34 x 106 m
0.34 Mm
Example 3
0.00567 m
5.67 x 10-3 m
5.67 mm
Example 4
0.000089 m
89.0 x 10-6 m
89 m
1.
0.05 m
x 10
m
m
2.
0.1 m
x 10
m
m
3.
0.00000007 m
x 10
m
m
4.
9000000 m
x 10
m
m
5.
0.55 m
x 10
m
m
6.
60000 m
x 10
m
m
7.
0.0000000059 m
x 10
m
m
8.
78200 m
x 10
m
m
9.
4500000 m
x 10
m
m
10.
0.00009 m
x 10
m
m
Problem Set 7
Rewrite the following physical quantities in the units shown.
Example:
1.1 x 103 m
=
1100 m
=
1.1 km =
110 000 cm
1.
3.7 x 1011 m
=
m
=
Tm =
Gm
2.
5.6 x 108 m
=
m
=
Gm
=
Mm
3.
2.3 x 107 m
=
m
=
Mm
=
km
4.
0.69 x 102 m
=
m
=
km =
cm
5.
74 x 10-2 m
=
m
=
cm
=
mm
6.
2.5 x 10-5 m
=
m
=
mm
=
μm
7.
4.5 x 10-7 m
=
m
=
μm
=
nm
8.
1800 x 10-13 m
=
m
=
nm =
pm
Orders of Magnitude
Physics seeks to explain the universe and all that is contained within it and in attempting to do so the
range of magnitudes of physical quantities will be vast.
10
In order to compare physical quantities and for that comparison to be meaningful, we must get a feel for
their relative sizes.
For example, how does the mass of the sun compare with the mass of earth?
mass of the earth
mass of the sun
=
=
5.98 x 1024 kg
1.99 x 1030 kg
5 980 000 000 000 000 000 000 000 kg
1 990 000 000 000 000 000 000 000 000 000 kg
To answer this question the ratio of these masses must be calculated:
mass of Earth
5 980 000 000 000 000 000 000 000
:
:
mass of sun
1 990 000 000 000 000 000 000 000 000 000
Divide both sides by 5 980 000 000 000 000 000 000 000:
mass of Earth
1
:
:
mass of sun
332 776
The mass of the sun is 332 776 times greater than that of the Earth.
Ratios of quantities as differences of orders of magnitude
A much simpler way of doing this without getting lost amongst the numbers is by stating the physical
quantity to its nearest order of magnitude or power of 10 (either by rounding up or down as appropriate)
and then comparing them by working out the ratio.
The mass of the earth is rounded up to 10 x 10 24 kg = 1 x 1025 kg. This order of magnitude is
expressed as 1025 kg.
The mass of the earth is rounded down to 1 x 1030 kg. This order of magnitude is expressed as 10 30 kg.
mass of earth
mass of sun
~
~
1025 kg
1030 kg
mass of earth
10 000 000 000 000 000 000 000 000
:
:
mass of sun
1 000 000 000 000 000 000 000 000 000 000
Divide both sides by 10 000 000 000 000 000 000 000 000:
mass of earth
1
100 000
:
:
=
mass of sun
100 000
1 x 105
The mass of the sun is 100 000 times (5 orders of magnitude (10 5)) greater than that of the earth. It’s
not as accurate as the previous method, but it’s easier and quicker to do to get a feel for their relative
magnitudes.
Comparing these 2 values is easy because working out the ratio between 2 powers of 10 is just a
matter of subtracting the values for the powers of 10.
1030 - 1025 = 105
Compare the mass of an electron (10-30 kg) with the total mass of the observable Universe (10 52 kg).
The total mass of the observable Universe is 82 orders of magnitude (1082 kg) greater than the
mass of an electron.
11
Compare the diameter of a proton (10-15 m) with the radius of the observable Universe (1027 m).
The radius of the observable Universe is 42 orders of magnitude (10 42 m) greater than the
diameter of a proton.
Compare the length of time taken for the passage of light across a nucleus (10 -23 s) to the age of the
Universe (1019 s).
The age of the Universe is 42 orders of magnitude (10 42 m) greater than the length of time taken
for the passage of light across a nucleus.
Significant Figures
It is important to be honest when reporting a measurement, so that it does not appear to be more
accurate than the equipment used to make the measurement allows. We can achieve this by controlling
the number of digits, or significant figures, used to report the measurement. All answers should contain
the same number of significant figures as the least precise reading or contain the greatest number of
significant figures as those given in any question.
The rules for significant figures are as follows:
▪
▪
▪
▪
The leftmost non-zero digit is significant and is the most significant digit in the number.
If the number has no decimal point, the rightmost non-zero digit is significant and is the least
significant digit in the number.
If the number has a decimal point, the least significant digit is the rightmost digit (which may be
zero).
The number of significant digits or figures is the number of digits from the most to the least
significant.
Examples:
▪
▪
▪
▪
▪
▪
▪
▪
▪
▪
▪
▪
▪
▪
▪
300 000
5460
4308
54
45.0
40.05
30
6.791
3.450
2.53
8.6
3
0.345
0.000 500
0.000 000 6
1 significant figure
3 significant figures
4 significant figures
2 significant figures
3 significant figures
4 significant figures
1 significant figure
4 significant figures
4 significant figures
3 significant figures
2 significant figures
1 significant figure
3 significant figures
3 significant figures
1 significant figure
3 is the most and least significant
5 is the most significant and 6 is the least significant
4 is the most significant and 8 is the least significant
5 is the most significant and 4 is the least significant
4 is the most significant and 0 is the least significant
4 is the most significant and 5 is the least significant
3 is the most and least significant
6 is the most significant and 1 is the least significant
3 is the most significant and 0 is the least significant
2 is the most significant and 3 is the least significant
8 is the most significant and 6 is the least significant
3 is the most and least significant
3 is the most significant and 5 is the least significant
5 is the most significant and 0 is the least significant
6 is the most and least significant
Estimating approximate values of everyday quantities
When estimating the approximate values of everyday quantities, an element of common sense must
prevail! For example, an apple is unlikely to have a mass in kilograms or indeed a few milligrams. A
typical apple has a mass of 100g. For example, the area of a shoe sole may be 0.1 m x 0.2 m = 0.02
m2. The volume of air in a classroom may be 10m x 10m x 2m = 200 m3. Some simplifications will be
made here both in the estimated values and in other aspects (the shoe is estimated to have a
rectangular area, the classroom to have a box-like shape where the volume of persons and objects in
the room is ignored).
12
3.1.2 Limitations of physical measurement
Uncertainty and Error in Measurement
A measurement should always be regarded as an estimate. The precision of the final result of an
experiment cannot be better than the precision of the measurements made during the experiment, so
the aim of the experimenter should be to make the estimates as good as possible.
There are many factors which contribute to the accuracy of a measurement. Perhaps the most obvious
of these is the level of attention paid by the person making the measurements: a careless experimenter
gets bad results! However, if the experiment is well designed, one careless measurement will usually
be obvious and can therefore be ignored in the final analysis. In the following discussion of errors and
level of precision we assume that the experiment is being performed by a careful person who is making
the best use of the apparatus available.
Any measured physical quantity is always subject to a degree of uncertainty i.e. there is a difference
between the recorded value and the actual or ‘true’ value. A ‘true’ value can never be obtained
because all measurements suffer from one or more types of uncertainty.
There are 3 main types of uncertainties:
1.
2.
3.
random errors;
systematic errors;
reading errors.
Random errors tend to be the fault of the experimenter.
Systematic errors tend to be the fault of both the experimenter and the apparatus being used.
Reading errors tend to be the fault of the apparatus being used.
In practice, all uncertainties are a combination of two or all of the types of uncertainty that exist.
Random and Systematic Errors and How They Can be Reduced
Random Errors
This type of uncertainty occurs when physical quantities change whilst being measured e.g. resistance,
temperature, pressure, etc. Random fluctuations can affect measurements from reading to reading
which can be out with the control of the experimenter. The best estimate of the ‘true’ value can be
obtained by repeating the measurements and then calculating the mean value. When a large number
of readings are taken, the difference between the mean value and the ‘true’ value is reduced. Repeat
measurements help to eliminate random errors. When taking measurements, it can be expected that
some will be larger than expected and some will be smaller than expected. Taking the mean value
cancels out this error.
Mean Value
=
SumOfAllVa luesObtained
NumberOfValuesObtained
Manufacturers of apparatus quote an equipment calibration error which gives an indication as to how
accurate their product is. For example, resistors with a gold strip are accurate to within + 5 % of their
quoted value.
Other sources of random error include the readability of the apparatus being used and the experimenter
being less than perfect.
Approximate Random Error =
MaximumVal ue − MinimumVal ue
NumberOfMeasurementsTaken
13
Systematic Errors
This is the most difficult to detect and can occur when the measurements are affected all in the same
way. There are two main sources of systematic error:
1.
Faulty or un-calibrated apparatus.
Examples include:
▪
▪
▪
A metre stick may have ‘shrunk’ giving consistently incorrect readings.
Timers that are running too fast or too slow that will consistently underestimate or overestimate
the actual time taken.
Apparatus that has not been zeroed or does not read zero when it’s not being actively used in
experiments (zero error) which will give rise to consistently incorrect readings.
Apparatus used must be calibrated (checked/compared) against a known standard.
2.
Using bad or inappropriate experimental techniques.
Examples include:
▪
When consistently not standing directly overhead a ruler to take measurements i.e. standing to
the side.
A systematic error might be present if the experimenter is making the same mistake each time when
making and recording measurements.
Repeating readings does not reduce systematic errors.
Reading Errors
This value indicates how well an instrument scale can be read.
There exist two types of scale:
▪
▪
analogue (e.g. liquid-in-glass thermometers);
digital (e.g. stopwatch).
The reading error for an analogue scale is + the smallest subdivision of the scale.
For example:
▪
▪
The reading error using a ruler
The reading error using a protractor
=
=
+ 1 mm
+1o
The reading error for a digital scale is + the smallest division of the instrument (+ 1 of the last digit i.e.
the least significant figure).
For example:
1.15s
14
The reading on a stopwatch is as shown:
The reading error of the time measured using the stopwatch
=
1.15 + 0.01 s
The ‘true’ value of the time lies between 1.14 and 1.16 seconds.
Reading errors can be thought of as random errors. If you were asked to measure the lengths of 20
rods that were identical in length but this was not made known to you then you are most likely to have
20 measurements that would not be the same but which would not differ from one another by much.
The chances are that the reading error in your measurements will end up being + 0.05 m. If you are not
careful when taking you measurements then they can differ substantially from one another in which
case the error is treated as a random error as opposed to a reading error.
Precision and Accuracy
A precise experiment is one which has a small random error and an accurate experiment is one which
has a small systematic error. Another way to think about this is to say that a precise experiment is one
in which repeated experimental values are all close to one another and an accuracy can be thought of
as a reading close to the accepted value. An accurate reading will have a low relative error.
Generally, uncertainties in experimental work will be caused by a lack of precision in measurement.
The greater the accuracy and precision in carrying out measurements, the lower the degree of
uncertainty.
(random error)
Determining relevant uncertainties in results/data
Uncertainties can be represented in 1 of 3 ways:
▪
▪
▪
using absolute uncertainties;
using fractional uncertainties;
using percentage uncertainties.
If a quantity x is measured then the absolute uncertainty is expressed as:
x + x
The fractional uncertainty is calculated and expressed as:
x+
x
x
15
For example:
1.15s
The reading on a digital stopwatch is as shown:
The reading error of the time measured using the digital stopwatch
=
1.15 + 0.01 s
The ‘true’ value of the time lies between 1.14 and 1.16 seconds. It is expressed as the absolute
uncertainty.
Expressed as a fractional uncertainty:
Fractional uncertainty
=
0.01
1.15
Fractional uncertainty
=
0.0087
The reading error of the time measured in fractional form
(NB: uncertainty always given to 1 sig fig)
=
1.15 + 0.01 s
The reading error could also be expressed as a percentage uncertainty:
% reading error in seconds
=
% reading error in seconds
=
The reading error of the time measured in percentage form =
0.01
 100
1.15
0.87 %
1.15 s + 0.9 %
When 2 or more numbers are multiplied or divided, the overall uncertainty is found by adding the
fractional uncertainties
X=YxZ
or
X=
Y
Z
then:
X Y Z
=
+
X
Y
Z
Putting this into practice, if the mass of a block was measured as 10.0 + 1 kg and the volume was
measured as 5.0 + 0.2 m3, calculate both the fractional and the percentage uncertainty in each
measurement.
Fractional uncertainty:
Mass
Volume
1
10 .0
=
10.0 + 0.1
As
density
0 .2
5 .0
0.1
=
0.04
10 .0
5 .0
=
5.0 + 0.04
=
mass
volume
=
2.0 kg m-3
The total uncertainty in the result is 2.0 kg m-3 + 0.3 kg m-3.
16
When numbers have powers associated with them, the overall uncertainty is found by multiplying the
fractional or percentage uncertainties by the power itself.
If X = YZ then:
X
 Y 
= Z

X
 Y 
For example, if a cube is measured to be 4.0 + 0.1 m in length along each side then:
% uncertainty in length
Volume
=
length3
% uncertainty in volume
% uncertainty in volume
Volume
=
=
0.1
 100 %
4.0
=
2.5 %
=
4.03
=
64.0 m3
=
=
3 x % uncertainty in length
+ 7.5 %
=
3 x 2.5
64.0 m3 + 7.5 %
Absolute uncertainty in volume
=
64.0 + 4.8 m3
Identify uncertainties and represent them graphically as error bars
The results of an experiment are often used to plot a graph. A graph can be used to verify the relation
between two variables and, at the same time, give an immediate impression of the precision of the
results. When we plot a graph, the independent variable is plotted on the horizontal axis. (The
independent variable is the cause and the dependent variable is the effect.)
If one variable is directly proportional to another variable, then a graph of these two variables will be a
straight line passing through the origin of the axes. So, for example, Ohm's Law has been verified if a
graph of voltage against current (for a metal conductor at constant temperature) is a straight line
passing through (0,0). Similarly, when current flows through a given resistor, the power dissipated is
directly proportional to the current squared. If we wanted to verify this fact we could plot a graph of
power (vertical) against current squared (horizontal). This graph should also be a straight line passing
through (0,0).
The "best-fit" line is the straight line which passes as near to as many of the points as possible. By
drawing such a line, we are attempting to minimize the effects of random errors in the measurements.
For example, if our points look like this:
our best-fit line should look like this:
17
Notice that the best-fit line does not necessarily pass through any of the points plotted.
Any experimental data plotted on a graph will have an associated uncertainty. This uncertainty can be
represented as an error bar on a graph as shown below:
The error bars represent the ± uncertainty in the data and so extend above and below for y-axis plots
and left and right for x-axis plots. Error bars are used to represent random uncertainty; systematic
uncertainty should have been reduced so as to not affect data. If the uncertainty is insignificant, error
bars are not required. A line of best fit is added.
Uncertainties in the Gradient and Intercepts of a Straight-Line Graph
The gradient of a straight line graph is calculated by dividing the change in y-axis values by the
corresponding change in x-axis values; more commonly remembered as rise over run.
In order to determine the range of uncertainty in the gradient of a graph, lines of maximum and
minimum fit are drawn using the error bars.
For the line of minimum fit, the line is drawn as shown below and the gradient calculated:
18
For the line of maximum fit, the line is drawn as shown below and the gradient calculated accordingly:
Uncertainty in best fit gradient = maximum difference between best fit gradient and maximum and
minimum gradient
Best fit gradient = 6.8
Max fit gradient = 7.1
Min fit gradient = 6.2
Uncertainty = 6.8 – 6.2 = 0.6
So best fit value = 6.8  0.6
For the axis intercepts, each of the lines of fit are extended to the relevant axis and the uncertainty in
the best fit axis intercept is found thus:
Uncertainty in best fit intercept = maximum difference between best fit intercept and maximum and
minimum intercept
Best fit intercept = 10.4
Max fit intercept = 10.9
Min fit intercept = 10.2
Uncertainty = 10.9 – 10.4 = 0.5
So best fit value = 10.4  0.5
19
AQA 3.1.1 SI Units
The Système Internationale d’Unités, abbreviated to SI, units was approved by an international conference
of weights and measures, at the International Bureau of Weights and Measures at Sèvres, in a suburb of
Paris, in 1960.
This was an attempt to introduce a system of units that could be used throughout the world. All SI units are
based on seven basic, base or fundamental units, which are defined:
Physical quantity
length
mass
time
electric current
temperature
amount of substance
luminous intensity
Unit
metre
kilogram
second
ampere
kelvin
mole
candela
Unit symbol
m
kg
s
A
K
mol
cd
Table 1
The seven fundamental SI units
Derived units
All other units, apart from those listed in table 1, are derived from these units. Derived units are formed by
multiplication and/or division of one or more basic units, without the inclusion of any numerical factors (e.g.
one coulomb = one ampere  one second). Derived units are named after people but the name of a derived
unit is always spelt with lower case letters whereas the symbols of the units have capital letters.
Find the equivalent version of the unit in terms of fundamental SI units. The first question has been done
for you.
Unit
Unit symbol
In fundamental SI units
newton
N
kg m s2
volt
V
coulomb
C
pascal
Pa
ohm

watt
W
joule
J
9
Unit symbol
In fundamental SI units
N C1
J s1
V m1
N m2
m
J C1
Ws
V A1
Ns
10
AQA 3.1.2 Experimental data handling 1
Name:_______________________
1) The resistance of a resistor was measured as 16(±2) Ω
a) What is the absolute uncertainty in the resistance?
b) What is the fractional uncertainty in the resistance?
c) What is the percentage uncertainty in the resistance?
d) What is the relative uncertainty in the resistance? N.B. relative uncertainty is another name for fractional uncertainty
e) Which of these uncertainties have a unit?
2) Giles recorded the following values for the resistance of a resistor:
509 Ω; 500 Ω; 480 Ω; 502 Ω; 503 Ω; 505 Ω; 497 Ω; 402 Ω; 501 Ω; 506 Ω; 490 Ω; 507 Ω
a) Mark and George checked the results for any anomalous values. Are there any anomalous values? If so, write
these values in the box below.
b) Should the anomalous results be included in the calculation of the mean value?
12
c) Calculate the mean of these values.
d) Calculate the range of these values.
e) Calculate range/2 for these values.
f) What is the absolute uncertainty in these values?
g) Write the mean value, with the absolute uncertainty and unit.
3) Libby recorded the following times:
12.6 s
12.5 s
12.6 s
12.4 s
12.7 s
15.2 s
12.4 s
12.4 s
12.4 s
12.3 s
12.3 s
6.3 s
12.1 s
12.2 s
12.2 s
12.4 s
12.6 s
12.6 s
a) Are there any anomalous values? If so, list them.
13
b) Should the anomalous values be included in the calculation of the mean value?
c) Calculate the mean of these values.
d) Calculate the range of these values.
e) Calculate range/2 for these values.
f) What is the absolute uncertainty in these values?
g) Write the mean value, with the absolute uncertainty and unit.
14
Experimental data handling 1
ANSWERS
1) a) 2
b)
2
16
1
= = 0.125
8
c) 12.5%
d) 0.125
e) absolute uncertainty
2 a) 402 
b) No
c) 500 
d) 29
e) 14.5 
f) 14.5   10 
g) 500(10) 
3)
a) Yes (6.3 s and 15.2 s)
b) No
c) 12.4 s
d) 0.6 s
e) 0.3 s
f) 0.3 s
g) 12.4(0.3) s
15
Uncertainties and errors A
Name:______________________________________
If the resistance, R is given by: R = 624(3) ,
The absolute uncertainty in is given by R = 3 ,
The fractional or relative uncertainty in R is given by:
The percentage uncertainty in R is given by:
∆𝑅
𝑅
∆𝑅
𝑅
3
= ± 642 = ±4.67 × 10−3
3
× 100% = ± 642 × 100% = 4.67 × 10−3 × 100% = 0.467%
1) Complete the Unit of absolute uncertainty column in the table (the absolute uncertainty should only be quoted to one significant digit).
2) Calculate the missing values (a) to (l), showing all working overleaf.
Quantity
Unit of quantity
Value of quantity
Absolute uncertainty
R

20
4
V
V
30
I
A
L
Unit of absolute uncertainty


Fractional or relative uncertainty
Percentage uncertainty
(a)
20
6
0.2
(b)
0.5
(c)
0.04
4
m
(d)
20
0.125
12.5
E
J
700
70
(e)
10
P
W
250
100
0.4
(f)
i

4800
(g)
0.005
0.5
r

(h)
12
0.005
0.5
F
N
64
4
(i)
6.25
a
m s2
880
11
0.0125
(j)
m
kg
1200
(k)
0.02
2
D
kg m3
(l)
200
0.4
40




16


(a)
(b)
(c)
(d)
(e)
(f)
(g)


(h)


(i)
(j)
(k)
(l)
17
Uncertainties and errors A ANSWERS
Quantity
R
V
I
L
E
P
i
r
F
Unit of
quantity

V
A
m
J
W


N
Value of
quantity
20
30
0.5
160
700
250
4800
2400
64
Absolute
uncertainty
4
6
0.02
20
70
100
24
12
4
Unit of absolute
uncertainty

V
A
m
J
W


N
Fractional or relative uncertainty
0.2
0.2
0.04
0.125
0.1
0.4
0.005
0.005
0.0625
percentage uncertainty
20
20
4
12.5
10
40
0.5
0.5
6.25
a
m
m s-2
kg
880
1200
11
24
m s-2
kg
0.0125
0.02
1.25
2
D
kg m-3
500
200
kg m-3
0.4
40
18


(a)
(b)
(c)
(d)
(e)
(f)
(g)


(h)


(i)
(j)
(k)
(l)
20
Uncertainties and errors B ANSWERS
Quantity
Unit of quantity
Value of quantity
Absolute uncertainty
Unit of absolute uncertainty
Fractional or relative uncertainty
Percentage uncertainty
R
80
20
25
120
3

V
0.25
V

V
0.025
2.5
I
A
0.8
0.02
A
0.025
2.5
L
m
320
40
m
0.125
12.5
E
J
800
40
J
0.05
5
P
W
1600
200
W
0.125
12.5
i

4800
24

0.005
0.5
r

24000
120

0.005
0.5
F
N
64
4
N
0.0625
6.25
a
m s-2
900
20
m s-2
0.0222
2.22
m
kg
1400
70
kg
0.05
5
D
kg m-3
800
300
kg m-3
0.375
37.5
21
Uncertainties and errors C
Name:______________________________________
If the resistance, R is given by: R = 624(3) ,
The absolute uncertainty in is given by R = 3 ,
The fractional or relative uncertainty in R is given by:
The percentage uncertainty in R is given by:
∆𝑅
𝑅
∆𝑅
𝑅
3
= ± 642 = ±4.67 × 10−3
× 100% = ±
3
642
× 100% = 4.67 × 10−3 × 100% = 0.467%
1) Complete the Unit of quantity and Unit of absolute uncertainty columns in the table (the absolute
uncertainty should only be quoted to one significant digit).
2) Calculate the missing values (a) to (l), showing all working overleaf.
Quantity
Unit of
quantity
a
D
kg m
E
-3
Value of
quantity
Absolute
uncertainty
Unit of absolute
uncertainty
(a)
20
ms
800
(b)
400
40
-2
J
Fractional or relative
uncertainty
Percentage
uncertainty
0.0233
2.33
0.625
62.5
(c)
10
F
N
20
4
0.2
(d)
I
A
(e)
0.02
0.025
2.5
4800
(f)
0.0025
0.25
320
40
(g)
12.5
1400
70
0.05
(h)
i
L
m
m
°
kg
P
W
(i)
400
0.25
25
R
W
80
(j)
0.25
25
24000
60
(k)
0.25
120
6
0.05
(l)
r
V
V
°
22
(a)


(b)
(c)
(d)
(e)
(f)
(g)
(h)






(i)
(j)
(k)
(l)
23
Uncertainties and errors C ANSWERS
Quantity
a
Unit of quantity
m s-2
Value of quantity
860
Absolute uncertainty
20
Unit of absolute uncertainty
m s-2
Fractional or relative uncertainty
0.0233
Percentage uncertainty
2.33
D
kg m-3
800
500
kg m-3
0.625
62.5
E
F
I
i
L
m
P
R
r
V
J
N
A

m
kg
W


V
400
20
0.8
4800
320
1400
1600
80
24000
120
40
4
0.02
12
40
70
400
20
60
6
J
N
A

m
kg
W


V
0.1
0.2
0.025
0.0025
0.125
0.05
0.25
0.25
0.0025
0.05
10
20
2.5
0.25
12.5
5
25
25
0.25
5
24
AQA 3.1.2 Uncertainty calculations AA
Name:_______________________________
1) For the length L = 32 (2) cm, the absolute uncertainty in the length, ΔL is given by ΔL=2 cm
The fractional uncertainty in L is given by:
Where:
𝛥𝐿
𝐿
=
2
32
𝛥𝐿
𝐿
= 0.0625
The percentage uncertainty in L is given by:
where
𝛥𝐿
𝐿
,
𝛥𝐿
𝐿
 100%,
2
 100% = 32 × 100% = 6.25%,
If the fractional uncertainty of a quantity is known, the absolute uncertainty can be calculated:
Absolute uncertainty in a quantity = fractional uncertainty in quantity  quantity
Absolute uncertainty in quantity = 0.0625  32 cm = 2 cm
During the multiplication or division of quantities, the fractional uncertainty in the result is given by the sum of the
fractional certainties of the quantities multiplied or divided.
2) For the lengths x = 30 (3) and y = 20 (4),
the fractional uncertainty in x is given by
the fractional uncertainty in y is given by
𝛥𝑥
𝑥
𝛥𝑦
𝑦
3
= 30 = 0.1
4
= 20 = 0.2
3) What is the value of xy?
xy = 30  20 = 600
What is the fractional uncertainty in xy?
fractional uncertainty in xy = fractional uncertainty in x + fractional uncertainty in y
𝛥(𝑥𝑦) 𝛥𝑥 𝛥𝑦
=
+
𝑥𝑦
𝑥
𝑦
fractional uncertainty in xy =
𝛥(𝑥𝑦)
𝑥𝑦
= 0.1 + 0.2 = 0.3
What is the absolute uncertainty in xy?
absolute uncertainty in xy = (fractional uncertainty in xy)  xy = (xy)  xy = 0.3  600 = 180
25
Examples
A = 20 (2), B = 40 (8), C = 10 (2), D = 30 (3)
1) a) Calculate the fractional uncertainty in A
b) Calculate the fractional uncertainty in B
c) Calculate the fractional uncertainty in C
d) Calculate the fractional uncertainty in D
2) a) Calculate the percentage uncertainty in A
b) Calculate the percentage uncertainty in B
26
c) Calculate the percentage uncertainty in C
d) Calculate the percentage uncertainty in D
3) a) Calculate the value of AB
b) Calculate the fractional uncertainty in AB
c) Calculate the absolute uncertainty in AB
4) Calculate the following
a)
CD
27
b)
The fractional uncertainty in CD
PTO
c)
The absolute uncertainty in CD
5) Calculate the following
a)
b)
c)
𝐶
𝐷
The fractional uncertainty in
The absolute uncertainty in
𝐶
𝐷
𝐶
𝐷
28
Uncertainty calculations AA
Answers
1)
a) 0.1
b) 0.2
c) 0.2
d) 0.1
2)
a) 10%
b) 20%
c) 20%
d) 10%
3)
a) 800
b) 0.3
c) 240
4)
a) 300
b) 0.3
c) 90
5)
a) 0.333
b) 0.3
c) 0.1
29
AQA 3.1.2 Uncertainty calculations 1
Name:_________________________________________________
All measurements are subject to an uncertainty. The uncertainty in a particular quantity can be given as
 an absolute uncertainty,
 a factional (or relative) uncertainty or
 a percentage uncertainty
Consider the following measurement of a length, L, where L = 22.3 (0.5) m
The absolute uncertainty in L = ΔL = 0.5 m
∆𝐿
𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑢𝑛𝑐𝑒𝑟𝑡𝑎𝑖𝑛𝑡𝑦 𝑖𝑛 𝐿
0.5
The fractional uncertainty in the L =  𝐿 = 
= ± 22.3 = ±0.022
𝐿
∆𝐿
𝐿
The percentage uncertainty in L = fractional uncertainty in L  100% =
2.2%
× 100% = 0.022  100% =
The absolute uncertainty in a measurement has the same unit as the measurement. The fractional and
percentage uncertainties do not have any unit. The absolute uncertainty is an estimate and should be
written to one significant digit. The fractional or percentage uncertainty in a measurement can be written
to two or three significant digits. The measurement should be written to the same precision as the
uncertainty i.e. to the same number of decimal places or order of magnitude as the uncertainty.
Write the following uncertainty values to one significant digit (include the unit with each value) e.g.
678.983 N becomes 700 N.
a)
67.838 J
b)
0.0978 N kg1
c)
2356 N m2 kg2
d)
345.87 J K1
Write the following measurements to the same precision as the uncertainty (include the uncertainty and the
unit) e.g. 828.98 (70) J kg1 becomes 830 (70) J kg1.
a)
78.348 (2) m
b)
0.003528 (0.0006 ) N m1
c)
5678 (500) 
d)
782.98 (90) J kg1
For each of the following measurements, write the uncertainty to one significant digit and then the
measurement, to the same precision as the uncertainty e.g. 7894 (578)  becomes 7900 (600) .
a)
678 (28) m
b)
0.073528 (0.0189 ) N m1
c)
43567 (567) 
d)
6789 (98) J kg1
e)
56789 (235) N m2 kg2
f)
456.879 (0.878) J K1
30
For each of the following measurements, write down or calculate the absolute, fractional and percentage
uncertainty. Show all steps in each calculation.
Measurement
Absolute uncertainty
Fractional uncertainty
Percentage uncertainty
P = 234 (6) N m1
L = 3.05 (0.05) m
R = 6.5 (0.7) 
Calculation of absolute uncertainty, given the fractional uncertainty in a measurement
Absolute uncertainty in a measurement = fractional uncertainty in measurement  measurement
Calculate the absolute uncertainties in the following values, showing all working.
Measurement
Fractional uncertainty in measurement
65 
0.23
356 m
0.06
17 N m1
0.32
657 J kg1
0.13
31
Absolute uncertainty in measurement
Compound uncertainties
The uncertainty in a value calculated using a number of variables, each subject to an uncertainty can be
calculated using a simple set of rules.
Adding or subtracting two or more measurements
When adding or subtracting two or more values, the total absolute uncertainty in the calculated value is
equal to the sum of the absolute uncertainties in the values which are added or subtracted.
e.g. A = 62 (4) m and B = 85 (5) m
A+B = 62 + 85 = 147
Δ(A+B) = ΔA + ΔB = 4+5 = 9
and
A+B = 147 (9) m
BA = 23 (9) m
Questions
Given that:
P = 68 (3)
Q = 25 (4)
R = 68 (5)
Calculate the values and the uncertainties for the following quantities:
a) P + Q
b) P  Q
c) P + Q  R
d) T  P
e) T + S  R
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S = 580 (20) T = 700 (100)
Multiplying or dividing two or more measurements
When multiplying or dividing two or more measurements, the fractional uncertainty in the result is equal to
the sum of the fractional uncertainties in the values which have been multiplied (or divided). The
percentage uncertainty in the result is obtained by adding the percentage uncertainties in the values which
have been multiplied (or divided).
Examples
A = 62 (4) m and B = 85 (5) m
1) Calculate the value of AB and the absolute uncertainty in that value.
AB =6285 = 5270 m2
4
5
The fractional uncertainty in AB = 62 + 85 = 0.123
The absolute uncertainty in AB = 5270  0.123 = 648 m2
As the absolute uncertainty should be given to one significant digit, the uncertainty in AB = 600 m2. The
calculation can only be given to the same precision as the uncertainty.
AB = 5300 (600) m2
𝐴
2) Calculate the value of 𝐵 and the absolute uncertainty in that value.
𝐴
𝐵
62
= 85 = 0.729
The fractional uncertainty in
The absolute uncertainty in
The value of
𝐴
𝐵
𝐴
𝐵
𝐴
𝐵
=
4
5
+
62
85
= 0.123 ×
with the uncertainty is
𝐴
𝐵
= 0.123
𝐴
𝐵
= 0.123 × 0.729 = 0.0897
= 0.12 (±0.09)
3) Calculate the absolute uncertainty in A3
A3 = A  A  A = 62  62  62 = 238328 m3
4
4
4
4
12
The fractional uncertainty in A3 or A  A  a = 62 + 62 + 62 = 3 × 62 = 62 = 0.194
The absolute uncertainty in A3 = (fractional uncertainty in A3)  A3 = 0.194  23832 = 4623 m3
N.B. Using this reasoning,
The fractional uncertainty in F2 = 2  fractional uncertainty in F
The fractional uncertainty in D4 = 4  fractional uncertainty in D
The fractional uncertainty in X1/2 = ½  fractional uncertainty in X
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Questions
Given that:
L = 6.7 (0.6)
M = 9.56 (0.06)
N = 12.8 (0.5)
4) Calculate the values and the uncertainties for the following quantities:
𝐿
a) 𝑀
b) LM
c)
𝑃𝑁
𝑀2
d) P2ML
𝑃𝑀 3
√𝑁
e) 𝐿
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P = 65 (7)
The uncertainty in measurements subject to a mathematical function
When measuring an angle  which is subject to an uncertainty and then finding sine , to find the
uncertainty in the value of sine , determine the range of values of sine , using the maximum and
𝑟𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑠𝑖𝑛𝑒 𝜃
minimum values of . The absolute uncertainty in sine  is given by ±
2
Examples
1) An angle was measured as  = 45.0 (0.5). Calculate the value of sine , with the absolute uncertainty
in that value.
𝜃𝑚𝑎𝑥 = 45.5
𝜃𝑚𝑖𝑛 = 44.5
sine  = sine 45.0 = 0.7071
sine max = sine 45.5 = 0.7133
sine min = sine 44.5 = 0.7001
𝑟𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑠𝑖𝑛𝑒 
0.7133−0.7001
0.0132
uncertainty in sine  = ±
=±
= ± 2 = 0.007
2
2
sine  = 0.707 (0.007)
Questions
5) Find the absolute uncertainty in ln x if x = 20.5 (0.5)
6) Find the absolute uncertainty in cos  if  = 60 (2)
7) Find the absolute uncertainty in M2 sin if M = 34 (5) N and  = 56 (1)
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