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Uncertainty in Practical Work

www.curriculum-press.co.uk Number 132

FactsheetPhysics

This Factsheet will consider the following questions:

How certain are you of your result?

How accurate was the measuring device?

How can you determine the uncertainty in your measurement?

What is the difference between absolute and percentage errors?

If a quantity is the product or sum of other measurements, how

are the uncertainties combined?

How many significant figures should you use when quoting a

result?

What is the difference between random and systematic errors?

You have two pieces of wire, 761mm and 481mm in length.

761mm

481mm

How long are they in total?

The absolute error in a measurement

gives an upper and lower limit,

expressed in the units of the

measurement.

The combined length is 76mm + 48mm = 124mm. But the uncertainty

in each measurement means the total length could be as short as

75mm + 47mm = 122mm or as a long as 77mm + 49mm = 126mm. The

total length is 1242mm. Compare the uncertainty for each length

with the uncertainty in the total length.

Exam Hint: The absolute error in the measurement 761mm is

1mm. It must be expressed in the units of the actual

measurement.

Whenaddingquantities, the absolute error in the result

is the sum of the absolute errors in each quantity.

So how are absolute errors combined when one quantity is

subtracted from another?

Worked exampleTwo wires are 0.230.01m and 0.420.01m in length. What is

the difference between them? What is the uncertainty in this

measurement? How are absolute errors combined when

quantities are subtracted from one another?

Answer: 0.42m - 0.23m = 0.19m.

The minimum possible value is 0.41m-0.24m = 0.17m

The maximum possible value is 0.43m-0.22m = 0.21m

Therefore the answer is 0.190.02m.

The overall absolute error due to subtraction is thesum of the individual absolute errors for each quantity.

Which is a more precise measurement: 761mm or 481mm? Both

values have the same absolute error but 481mm could be as small

as 47mm or as large as 49mm. 1/48 is 0.021 or 2.1%.

761mm could be as small as 75mm or as large as 77mm. 1/76 is 0.013

or 1.3%. The relative error in 481mm is 0.021 and the percentage

error is 2.1%. The 761mm has the same absolute error as 481mm,

but a smaller relative error.

The relative error in a measurement is given by Absolute

error / Measurement. The percentage error is the relative error

expressed as a percentage.

What about measuring the area of a rectangle with sides 761mm

and 481mm? The area of the rectangle is 76mm48mm = 3648mm2.

However, the area could be as small as 75mm 47mm = 3525mm2or

as large as 3773mm2. The area is 3648124mm2. How does the

uncertainty of each length relate to the uncertainty in the area?

The uncertainty in the first side as a percentage is 1/76 100% = 1.3%

The uncertainty in the second side as a percentage is 1/48 100% = 2.1%

The uncertainty in the area as a percentage is 124/3648 = 3.4%

When multiplying quantities together, the percentage

uncertainty in the product is equal to the sum of the percentage

errors in each quantity.

Worked example

A br ick has dimensions

10.00.1cm, 20.00.1cm

and 7.00.1cm.

a) Calculate the volume

of the brick.

b) Find the uncertainty in

this volume.

200.1cm 100.1cm

70.1cm

c) Calculate the density of the brick if the mass is 18731g.

d) Calculate the largest and smallest possible brick densities.

e) Calculate the percentage error in these quantities: Volume,mass, density.

f) Thus state how uncertainties are combined when quantities

are divided.

Answer:

a) Volume of the brick is 10.0cm 20.0cm 7.0cm = 1400.0cm3

b) Add the percentage error in each quantity:

0.1/10.0 100% = 1.0% 0.1/20.0 100% = 0.5%

0.1/7.0 100% = 1.4(3)%

Therefore the volume is 1400.0cm32.93%

c) Density of the brick = 1873g / 1400.0cm3 = 1.338g/cm3

d) Largest possible density = 1874g/1359cm3 = 1.379g/cm3

Smallest possible density = 1872g/ 1441cm3

= 1.299g/cm3

e) Percentage error of the volume = 2.93%

Percentage error of the mass = 1/1873g 100% = 0.05%

Density = 1.3380.040 g/cm3

Percentage error of the density = 0.040 / 1.338 100% =

2.99%

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132. Uncertainty in Practical Work Physics Factsheet

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Worked exampleThe potential difference of two cells is found to be 1.50.1V

and 1.30.2V.

When quantities are divided, the sum of the

individual percentage errors gives the overall percentage

error.

1.50.1V 1.30.2V

a) Find the combined p.d. of these cells in series.

A current of 2.20.1A was measured when these cells were

put into a circuit.

b) Calculate the resistance of the circuit.

c) Find the uncertainty in this measurement.

Answer:

a) Potential difference = 1.50.1V + 1.30.2V = 2.80.3V

b) Resistance = P.d./ Current = 2.80.3V / 2.20.1A = 1.27

c) Sum the individual percentage errors:

0.3 / 2.8V 100% = 10.7% 0.1 / 2.2A 100% = 4.5%

Resistance = 1.2715.2%

frequency Mean

68% of measurements

are within one

standard deviation of

the mean

-1 +1

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Acknowledgements:

This Physics Factsheet was researched and written by J Carter

The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU

ISSN 1351-5136

AveragesHow does taking more than one reading affect uncertainty? Make

ten measurements of a length, each with an absolute error of 1mm.The absolute error of the average will be 1mm.

Exam Hint: The absolute error of a measurement is not

dependent on the number of readings.

However the 10 readings will have a bell shaped distribution as

shown in the diagram. About two thirds of the measurements willbe within one standard deviation of the mean. The standard

deviation is reduced by a large number of readings

Exam Hint:The relative or percentage error of a measurement

is dependent on the number of readings.

Random and systematic errorsRandom errors are variations in your measurement due to the

precision of your measuring device or small changes in conditions.

For example, three pendulum period readings might be 3.45s, 3.51s

and 3.48s.

An example of a systematic error would be an electronic balance

that has not been correctly reset to zero or a metre rule with a

rounded end. With a systematic error, the measuring device is

consistently out by a particular amount.

Pactice Questions1. (a) Define the terms absolute error and percentage error.

(b) State the precision in these measuring devices: 15cm ruler,

Vernier callipers, micrometer.

(c) An electrical component has dimensions of approximately

2.0cm by 1.4cm.

Estimate the percentage error in a measurement of area made

with (i) 15cm ruler (ii) Micrometer

2. (a) Give an example of a systematic error.(b) Explain how random errors can be reduced.

(c) Calculate the volume of a cube with 3.4cm sides. Include the

uncertainty in the volume.

3. A student measures the current in each resistor 1, 2 and 3. The

measurements are 0.80.1A, 1.20.1A and 2.70.1A respectively.

(a) Calculate the total current, Itpassing through the networkand state the absolute error in this measurement.

(b) Find the percentage errors for I1, I

2, I

3and I

t.

The student then measures a current of 4.70.1A for Itdirectly

in the circuit.

(c) Now calculate the percentage error in this measurement of It.

(d) Explain the difference in the percentage error for Itin b) and c).

Answers1. (b)1mm,0.1mm, 0.01mm

(c) 12%, 0.12%

2. (c) 39.3cm38.8%, although a more realistic figure is 39cm39%

3, (a) 4.70.3A(b) I

1, 12.5%, I

2, 8.3% , I

3, 3.7%, I

t6.4%

(c) 2.1%

The percentage error of the resistance measurement is approximately

15%. How many significant figures should we express in our answer?

1 Resistance could be between 0.5 and 1.5?, an uncertainty of50%.

1.3 Resistance could be between 1.25 and 1.35, an uncertainty

of 4%.

1.27 Resistance could be between 1.265 and 1.275, an

uncertainty of 0.4%.

If the uncertainty in the resistance measurement is 15%, what is the

most appropriate number of significant figures? An uncertainty of

50% is far less precise than 15%, and 0.4% is far more precise. Two

significant figures are enough in this case.