1© Manhattan Press (H.K.) Ltd. Measurements and errors Precision and accuracy Significant figures...

© Manhattan Press (H.K.) Ltd. 1

Measurements and Measurements and errorserrors

• Precision and accuracyPrecision and accuracy• Significant figuresSignificant figures

• Scientific notationcientific notation• MeasurementsMeasurements• Types of errorsTypes of errors

• Combination of errorsCombination of errors


Measurements and errors

Reading:

• single determination of the value of an unknown quantity

• actual reading taken during an experiment

Measurement and errors (SB p. 13)

Measurement:

• final result of the analysis of a series of readings


Precision and accuracy

Precision:

indicates the agreement among repeated measurements


• take readings repeatedly

• calculate mean ( )

• deviation (d) =

• measure the precision: by the mean deviation

x

ixx



Improvement:

e.g.

• Use a hand lens when reading the scale of a meter

• Use a plane mirror behind the pointer

• Read the scale when the pointer is directly on top of its image



Accuracy



A measurement is said to be accurate if it is close to the actual value.

Accuracy:

Indicates how correct the result is


Accuracy



e.g.

1. Using a metre rule

Length recorded = 34.7 cm

length accurate to 0.1 cm

34.7 0.1 cm


Accuracy



2. Using a micrometre screw gauge

Length recorded = 3.62 mm

3.620 0.005 mm maximum possible error

Fractional error = 62030050

Mean valueerror possible Maximum

.

.


Accuracy



3. Percentage error

%.%..

%

2760100623010

100Mean value

error possible Maximum

Smaller the percentage error,

higher the accuracy

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More to Know 4More to Know 4

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Common Error


(a) Most significant digit:

the leftmost non-zero digit

(b) Least significant digit:

the rightmost non-zero digit for the number without decimal point, or

the rightmost digit (including zero) for the number with decimal point

Significant figures:

No. of digits except for zeros at the beginning


Significant figures


(a) ruler of smallest division 0.1 cm

Length = 8.074 6 cm is not consistent

Length = 8.1 0.1 cm

Significant figures:

consistent with the accuracy of the measurement


Significant figures


(b) Mass of X = 0.376 kg (3 sig. fig.)

Mass of Y = 0.056 2 kg (3 sig. fig.)

Combined weight

mg = (0.376 + 0.056 2) ×9.8

= 0.432 2 ×9.8

= 4.235 56 N

Combined weight = 4.2 N (2 sig. fig.)


Significant figures

g (2 sig. fig.)


No. of significant figures

= no. of significant figures in the quantity which has the least no. of significant figures


Significant figures



Significant figures

No. of significant figures reflects a measurement’s order of accuracy

e.g.

Length of classroom = 8 m (1 sig. fig.)

Actual measured length

= 7.5 m to 8.5 m



Significant figures

e.g.

Length of classroom = 8.0 m (2 sig. fig.)

Actual measured length

= 7.95 m to 8.05 m



Scientific notation

Scientific notation:

represent extremely large or extremely small numbers

e.g.

800 000 (up to 1, 2 or 3 sig. fig?)

8.00 x 105 (3 sig. fig)



Measurements

1. Metre rule

Maximum possible error

= Half the smallest division

=

= 0.05 cm210.



Measurements

e.g.

At “8 cm” mark, reading = 8.00 0.05 cm

7.95 cm – 8.05 cm

1. the first reading from “0 cm”

2. the second reading from “8 cm”

Length = 8.00 0.1 cmMaximum

possible error

2 reading errors



Measurements

2. Vernier caliper

external diameter

internal diameter

depth of container


2. Vernier caliper



=

= 0.05 mm210.


Measurements



Measurements

e.g.1.30 cm to 1.40 cm

The 7th mark is exactly opposite a mark on the main scale. Reading = 1.370 cm

Reading

= 1.370 0.005 cm

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Measurements

3. Micrometre screw gauge

Measure outer dimension of object up to accuracy of at least 0.01 mm



Measurements

Length of division in main scale = 0.5 mm

Each smallest division = = 0.01 mm5050.



=

= 0.005 mm2010.



Measurements

e.g.

Reading

= 2.480 0.005 mm

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Measurements

4. Computer data-logging system

Data collection and storage for data processing later

InterfaceSensor



Types of errors

1. Systematic errors

Systematic errors are errors in the measurement of physical quantities due to instruments, faults in the surrounding conditions or mistakes made by the observer. An experiment with small systematic error is said to be accurate.



Types of errors

Sources of systematic error

(a) Zero errors

reading on instrument is not zero when it is not used

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Types of errors

(b) Personal errors of the observer

From physical constrains or limitation of an individual (reaction time)



Types of errors

(c) Errors due to instruments

e.g.

(i) A watch which is fast

(ii) An ammeter which is used under different conditions from which it had have calibrated



Types of errors

(d) Errors due to wrong assumption

e.g.

acceleration due to gravity (g) is assumed to be 9.81 m s-2

Systematic errors cannot be reduced by repeating measurements using the same method, same instrument and by the same observer.



Types of errors

Systematic errors can reduced by taking measurements carefully, or varying conditions of measurements.

More accurate Less accurate



Types of errors

2. Random errors

Random errors are errors in a measurement made by the observer or person who takes the measurement. An experiment with small random error is said to be precise.



Types of errors

Sources

(a) Due to parallax

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Types of errors

(b) Due to temperature change

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More precise Less precise



Combination of errors

1. Addition or subtraction

(a) If U = x + y

U = (x + y)

(b) If V = x - y

V = (x + y)

x, y are errors

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Example 3Example 3


2. Product

If U = xyz

][zz

yy

xx

UU



x, y, z are errors

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Example 4Example 4


3. Quotient

If U =

][yy

xx

UU

yx



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Example 5Example 5




4. Constant power

If U = xp (p is constant)

xxp

UU


5. General case

If U = (c, p, q, r are constant)r

qp

z

ycx

][zzr

yy

qxxp

UU



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Example 6Example 6Go to

Example 7Example 7


End



If a metre rule is graduated in mm, the maximum possible error of a measurement is equal to the half of the smallest division (0.05 cm or 0.5 mm). Why should the maximum possible error of the measurement of the metal rod be 0.1 cm? For details, please refer to the Section D of Metre rule.

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A common zero error arisesfrom using a metre rule fromone end, which may be worn.It is a better practice to usethe centre of the rule, insteadof measuring from one end.

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Most vernier calipers will read zero when the jaws are closed, without an object in place. However, as a result of misuse or wear, the instrument may not read zero. In these cases, a zero reading error must be added or subtracted. The maximum possible error becomes ± 0.05 mm × 2 = ± 0.1 mm.

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Choice of measuring instrument

When we choose a measuring instrument, we should consider:

1. its convenience to be used,2. its precision, and3. the range of the reading we need.

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Zero reading error and zero error

Zero reading error is due to the limitation of measuring device at the “0” mark and is equal to the half of the smallest division of the device.Zero error of the device is due to the deviation from “0” value at the “0” mark.

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1. Reading can be taken with great precision but not accurate if there is a systematic error.

2. Reading can be accurate but not precise when there is a random error.

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Random errors can be reduced by repeated measurements while systematic errors cannot.

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Q: Q: The internal diameter d1 and the external

diameter d2 of a metal tube are d1 = 45 ± 1

mm and d2 = 60 ± 2 mm. What is the

maximum percentage error in the total thickness of the tube when it is pressed together?

Solution



Solution:Solution:

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TextText

Thickness of the tube (t)

= d2 – d1 = 60 – 45 = 15 mm

Error in t (δt)

= ±(δd2 + δd1) = ±2 + 1 = ±3 mm

∴ Thickness of the tube = 15 ± 3 mm

∴ Maximum percentage error in t :

= 20%

%%tt 100

153100



Q:Q: The dimensions of a box are recorded as follows:Length () = 5.0 ± 0.2 cmWidth (b) = 4.0 ± 0.1 cmHeight (h) = 8.0 ± 0.2 cmWhat is the maximum percentage error in the volume of the box?

Solution



Solution:Solution:

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Volume of box ( V )

= bh = 5.0 4.0 8.0 = 160 cm3

∴Maximum percentage error in V:

%hh

bb%

VV 100][100

%..

.

... 100]

0820

0410

0520[

= 9%



Q:Q: The mass of a metal block is 11.5 ± 0.5 kg and its volume is 1 000 ± 20 cm3. How would you express the density of the metal?

Solution



Solution:Solution:

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Density ()

32 cm kg 101511000

511)( Volume

)( Mass

..V

M

Maximum fractional error in :

)10(1.15]1000

2051150[

]1000

2051150[

][

2-

....

VV

MM

= 0.07 x 10-2 kg cm-3

Density of metal () = (1.15 0.07) x 10-2 kg cm-3

Note: The error is calculated to only one sig. fig.



Q:Q: In an experiment, the external diameter D and internal diameter d of a metal tube werefound to be 64 ± 2 mm and 47 ± 1 mm respectively. What is the maximum percentage error in the cross-sectional area of the metal tube as shown in the figure?

Solution



Solution:Solution:

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Area of the shaded region (A)

%

%

%dDdD

dDdD

%AA

20

100]4764

34764

3[

100])()(

)()(

[100

))((4

)(4

22

dDdD

dD

Maximum error in (D + d):

δ(D + d) = δD + δd = 2 + 1 = 3 mm

Maximum error in (D – d):

δ(D – d) = δD + δd = 2 + 1 = 3 mm

Maximum percentage error in the cross-sectional area:



Q:Q: (a) Explain the principle of vernier calipers.(b) The figure shows a steel vernier caliper which was used to measure the diameter of a cylinder. What is the reading for the diameter?

(c) Assuming that the zero reading is correct, what is the percentage error in this measurement?(d) Calculate the percentage error in the cross-sectional area of the cylinder.Solution



Solution:Solution:

(a) Vernier calipers consist of two scales.1. The main scale has 1.0 cm divided into 10 equal divisions.

Size of each division = 0.10 cm2. The vernier scale has the length of 10 divisions adding up to 0.9 cm.

Size of each division = 0.09 cm∴Difference in size of one division on the main scale and one division on the vernier scale is:

(0.10 – 0.09) cm = 0.01 cmIn the figure above, the distance between the 1.2 cm mark on the main scale and the 0 mark on the vernier scale is 3 0.01 = 0.03 cm.∴The vernier reading is 1.23 cm.



Solution (cont’d) :Solution (cont’d) :

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TextText

(b) From the figure, diameter of the cylinder (d) = 2.53 cm

(c) Error in d (δd) = 0.005 cm

∴ Percentage error:

%

%.

.%dd

.20

100532

0050100

(d) Cross-sectional area (A) =

∴ Percentage error:4

2d

0.4% .202

1002100

%

%dd%

AA


Date post:	04-Jan-2016
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1© Manhattan Press (H.K.) Ltd. Measurements and errors Precision and accuracy Significant figures...

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