ErrorsCourse: Diploma

Subject: Applied Science Physics

Unit: I

Chapter: II

oErrors Since all measurements contain an estimated digit,

all measurements contain some uncertainty (error).

Scientists try to limit the uncertainty (error) as

much as possible but they cannot eliminate it.

There are three main reasons for uncertainty in

measurements:

i. instrumental error

ii. observer error

iii. procedural error

oErrorsi. Instrumental Error:

All measuring instruments have error. The more

sensitive and precise the instrument is, the lower

the amount of error will be.

A more sensitive instrument will give more

significant figures than a less sensitive one.

A more precise instrument will give the same

reading more often than a less precise one.

oErrorsii. Observer Error:

An instrument is only as good as the person using

it! Persons who have more experience and who

take more precautions will generally record

measurements with less error.

iii. Procedural Error:

Measurements can have error due to faulty

experimental procedure.

oAccuracy and Precision

In an experiment, it is important to be able to state the

level of confidence of one’s data.

In this course, you will analyze the accuracy and the

precision of data.

Accuracy measures how close a measured value is to

the accepted value

Precision measures how close together several

measured trials are to one another.

oAccuracy and Precision

In this course you will use percent error to measure accuracy.

%Error = Measured Value – Accepted Value 100

Accepted Value

%Error can be positive or negative!

%Error < than |5%| = high accuracy.

|5%| ≤ %Error ≤ |10%| = moderate accuracy.

%Error > |10%| = low accuracy.

oAccuracy and Precision

In this course, precision will be measured by the

“eyeball test”.

high

precision

high

precision

low

precision

moderate

precision

oErrors

Errors can be divided into 2 main classes

Random errors

Systematic errors

oSystematic Errors Positive/Negative

Known

So it Can Reduced

1. Instrumental Error• Due to instrumental defect

• Or Defect in Calibration of Scale

e.g. one end of meter scale is Broken

Thermometer Calibration

2. Error due to procedure

1

oSystematic Errors3. Personal Error

• Due to improper arrangement of

instruments

4. External Errors

• Due to External Properties

e.g. Temperature, Pressure, Air Flow, Velocity,

Medium

oRandom Errors Are due to variations in performance of the

instrument and the operator

Even when systematic errors have been allowed

for, there exists error.

oAnother types of Error

Three other ways of defining the error are:

Absolute error

Relative error

Percentage error

oCalculation the Absolute Error

Absolute error.

ea = |True value – Approximate value|

14

oRelative Error

The relative error is the ratio of the absolute error to the

true value. For the argument above we can calculate

the relative error as:

Relative error = (absolute error/true value)

= (-1.0/12.1) = -0.0826ZZ

oPercentage error Percentage error = (absolute error/true value)x100%

= (-1.0/12.1)x 100% = -8.26%

oMinimization of errors The accuracy of measuring instrument is guaranteed

within a certain percentage (%) of full scale reading

E.g manufacturer may specify the instrument to be

accurate at 2 % with full scale deflection

For reading less than full scale, the limiting error increases

LIMITING ERROR

Example

Given a 600 V voltmeter with accuracy 2% full scale.

Calculate limiting error when the instrument is used to measure

a voltage of 250V?

Solution

The magnitude of limiting error, 0.02 x 600 = 12V

Therefore, the limiting error for 250V = 12/250 x 100 = 4.8%

LIMITING ERROR

Example

Given for certain measurement, a limiting error for voltmeter at 70V is

2.143% and a limiting error for ammeter at 80mA is 2.813%. Determine

the limiting error of the power.

Solution

The limiting error for the power = 2.143% + 2.813%

= 4.956%

Example

What is the relative error & Percentage Error in the

approximation X = 2.0 to x = 1.98?

Solution

Relative error = X/x = 2/1.98 = 0.01010

Percentage Error = X/x*100% = 1.01%

oSignificant Figures

Significant figures reflect precision. Two students may have calculated the

free-fall acceleration due to gravity as 9.625 ms-2 and 9.8 ms-2 respectively.

The former is more precise – there are more significant figures – but the

latter value is more accurate; it is closer to the correct answer.

General Rules:

1. The leftmost non-zero digit is the most significant

figure.

111

2. If there is no decimal point, the rightmost non-zero digit is

the least significant.

4. All digits between the most significant digit and the least

significant digit are significant figures.

0.100001

3. If there is a decimal point, the rightmost digit is the least

significant digit, even if it is a zero.

0.022

5. The zeros on left side are not significant ,zeros on right

side are significant figures

0.032, 1.2300

6.When no is to be round off then if last figure is greater

than 5 the last figure retained is increased, if less than 5

then it remains un changed

oArithmetic with Significant Figures

When adding or subtracting measured quantities the recorded answer cannot be more precise

than the least precise measurement.

Add: 24.686 m + 2.343 m + 3.21 m Answer: 30.24 m

3.21m has the least dp’s

When multiplying or dividing measurements the factor with the least number of significant

figures determines the recorded answer

Multiply: 3.22 cm x 2.1 cm Answer: 6.8 cm2

2.1 cm has only 2sf

Note significant digits are only considered when calculating with measurements; there is

no uncertainty associated with counting. If you counted the time for ten back and forth

swings of a pendulum and you wanted to find the time for one swing, the measured time

has the uncertainty but the number of swings does not.

Tutorial on the Use of Significant Figures

1. 37.76 + 3.907 + 226.4 = ...

2. 319.15 - 32.614 = ...

3. 104.630 + 27.08362 + 0.61 = ...

4. 125 - 0.23 + 4.109 = ...

5. 2.02 × 2.5 = ...

6. 600.0 / 5.2302 = ...

7. 0.0032 × 273 = ...

Tutorial on the Use of Significant Figures

1. 37.76 + 3.907 + 226.4 = 268.1

2. 319.15 - 32.614 = 286.54

3. 104.630 + 27.08362 + 0.61 = 132.32

4. 125 - 0.23 + 4.109 = 129

5. 2.02 × 2.5 = 5.0

6. 600.0 / 5.2302 = 114.7

7. 0.0032 × 273 = 0.87

Example

Given expected voltage value across a resistor is 80V.

The measurement is 79V. Calculate,

i. The absolute error

ii. The % of error

iii. The relative accuracy

iv. The % of accuracy

Solution (Example)

Given that , expected value = 80V

measurement value = 79V

i. Absolute error, e = = 80V – 79V = 1V

ii. % error = = = 1.25%

iii. Relative accuracy, = 0.9875

iv. % accuracy, a = A x 100% = 0.9875 x 100% = 98.75%

Y n− Xn

10080

7980

100

n

nn

Y

XY

n

nn

Y

XY=A 1

Example

From the value in table 1.1 calculate Table 1.1

the precision of 6th measurement?

Solution

the average of measurement value

the 6th reading Precision =

No Xn

1 98

2 101

3 102

4 97

5 101

6 100

7 103

8 98

9 106

10 99100.510

1005

10

99....10198==

+++=X n

100.5

0.51

100.5

100.51001

= = 0.995

Example:

(Not to scale!)

Area of board, dimensions 11.3 cm 6.8 cm

Area = (11.3) (6.8) = 76.84 cm2

11.3 has 3 sig figs , 6.8 has 2 sig figs

76.84 has too many sig figs!

Proper number of sig figs in answer = 2

Round off 76.84 & keep only 2 sig figs

Reliable answer for area = 77 cm2

2

Q: For temperatures close to 0 K, the specific heat

capacity (c) for a particular solid is given by c = aT 3,

where T is its temperature and a is a constant. What

is the unit for a (in terms of the base units in SI)?

© Manhattan Press (H.K.) Ltd.

Solution:

Unit for specific heat capacity (c)

= J kg-1 K-1

= (N m) kg-1 K-1

= (kg m s-2) m kg-1 K-1

= m2 s-2 K-1

From c = aT3

a =

Unit for a =

= m2 s-2 K-4

3T

c

3

12-2

K

K s m

Units and dimensions (SB p. 10)

3

Q: The length l, Young modulus (E) and density (ρ) of

a tuning fork determine its period (T). Which of the

following equations is / are homogeneous?

(a) T = (b) T =

(c) T =

where y is a constant and g is the acceleration due to

gravity. Given that the unit for Young modulus (E) is

kg m–1 s–2.

3gE

yE

y

g

yE

4

© Manhattan Press (H.K.) Ltd.

Solution:

Unit for period (T) = s

(a) Unit for = s

The equation is homogeneous.

21

23

21

)s m

m(

m kg

s m kg

g

yE

21

32

21

33 )ms (m

s m kg

m kg

gE

y

(b) Unit for = s

The equation is homogeneous.

21

21

3

)s m kg

m kg( m

E

y

(c) Unit for = m2 s-1

The equation is not homogeneous.

Example: You are given a resistor with a resistance of

1500 ohms and a tolerance of 6%.What is the absolute

error, relative error in the resistance?

Solution: The absolute error is 6% of 1500 ohms = 90

ohms.

The relative error is 90/1500 = 6%

EXAMPLE 1: Refractive indices of glass are :

1.36, 1.29, 1.33, 1.34, 1.35, 1.32, 1.31, 1.34

Find the average reflective index

Absolute errors ,average absolute, Relative error &

Percentage error

EXAMPLE 2: R1= 200 ± 3 Ω, R2 = 500± 5 Ω are in

series. Find out max. absolute error, percentage error.

EXAMPLE 3:Two cylinders have length

l1= 6.52 ± 0.01 & l2 =4.48 ± 0.02. Find the difference of

length with percentage error.

Find significant figures:

1) 48956 2) 0.00003 3) 0.01270 4) 3.531

5) 8.3000 6) 9.1 x 10-31 7) 03.531 8) 0.02030

9) 6.23 x 1023 10) 1.457 + 83.2

11) 0.0367 – 0.004322 12) 4.36 x 0.00013

13) 12.300 ÷ 0.0230

EXAMPLE 4: A cube has length 7.206 m find area &

volume considering S.F.

REFERENCE BOOKS AUTHOR/PUBLICATION

ENGINEERING PHYSICSS S PATEL (ATUL

PRAKASHAN)

MODERN ENGINEERING

PHYSICSA S VASUDEVA

ENGINEERING PHYSICS K. RAJGOPALAN

Figure references links

1. http://www.imagesup.net/?di=814204410783

2. http://imgur.com/ZiSklF1

3. http://www.pixentral.com/show.php?picture=1iAs9Y

OV0erddkcxO5waMFWsi3Xx51

4. http://imgur.com/qEwxHuV

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