1
Measurements
and Errors
If you are asked to measure the same object two different times, there is always a possibility
that the two measurements may not be exactly the same. Then the difference between these two
measurements is called a variation or error in the measurements. In general, there are two types of
errors in measurement. These are called static error and dynamic error.
1.1 Static Error and its Classification
Static error of a measuring instrument refers to the numerical difference between the true value
of a quality and its value of quantity. This gives different indications when the same quantity is
measured repeatedly. On the other hand, dynamic error is the difference between the true value of
a quantity changing with time and the value indicated by the instrument.
Static errors are classified into three categories, namely, gross errors or human errors, systematic
errors and random errors.
1.1.1 Gross Errors
Gross errors include all the human mistakes while reading and recording. Mistakes carried out in
calculating the errors also fall within this category. For example, while taking the reading from the
meter of the instrument, a person may read 21 as 27 or 31. Gross errors can be avoided if proper care
is taken in reading, recording the data and doing calculations accurately. We can also reduce such
errors by increasing the number of experimenters and by taking the average of more readings.
1.1.2 Systematic Errors
Systematic errors are the errors which tend to be in one direction (either positive or negative).
Systematic errors include instrumental, environmental and personal errors. Instrumental error may
be due to wrong construction or calibration of the measuring instruments. These errors also include
the loading effect, misuse of the instruments and zero error in the instrument. Environmental error
arises due to external conditions, which include temperature, pressure, humidity, external magnetic
field, etc. We can minimize the environmental errors by maintaining the temperature and humidity
of the laboratory constant through some arrangements, and ensuring that there is no external
magnetic or electrostatic field around the instrument. On the other hand, personal errors are due to
Engineering Physics_01.indd 1
8/12/2015 4:24:57 PM
1.2
Engineering Physics
wrong observations, which may be due to lack of proper setting of the apparatus or individual carelessness
in taking observations.
1.1.3 Random Errors
After calculating all the systematic errors, it is observed that there are still some more errors in the
measurement. These random errors are those errors which occur irregularly and are random with respect
to sign and size. Random and unpredictable fluctuations in temperature, voltage supply and mechanical
vibrations of experimental set-up may lead to random errors. The important property of a random error is that
it adds variability to the data but does not affect average performance for the group. For this reason, random
error is sometimes referred to as noise.
1.2 Dynamic Error
Dynamic errors are caused by the inertial properties of measuring instruments or equipment. Consider that
a varying quantity is recorded with the help of a recording device. Then the difference between the obtained
function and the actual process of change of the recorded quantity in time is called the dynamic error of the
given dynamic instrument. This is clear that these errors are caused by the time variation in the measured
quantity.
1.3 Sources of Errors
Sources of error, other than the inability of a piece of hardware to provide true measurements, are as under:
1. Insufficient knowledge of process parameters and design conditions
2. Errors caused by person operating the instrument
3. Change in process parameters, irregularities, etc.
4. Poor maintenance
5. Poor design
6. Certain design limitations
1.4 Accuracy and Precision
If you obtain a weight measurement of 6.5 kg for a given substance, and the actual or known weight is
10 kg, then your measurement is not accurate. It means your measurement is not close to the known value.
The closeness of a measured value to a standard or known value is referred to as accuracy. On the other hand,
precision refers to the closeness of two or more measurements to each other. The precision tells us to what
resolution (or limit) the quantity is measured. In the above example, if you weigh a given substance six times
and get 6.5 kg each time, then your measurement is very precise. It means the precision is independent of the
accuracy. You can be very precise but inaccurate, and also you can be accurate but imprecise. Moreover, you
can have accuracy without precision. For example, if on average your measurements for a given substance
are close to the known value, but the measurements may be far from each other.
Engineering Physics_01.indd 2
8/12/2015 4:24:57 PM
Measurements and Errors
1.3
1.5 Resolution
Resolution is the fineness to which an instrument can be read. We can take the example of two stopwatches,
out of which one is analog and the other is digital. Both are manually actuated and are looked for resolution.
The analog stopwatch has to be viewed on its dial. If we look closely, we can relate the big hand to the
smallest tick mark on the big dial. That tick mark is a tenth of a second. It means the best a good eye can do is
resolve a reading to 1/10 second. Hence, this is the resolution of the stopwatch. On the other hand, the digital
stopwatch has two digits beyond the seconds. So it subdivides time in hundredths of a second. Since it is easy
to read to 1/100 of a second, the resolution of the digital stopwatch is 1/100 second.
1.6 Measurement Uncertainty
Certainty is perfect knowledge which has total security from error.
Certainty is also the mental state of being without doubt. Every
experiment is approximate due to error in measurement. When a
number of measurements are done to a stable voltage (or other
parameter), the measured values will show a certain variation. This
variation is caused by thermal noise in the measuring circuit of the
measuring equipment and the measurement set-up. These variations
or the uncertainties are shown in Fig. 1.1.
Fig. 1.1
1.7 Standard Deviation and Variances
The uncertainty is estimated by calculating the standard deviation. Let x1, x2, x3 …xN be the results of an
experiment repeated N times. Then the standard deviation s is defined as
s=
Ê 1
where, x Á ∫
Ë N
ˆ
 xi ˜¯
1
N
N
 ( xi - x )2
i =1
represents the average of all the values of x. In this case, the uncertainty is of the order
i
of ±s. The standard deviation is defined in terms of square of the deviations from the mean, which is clear
N
from the term
 ( xi - x )2
in the above formula. Moreover, s2 is known as the variance of the data. The
i =1
standard deviation s is the root mean square deviation of the data, measured from the mean.
Engineering Physics_01.indd 3
8/12/2015 4:24:58 PM
1.4
Engineering Physics
1.8 Absolute Error
Absolute error is defined as the magnitude of difference between the actual and the approximated values of
any quantity. For example, we measure a given quantity n times and a1, a2, a3, … an are the individual values.
Then the arithmetic mean (say am) can be found as
a1 + a2 +  + an
n
This can also be written as
am =
(i)
n
 ai
am =
i =1
n
The absolute error can now be obtained from the following formula:
Dan = am – an
(ii)
(iii)
In simple words, absolute error is the amount of physical error in a measurement. If we use a metre stick
to measure a given distance, then the error may be ±1 mm or ±0.001 m. This is the absolute error of the
measurement. For the measurement of a quantity x, the absolute error is Dx.
1.9 Relative Error
Relative error gives us an idea of how good a measurement is relative to the size of the object being measured,
For example, we measure the height of a room and the length of a small table by using a metre stick. We find
the height of the room as 3.256 m ± 1 mm and the length of the table as 0.082 m ± 1 mm. Then
Relative error =
Absolute error
Value of thing measured
(i)
\ relative error in measuring the height of the room
=
0.001 m
= 0.0003
3.256 m
Relative error in measuring the length of the table
=
0.001 m
= 0.0122
0.082 m
Here, it is clear that the relative error in measuring the length of the table is larger than the relative error in
measuring the height of the room. In both the cases, however, the absolute error is the same.
1.10 Percentage Error
If the relative error is represented in percent, then the error is called percentage error. For example, in the
above example, the percentage error in measuring the height of the room is 0.0003 ¥ 100 = 0.03%, while the
percentage error in measuring the length of the table is 0.0122 ¥ 100 = 1.22%.
Engineering Physics_01.indd 4
8/12/2015 4:24:58 PM
Measurements and Errors
1.5
The absolute error, relative error and percentage error can be summarized as follows. If we represent the
given or actual value by a and its approximated value as aapp, then
Absolute error = a - aapp
Relative error =
a - aapp
a
= 1-
aapp
Percentage, or percent error = 1 -
a
aapp
a
¥ 100
The important point is that the actual value a π 0.
1.11 Errors Occurring in Arithmetic
Operations
1.11.1 Addition and Subtraction
Let us consider two measured values a ± da and b ± db in which a and b are actual values, whereas da and db
are absolute errors of a and b respectively.
The error obtained in the sum of these quantities is given by
Q ± dQ = (a ± da) + (b ± db)
= (a + b) + (±da ± db)
fi
dQ = da + db
(i)
Similarly, the error obtained in their difference is given by
Q ± dQ = (a ± da) – (b ± db)
= (a – b) + (±da ± db)
fi
dQ = da + db
(ii)
Hence, in arithmetic operations of addition and subtraction, the absolute error in the resultant is the sum of
the absolute errors of individual quantities. So errors are always added in these operations.
1.11.2 Multiplication and Division
Let us consider the two measured quantities a ± da and b ± db, where a and b are actual quantities and da and
db are the absolute errors in a and b, respectively. The errors occurring in multiplication and division can be
estimated as given below.
Q ± dQ = (a ± da) (b ± db) = ab ± bda ± adb ± dadb
(iii)
On dividing by Q on the LHS and by ab on the RHS, we obtain
1±
Engineering Physics_01.indd 5
dQ
da db
=1±
±
Q
a
b
(iv)
8/12/2015 4:24:58 PM
1.6
Engineering Physics
In terms of percentage error, we have
dQ
da
db
¥ 100 =
¥ 100 +
¥ 100
Q
a
b
For general treatment, we can consider a quantity
Q = kax bycz
(v)
Then error can be determined as
dQ
da
db
dc
=x
+y
+z
Q
a
b
c
(vi)
It means, if Q is a measure of x power of a, y power of b and z power of c, then the fractional error will be the
sum of x times of fractional error in a, y times of fractional error in b and z times of fractional error in c. This
also infers that maximum error will be encountered due to the quantity carrying highest power. Therefore, in
experiments, that quantity should be determined with more precision in order to minimize the error.
Solved Examples
Example 1 Find the standard deviation of the numbers 1, 5, 6, 7, 8, 10, 12.
Solution The mean of these numbers is found to be 7. The deviations are –6, –2, –1, 0, 1, 3, 5, respectively. Now
using the formula
s =
1
N
N
 ( xi - x )2
i =1
=
1
[36 + 4 + 1 + 0 + 1 + 9 + 25]
7
=
76
7
So,
s =
10.86
= 3.295
Example 2 If two resistances given as R1 = (50 ± 5) W and R2 = (150 ± 2) W are connected in series, then
find the equivalent resistance.
Solution R = (50 ± 5) + (150 ± 2)
= (50 + 150) ± (5 + 2)
= (200 ± 7) W
Engineering Physics_01.indd 6
8/12/2015 4:24:58 PM
Measurements and Errors
1.7
Example 3 If the mass of a bulb with air is 98.625 ± 0.002 g and the mass of an empty bulb is 98.305 ±
0.002 g, then find the mass of air.
Solution Error in difference = (a ± da) – (b ± db)
= (a – b) + (±da ± db)
= (98.625 – 98.305) ± (0.002 + 0.002)
= 0.320 ± 0.004 g
Example 4 If the capacity of a capacitor is C = 2 ± 0.4 F and the applied voltage is V = 20 ± 0.2 V, then
find the charge on the capacitor.
Solution Charge on capacitor, Q = CV = 2 ¥ 20 = 40 C
Percentage error in C =
0.4
¥ 100 = 20%
2
0.2
¥ 100 = 1%
20
\ percentage of error in Q = 20 + 1 = 21%
Percentage error in V =
or error in Q = 40 ¥
21
= 8.4 C
100
Hence, charge on the capacitor Q = 40 ± 8.4 C
Example 5 The volumes of two bodies are measured to be V1 = (10.2 ± 0.02) cm3 and V2 = (6.4 ± 0.01) cm3.
Calculate the sum and difference in volumes with error limits.
Solution V1 = (10.2 ± 0.02) cm3
V2 = (6.4 ± 0.01) cm3
DV= ± (DV1 + DV2)
= ± (0.02 + 0.01) cm3
= 0.03 cm3
Example 6 The mass and density of a solid sphere are measured to be (12.4 ± 0.1) kg and (4.6 ± 0.2) kg/m3.
Calculate the volume of the sphere with error limits.
Solution Here, m ± Dm = (12.4 ± 0.1) kg
r ± Dr = (4.6 ± 0.2) kg/m3
Volume V =
m 12.4
= 2.69 m3 = 2.7 m3
r
4.6
Ê Dm Dr ˆ
DV
=±Á
+
V
r ˜¯
Ë m
Engineering Physics_01.indd 7
8/12/2015 4:24:58 PM
1.8
Engineering Physics
Ê Dm Dr ˆ
DV = ± Á
+
V
r ˜¯
Ë m
Ê 0.1 0.2 ˆ
= ±Á
+
¥ 2.7 = ± 0.14
Ë 12.4 4.6 ¯˜
V ± DV = (2.7 ± 0.14) m3
Example 7 A current of (3.5 ± 0.5) A flows through a metallic conductor and a potential difference of
21 ± 1 volts is applied. Find the resistance of the wire.
Solution Given V = 21 ± 1 volts, DV = 1, I = 3.5 ± 0.5 A
DI = 0.5 A
Resistance R =
V
(21 ± 1)
= 6.01 ± DR
=
I (3.5 ± 0.5)
DR
DV DI
= error in measurement =
+
R
V
I
1 0.5
+
21 3.5
=
= 0.048 + 0.143 = 0.19
fi DR = 0.19 ¥ R = 0.19 ¥ 6 = 1.14
Effective resistance R = 6 ± 1.14 W
Example 8 A rectangular board is measured with a scale having an accuracy of 0.2 cm. The length
and breadth are measured as 35.4 cm and 18.4 cm, respectively. Find the relative error and
percentage error of the area calculated.
Solution l = 35.4 cm, Dl = 0.2 cm
w = 18.4 cm and Dw = 0.2 cm
Area (A) = l ¥ w = 35.4 ¥ 18.4 = 651.36 cm2
Relative error in area (dA) =
DA Dl Dw
=
+
A
l
w
Percentage error =
=
0.2
0.2
+
= 0.006 + 0.011 = 0.017
35.4 18.4
DA
¥ 100 = 0.017 ¥ 100 = 1.7%
A
Example 9 A physical quantity Q is related to four observable a, b, c, d as follows:
Q=
Engineering Physics_01.indd 8
a3 b4
d2 c
8/12/2015 4:24:58 PM
Measurements and Errors
1.9
The percentage errors of measurement in a, b, c and d are 1%, 3%, 4% and 3% respectively.
What is the percentage error in the quantity Q? If the value of Q calculated using the given
relation is 8.768, to what value should the result be rounded?
Solution Given
Q=
a3 b4
d2 c
Percentage error in Q is given by
DQ
Da
Db 1 Dc
Dd
=3
+4
+
+2
Q
a
b
2 c
d
Da
Db
Dc
Dd
= 1%,
= 3%,
= 4%,
= 3%
a
b
c
d
Since
DQ
1
= 3 ¥ 1% + 2 ¥ 3% + ¥ 4% + 2 ¥ 3%
Q
2
= 3% + 6% + 2% + 6%
= 17%
\ percentage error in Q = 17%
If the calculated value of Q is 8.768, the roundoff value is 8.8.
Example 10Find absolute error, relative error and percentage of error of the approximation 3.14 to the
value p.
Solution Absolute error = 3.14 - p = 0.0015926536
3.14 - p
Relative error =
Percentage error =
p
= 0.000506957383
3.14 - p
p
◊ 100% = 0.0506957383%
Example 11The refractive index (m) of water is found to have the values 1.29, 1.33, 1.34, 1.35, 1.32, 1.36,
1.30 and 1.33. Calculate the mean value, absolute error, the relative error and percentage
error.
Solution mmean =
Absolute errors are
1.29 + 1.33 + 1.34 + 1.35 + 1.32 + 1.36 + 1.30 + 1.33
= 1.3275  1.33
8
Dm1 = mmean – m1 = 1.33 – 1.29 = 0.04
Dm2 = mmean – m2 = 1.33 – 1.33 = 0.00
Dm3 = mmean – m3 = 1.33 – 1.34 = –0.01
Engineering Physics_01.indd 9
8/12/2015 4:24:58 PM
1.10
Engineering Physics
Dm4 = mmean – m4 = 1.33 – 1.35 = –0.02
Dm5 = mmean – m5 = 1.33 – 1.32 = 0.01
Dm6 = mmean – m6 = 1.33 – 1.30 = 0.03
Dm7 = mmean – m7 = 1.33 – 1.33 = 0.00
Mean absolute error,
Dmmean =
(| D m1 | + | D m2 | + | D m3 | + | D m 4 | + | D m5 | + | D m6 | + | D m7 | + | D m8 |
8
= (0.04 + 0.00 + 0.01 + 0.02 + 0.01 + 0.03 + 0.03 + 0.00) ∏ 8
= 0.14 ∏ 8 = 0.0175  0.02
Relative error (dm) = Dmmean/mmean = 0.02 ∏ 1.33 = ±0.015  0.02
Percentage error = Dmmean/mmean ¥ 100% = ±0.015 ¥ 100 = ±1.5
l2 h
+ , where l and h are given
6h 2
as 2 cm and 0.064 cm, respectively. Find the error in measuring the radius of curvature.
Example 12The radius of curvature of a concave mirror is given as R =
Solution l = 2 cm, Dl = 0.1 cm (LC of metre scale)
h = 0.064 cm, Dh = .001 cm (LC of spherometer)
R=
l2 h
+
6h 2
fi
DR 2 Dl
-Dh
Dh
=
+
+
R
l
h
h
2 ¥ 0.1 2 ¥ 0.001
DR 2 Dl 2 Dh
=
+
= 00.1 + 0.03 = 0.131
=
+
2
0.064
R
l
h
Example 13The time of 30 oscillations of a simple pendulum whose length is 90 cm was observed to be
60s. According to given data, find the value of g and determine percentage error in the value
of g.
Solution T = 2p
l
l
60
fi g = 4p 2 ¥ 2 , T =
= 2.00
g
30
T
g = 4 ¥ 3.142 ¥
90
(60 ¥ 30)2
= 887.364 cm/sec2
Maximum error in the value of g
g = 4p 2
2l
T2
= 4p 2
l
(t / 30)2
=
4p 2l
t2
¥ 302
Taking log on both sides,
log g = log 4 + 2 log p + log l – 2 log t + 2 log 30
Engineering Physics_01.indd 10
8/12/2015 4:24:58 PM
Measurements and Errors
1.11
Differentiating both sides, we get
Dg Dl
Dt
=
+2
g
l
t
l = 90 cm, Dl = 0.1 cm (LC of metre scale)
t = 60 sec, Dt = 0.1 sec (LC of stopwatch)
Dg 0.1 2 ¥ 0.1
=
+
= 0.0011 + 0.0033 = 0.0044
g
90
60
= 0.0044 ¥ 100% = 0.4%
Example 14In a measurement of the viscous drag force experienced by spherical particles in a liquid, the
force is found to be proportional to V1/3 where V is the measured volume of each particle. If V
is measured to be 30 mm3, with an uncertainty of 2.7 mm3, what will be the resulting relative
percentage uncertainty in the measured force?
2
Solution The relative percentage uncertainty in the measure of fore is 6 F 2 = ÊÁ ∂E ˆ˜ 6V 2
Ë ∂V ¯
Ê ∂E ˆ
6F = Á
6V , 6V Æ uncertainty in measurement of volume
Ë ∂V ¯˜
F µ V1/3
∂F 1 -2/3
µ V
∂V 3
fi 6 F =
1
3V 2/3
¥ 6V =
1
3(30)2 /3
¥ 2.7 =
1
¥ 2.7
3 ¥ 9.7
fi 6F = 0.09
Example 15One gram of salt is dissolved in water that is filled to a height of 5 cm in a beaker of 10 cm
diameter. The accuracy of length measurement is 0.01 cm while that of mass measurement is
0.01 mg. When measuring the concentration c, what is the fractional error Dc/c?
Solution c = mass/volume
V = pr2h =
p d2
h
4
Fractional error =
2
DV
Ê Dd ˆ
Ê Dh ˆ
= Á
+Á ˜
Ë d ˜¯
Ë h ¯
V
Engineering Physics_01.indd 11
2
Ê Dy ˆ
Ê Dx ˆ
ÁË x ˜¯ + ÁË y ˜¯
2
2
Dd 0.01
=
= 10 -3
d
10
8/12/2015 4:24:58 PM
1.12
Engineering Physics
DV
= 2 2 ¥ 10 -3
V
2
Dh 0.01
=
- 2 ¥ 10 -3
h
5
2
Dc
Ê Dm ˆ
Ê DV ˆ
= Á
+Á
= 10 -10 + (8 ¥ 10 -6 ) = .2 2 ¥ 10 -3 = 0.28%
Ë m ˜¯
Ë V ˜¯
c
Example 16A battery powers two circuits C1 and C2 as shown in Fig. 1.2.
Fig. 1.2
The total current I drawn from the battery is estimated by measuring the currents I1 and I2
through the individual circuits. If I1 and I2 are both 200 mA and if the errors in the measurement
are 3 mA and 4 mA respectively. What is the error in the estimate of I?
Solution I1 = (200 ± 3) mA
I2 = (200 ± 4) mA
I = 400 ± DI
I = I1 + I2 = (400 ± 7) mA
DI = 7 mA
Example 17A resistance is measured by passing a current through it and measuring the resulting voltage
drop. If the voltmeter and ammeter have uncertainties of 3% and 4% respectively, then
(a) Find the uncertainty in resistance.
(b) Find the uncertainty in the computed value of power dissipated across the resistance.
Solution (a) V = IR
Taking log on both sides,
dV d I d R
=
+
V
I
R
Engineering Physics_01.indd 12
8/12/2015 4:24:58 PM
Measurements and Errors
±0.03 = ±0.04 +
1.13
dR
R
dR
= ±0.07 (max.) = 7 %
R
(b) P = I2R
Taking log on both sides,
d P 2d I d R
=
+
P
I
R
= 2 ¥ 0.04 + 0.07
= 0.15 = 15%
Engineering Physics_01.indd 13
8/12/2015 4:24:58 PM
Engineering Physics_01.indd 14
8/12/2015 4:24:58 PM