11873_engineering-physics_01 - · PDF fileEngineering 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

1Measurements

and ErrorsIf 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 ErrorsGross 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 ErrorsSystematic 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

Engineering Physics1.2

wrong observations, which may be due to lack of proper setting of the apparatus or individual carelessness in taking observations.

1.1.3 random ErrorsAfter 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:

Insufficient knowledge of process parameters and design conditions1.

Errors caused by person operating the instrument2.

Change in process parameters, irregularities, etc.3.

Poor maintenance4.

Poor design5.

Certain design limitations6.

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.


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.

1.7 Standard dEviation and variancES

The uncertainty is estimated by calculating the standard deviation. Let x1, x

2, x

3 …x

N be the results of an

experiment repeated N times. Then the standard deviation s is defined as

2

1

1( )

N

ii

x xN

s=

= -Â

where, Ê ˆ∫Á ˜Ë ¯

Â1i

i

x xN

represents the average of all the values of x. In this case, the uncertainty is of the order

of ±s. The standard deviation is defined in terms of square of the deviations from the mean, which is clear

from the term =

-Â 2

1

( )N

ii

x x in the above formula. Moreover, s2 is known as the variance of the data. The

standard deviation s is the root mean square deviation of the data, measured from the mean.

Fig. 1.1



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 a

1, a

2, a

3, … a

n are the individual values.

Then the arithmetic mean (say am) can be found as

am =

+ + +1 2 na a a

n (i)

This can also be written as

am =

=Â

1

n

ii

a

n (ii)

The absolute error can now be obtained from the following formula:

Dan = a

m – a

n (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

0.00033.256 m

Relative error in measuring the length of the table

= =0.001 m

0.01220.082 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%.



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 a

app, then

Absolute error = - appa a

Relative error = -

= -app app

1a a a

a a

Percentage, or percent error = - ¥app1 100

a

a

The important point is that the actual value a π 0.

1.11 ErrorS occurrinG in arithMEtic oPErationS

1.11.1 addition and SubtractionLet 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 divisionLet 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

d d d± = ± ±1 1

Q a b

Q a b (iv)



In terms of percentage error, we have

100 100 100Q a b

Q a b

d d d¥ = ¥ + ¥

For general treatment, we can consider a quantity

Q = kax bycz (v)

Then error can be determined as

Q a b cx y z

Q a b c

d d d d= + + (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 = =

-Â 2

1

1( )

N

ii

x xN

= + + + + + +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 R

2 = (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



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

Percentage error in V = ¥ =0.2

100 1%20

\ percentage of error in Q = 20 + 1 = 21%

or error in Q = ¥ =21

40 8.4 C100

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 V

2 = (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 + DV

2)

= ± (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 = r

-12.4

4.6

m = 2.69 m3 = 2.7 m3

rr

Ê ˆD D D= ± +Á ˜Ë ¯

V m

V m



DV = r

rÊ ˆD D

± +Á ˜Ë ¯m

Vm

= Ê ˆ± + ¥ = ±Á ˜Ë ¯

0.1 0.22.7 0.14

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 (21 1)

(3.5 0.5)

VR

I

±= =

± = 6.01 ±DR

D D D= +error in measurement =

R V I

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) = A l w

A l w

D D D= +

= + = +0.2 0.2

0.006 0.01135.4 18.4

= 0.017

Percentage error = D¥ = ¥ =100 0.017 100 1.7%

A

A

Example 9 A physical quantity Q is related to four observable a, b, c, d as follows:

=3 4

2

a bQ

d c



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

=3 4

2

a bQ

d c

Percentage error in Q is given by

D D D D D= + + +

13 4 2

2

Q a b c d

Q a b c d

Since D D D D= = = =1%, 3%, 4%, 3%

a b c d

a b c d

D= ¥ + ¥ + ¥ + ¥

13 1% 2 3% 4% 2 3%

2

Q

Q

= 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 10 Find absolute error, relative error and percentage of error of the approximation 3.14 to the value p.

Solution Absolute error = p- =3.14 0.0015926536

Relative error = p

p-

=3.14

0.000506957383

Percentage error = p

p-

◊ =3.14

100% 0.0506957383%

Example 11 The 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 mean1.29 1.33 1.34 1.35 1.32 1.36 1.30 1.33

1.3275 1.338

m+ + + + + + +

= =

Absolute errors are

Dm1 = m

mean – m

1 = 1.33 – 1.29 = 0.04

Dm2 = m

mean – m

2 = 1.33 – 1.33 = 0.00

Dm3 = m

mean – m

3 = 1.33 – 1.34 = –0.01



Dm4 = m

mean – m

4 = 1.33 – 1.35 = –0.02

Dm5 = m

mean – m

5 = 1.33 – 1.32 = 0.01

Dm6 = m

mean – m

6 = 1.33 – 1.30 = 0.03

Dm7 = m

mean – m

7 = 1.33 – 1.33 = 0.00

Mean absolute error,

m m m m m m m mm

D + D + D + D + D + D + D + DD = 1 2 3 4 5 6 7 8

mean(| | | | | | | | | | | | | | | |

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

Example 12 The radius of curvature of a concave mirror is given as R = +2

6 2

l h

h, where l and h are given

as 2 cm and 0.064 cm, respectively. Find the error in measuring the radius of curvature.

Solution l = 2 cm, Dl = 0.1 cm (LC of metre scale)

h = 0.064 cm, Dh = .001 cm (LC of spherometer)

D D -D D= + fi = + +

2 2

6 2

l h R l h hR

h R l h h

D D D= +

2 2R l h

R l h =

2 0.1 2 0.00100.1 0.03 0.131

2 0.064

¥ ¥+ = + =

Example 13 The 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 p p= fi = ¥ = =22

602 4 , 2.00

30

l lT g T

g T

22

904 3.14

(60 30)g = ¥ ¥

¥ = 887.364 cm/sec2

Maximum error in the value of g

22 2 2

2 2 2

2 44 4 30

( /30)

l l lg

T t t

pp p= = = ¥

Taking log on both sides,

log g = log 4 + 2 log p + log l – 2 log t + 2 log 30



Differentiating both sides, we get

D D D= + 2

g l t

g l t

l = 90 cm, Dl = 0.1 cm (LC of metre scale)

t = 60 sec, Dt = 0.1 sec (LC of stopwatch)

D ¥= + = + =

0.1 2 0.10.0011 0.0033 0.0044

90 60

g

g

= 0.0044 ¥ 100% = 0.4%

Example 14 In 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?

Solution The relative percentage uncertainty in the measure of fore is ∂Ê ˆ= Á ˜Ë ¯∂

22 26 6

EF V

V

∂Ê ˆ= ÆÁ ˜Ë ¯∂6 6 , 6 uncertainty in measurement of volume

EF V V

V

F µ V1/3

2/31

3

FV

V-∂

µ∂

fi 2= ¥ = ¥ = ¥¥2/3 /3

1 1 16 6 2.7 2.7

3 9.73 3(30)F V

V

fi 6F = 0.09

Example 15 One 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 2

4

dh

Fractional error = Ê ˆD DÊ ˆ +Á ˜ Á ˜Ë ¯ Ë ¯

22x y

x y

2 230.01

1010

V d h d

V d h d-D D D DÊ ˆ Ê ˆ= + = =Á ˜ Á ˜Ë ¯ Ë ¯



-D= ¥ 32 2 10

V

V

30.012 10

5

h

h-D

= - ¥

2 210 6 310 (8 10 ) .2 2 10

c m V

c m V- - -D D DÊ ˆ Ê ˆ= + = + ¥ = ¥Á ˜ Á ˜Ë ¯ Ë ¯

= 0.28%

Example 16 A battery powers two circuits C1 and C

2 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 I

2

through the individual circuits. If I1 and I

2 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 + I

2 = (400 ± 7) mA

DI = 7 mA

Example 17 A 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


d d d= +

V I R

V I R



±0.03 = ±0.04 + d R

R

d R

R = ±0.07 (max.) = 7 %

(b) P = I2R


d d d= +

2P I R

P I R

= 2 ¥ 0.04 + 0.07

= 0.15 = 15%



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11873_engineering-physics_01 - · PDF fileEngineering 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