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AKASH MULTIMEDIA 40

h Accuracy and Precisionh Types of Errors

CHAPTER

h Methods to minimise Errorsh Significant Figures

2Chandrashekhara Venkata Raman was born on November 7, 1888.He carried out his experimental research in the laboratory of IndianAssociation for the Cultivation of Science at Calcutta with BurningDesire. He carried out research in acoustics and optics. On February28, 1928, through his experiments on the scattering of light, hediscovered the Raman effect. He was the first Asian and first non-White to receive any Nobel Prize in the sciences. In 1941 he wasawarded the Franklin Medal. In 1954 he was awarded the BharatRatna. He retired from the Indian Institute in 1948 and a year later heestablished the Raman Research Institute in Bangalore, where heworked till his death.

C.V. Raman1888-1970

MEASUREMENT OF PHYSICAL QUANTITIES

2.1 INTRODUCTIONWe can say that we know about a physical

quantity only if we are able to measure it.Measurement is comparison of a physical quantity

with a standard reference of same physical quantity. Inevery measurement final result is a number followedby a unit. For example length of a black board is2.5metre .

Here metre is the unit and 2.5 is numerical value,that is length of the black board is expressed as 2.5times the unit of length. An accurate measurementof physical quantities plays a very important role inthe world of science.

However inspite of great technologicaladvancement and sophistication in the developementof instrumentation, no measurement can be free fromerrors. Errors may be because of several reasonsboth controlable and uncontrolable.

Hence a knowledge of errors is essential tointerpret the results and evaluate the measurementson a scientific basis.2.2. ERROR

Error is the amount of uncer tainity that ispr esent in the measur ement made with ameasur ing instrument.

It is the difference between the measured valueand the true value.2.3. TRUE VALUE

We can never determine exactly the real or truevalue of a physical quantity due to the presence of

error. To minimize the errors we repeat theobservations a large number of times and take thearithmetic mean of all the readings.

The ar ithmetic mean of a large number ofrepeated measurements is taken as the true valueor actual value.2.4. ARITHMETIC MEAN

If X1, X2. . . . . . . . . XNare the results of anexperiment repeated N times, the arithmetic meanx is given by

i1 2 n xx x ............ xxn n

Arithmetic mean is the best value of x that canbe obtained from the N measurements. This is 'true'when N is very large.2.5. ACCURACY AND PRECISION

Accuracy is a measure of the closeness of themeasured value to the true value.

Smaller the uncertainty in the measured value,greater is its accuracy.

Precision refers to the agreement among agroup of measured values.

It does not imply anything about their relationto the true value.

A precise value doesn't necessarily mean anaccurate value.

Eliminating systematic error improves accuracyRepeating the experiment number of times

improves precision

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No measurement can be made with complete'accuracy.' For this , it must be shown by a valuewith infinite number of decimal places.Example :

The length of a straight line is measured anumber of times by a number of observers. Thefollowing are the results of these measurements.

Actual length = 3.785 cm 0.001cm1st set of measurements 3.8 cm, 3.9cm, 3.7cm2nd set of measurements 3.478 cm, 3.479 cm,

3.478 cm, 3.478 cm, 3.479 cm3rd set of measurements 3.55 cm, 3.65 cm,

3.45 cm, 3.35cm4th set of measurements 3.784 cm,3.785 cm,

3.784 cm, 3.785 cm, 3.784cmFirst set of measurements are accurate because

it is closest to the actual value but not precise.Second set of measurements are not accurate

but it is most precise because the readings arereproducible.

Third set of measurements are neither accuratenor precise.

Fourth set of measurements are accurate aswell as precise.2.6 CLASSIFICATION OF ERRORS

Errors creep in the observed or calculated valuesdue to various reasons. They are categorized undertwo heads.

(a) Systematic errors(b) Random errors

2.7. SYSTEMATIC ERRORSError that is allways unidirectional is called

systematic errorIf some error is occurring in the positive

direction only (or in the negative direction only) allthe time, it is systematic error.Example :

If measured value is allways 2 units more thantrue value it is systematic error. (positive) If measured value is allways 3 units less thantrue value it is systematic error. (negative)

i) Instrumental errors : These are inherenter rors of the appar atus and the measur inginsturements used.Example :

Zero-error of a measuring instrument. If a scaleis worn off at the starting end upto 0.3cm readingand we do not notice it and treat it as zero, then everylength we measure will be always 0.3cm more thanthe true value.

The instrumental error can be detected byinterchanging two similar instruments or by usingdifferent methods for measuring the same physicalquantity.ii) Constant errors : When the result of a ser iesof measur ements ar e in er ror by the sameamount, such an error is called constant error.

These are also instrumental errors. Zero error isa constant error.iii) Environmental errors : These arise due toexternal conditoins, that is, due to changes in theevnironment. During the experimnetal measurementthere maybe changes in external conditions i.e.,changes in temperature, pressure, humidity, windvelocity etc. For example if a metal scale is calibratedin winter and is used in summer as it expands insummer the measured values will be less than truevalues.iv) Personal errors : These are entirely due to thepersonal pecularities of the experimenter. Individualbias, lack of proper setting of the apparaturs,carelessness in taking observations (without takingthe required necesary precautions.) etc. are the causesfor this type of error. A perosn may be habituated tohold his eyes (head) always a bit too far to the right(or left) while taking the reading with a scale. Thiswill give rise to parallax error.

If a person keeps his eye-level below the levelof mercury in a barometer all the time, his readingswill have systematic error.

These errors can be mimimized by obtainingseveral readings carefully and then taking theirarithmetical mean.

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v) Imperfection in exper imental technique orprocedure : Sometimes, even when we know thenature of the error, it cannot be eliminated due toimperfection in experimental arrangement. Forexample, in calorimetry, loss of heat due to radiation,the effect on weighing due to buoyancy of air, etc.These errors will always exist.2.8. MINIMISATION OF SYSTEMATIC

ERRORSWe can minimise systematic errors.

1) by selecting better instruments with higherresolution.

2) by taking care to avoid personal bias.3) by improving the experimental technique.

2.9. RANDOM ERRORSThe errors which are ir regular whose cause

is not known, and random in nature in their signand size are called random errors

Unlike systematic errors these errors are atrandom with respect to the sign and magnitude. Theyoccur irregularly - some times increasing, some timesdecreasing with changing magnitudes.These errors arise due to the random fluctuations inexperimental conditions that cannot be predicted. Forexample, the random fluctuations in temperature,voltage supply and mechanicl vibrations ofexperimental sep up etc. are the causes for radnomerrors.

The diameter of a wire may be varying fromposition to position along its length due to nonuniform cross section.

Hence we get random errors in the measurementof diameter. At some places the value will be more,at some places less with different magnitudes.

The uneven tightening of the screw in a screwgauge at different places of a wire of uniform crosssection also results in random errors .

Thus, even when the same person repeats the sameobservation, he may get different reading each time.Example:Suppose true length of a body is 2.5cmSame person measures it three times as

2.3cm, 2.7cm, 2.9cm, then the errors are

0.2cm, +0.2cm, +0.4cm respectivelyRandom errors are also called indeterminate

errors.Random errors can not be eliminated but can

be minimized by increasing the number ofobservations. The total random errors are determined bystatistical methods.

Random errors can not be minimized bytaking precautions.

Note : Allmost all measurements contain bothsystematic and random errors.

2.10. GROSS ERRORSThe gross errors are the result of sheer

carelessness on the part of the observer.Causes for gross error are (a) neglect of the

sources of error and (b) reading the instrumentincorrectly.

Example :1) In a tangent galvanometer experiment, the

coil is to be placed exactly in the magnetic meridianand care should be taken so that no other magneticmaterials are present in the vicinity.

2) In the measurement of length of a pendulumthe obserever may read it as 52.4cm but may recordit as 25.4cm. This is gross error due to carelessrecording.

No corrections can be applied to these grosserrors.Taking all precautions that are required, wecan minimize gross errors.

2.11. TYPES OF ERRORS QUANTITATIVE APPROACHThe same error can be expressed in different

ways as an absolute error, a relative error or apercentage error.Error (a): Difference between measured valueand actual value (or true value) is called error

a = measured value actual value.So it can be +ve or ve.

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Example :Suppose true length of a body is 2.5cm

If it is measured as 2.7cm, then the error is +0.2cmIf it is measured as 2.2cm, then the error is -0.3cm

Absolute error (| a|) : The magnitude of thedifference between the true value of the measuredphysical quantity and the value of individualmeasurement is called the absolute error of thatmeasurement.

If amean = actual vale, and ai = ith observed value,

then the absolute error ( a ) in the ith observed value

is defined as : ia = i meana a

Absolute error is allways positive and has sameunits as that of the quantity being measured

Mean absolute error ( amean

)The ar ithmetic mean of all the absolute

errors is considered as the mean absolute erroror final absolute error of the value of the physicalquantity concerned.

Suppose a1 ,a2,a3,...an are measured values of aquantity and their mean value is amean

Each individual observed value has an errorattached with it, as follows.

a1 = 1 meana a

a2 = 2 meana a

a3 = 3 meana a

an = n meana a

amean =

n

ii 1

a

n

So we write actual value, a = amean amean.

Relative error. The relative (or propor tional) error of ameasured physical quantity is the ratio of themean absolute error ( ) meana to the mean value( )meana of the quantity measured.

Relative error = Absolute errorActual value =

mean

mean

aa

Relative error is a pure number having no units.Percentage error = Relative error x 100

100 meanmean

a

a

Problem : 2.1In an experiment, the values of refractive index of glasswere found to be 1.54, 1.53, 1.44, 1.54, 1.56 and 1.45in successive measurements. Calculate (i) Mean valueof refractive index of glass (ii) Absolute error in eachmeasurement (iii) Mean absolute error (iv) relative errorand (v) Percentage error.

Sol. (i) Mean value of refractive index,1.54 1.53 1.44 1.54 1.56 1.45 1.51

6meanm

(ii) Taking meanm as the true value, the errors in the sixmeasurements are1.54 1.51 = +0.03; 1.53 1.51 = +0.02;

1.44 1.51 = 0.071.54 1.51 = + 0.03; 1.56 1.51 = + 0.05and 1.45 1.51 = 0.06

The absolute errors are0.03, 0.02, 0.07, 0.03, 0.05 and 0.06(iii) Mean absolute error in the value of m is

0.03 0.02 0.07 0.03 0.05 0.066

meanm

= 0.26 0.04

6

(iv) Relative error in the valiue of m ,0.04 0.02649 0.031.51

meanmean

mm

(v) Percentage error in the value ofm = 0.03 100 = 3%

Problem : 2.2Readings of length of a pole are 2.63 m, 2.56 m, 2.42m, 2.71 m and 2.80 m. Calculate the absolute errorsand relative errors and percentage errors. What doyou think of the actual value of the length and itslimits?

Sol. The mean value of length

L = (2.63 2.56 2.42 2.71 2.80)m5

= 13.125

m = 2.624m = 2.62m

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As the lengths are measured to a resolution of0.01m, all lengths are given to the second place ofdecimal, it is proper to round off this mean lengthalso to the second place of decimal.In the first measurement Error = a1 = 2.63m 2.62m = + 0.01m Absolute error = 0.01m Relative error = 0.01/2.62 =0.0038 Percentage error = relative error x 100 = 0.38In the second measurement Error = a2 = 2.56m 2.62m = 0.06m Absolute error = 0.06m Relative error = 0.06/2.62 =0.023 Percentage error = relative error x 100 = 2.3In the third measurement Error = a3 = 2.42m 2.62m = 0.2m

Absolute error = 0.2m Relative error = 0.2/2.62 =0.076 Percentage error = relative error x 100 = 7.6In the fourth measurement Error = a4 = 2.71m 2.62m = +0.09m

Absolute error = 0.09m Relative error = 0.09/2.62 =0.034 Percentage error = relative error x 100 = 3.4In the fifth measurement Error = a5 = 2.80m 2.62m = +0.18m

Absolute error = 0.18m Relative error = 0.18/2.62 =0.068 Percentage error = relative error x 100 = 6.8Mean or final absolute error

51 ia

5

(0.01 0.06 0.20 0.09 0.18)m5

= 0.54m/5 = 0.108 m = 0.11m

This means that the length is (2.62 0.11m)i.e., it lies between (2.62 + 0.11 m) and (2.62 0.11m)i.e., between 2.73 m and 2.51 m.

2.11. COMBINATION OR PROPAGATION OF ERRORS Suppose we want to calculate acceleration due

to gravity using the formula, g = 2 2L4T

We can measure length L using scale and wecan measure timeperiod T using clock.

There may be errors in all measuringinstruments and measurements.

We should know the errors in individulmeasurements of length(L) and time (T) , then combinethem to find the error in the calculation of ' g '.

So we must know how errors combine(orpropagate) in addition,subtraction, multiplication anddivision.

i) Combination of errors in case of additionSuppose a physical quantity Z is sum of the

physical quantities X and YZ = X+YLet X and Y be the respective absolute

errors in X and Y. Then the values of X and Y willbe X X andY Y

Let the error in Z be ZSince Z = X+Y

( ) ( )Z Z X X Y Y ( )Z Z X Y X Y

Maximum possible error in Z is Z X Y

(If Z =X+Y then), Maximum possible error

= absolute error +absolute error

X Y

Percentage error in Z =X Y

100X Y

ii) Combination of errors in case of subtractionSuppose a physical quantity Z is difference of thephysical quantities X and Y Z = X Y

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Let X and Y be the respective absoluteerrors in X and Y. then the values of X and Y willbe X X and Y Y

Let the error in Z be ZSince Z = X YZ Z (X X) (Y Y)

Z Z (X Y) X Y

Maximum possible error in Z is Z X Y

(If Z =X-Y then), Maximum possible error

= absolute error +absolute error

X Y

Percentage error in Z =X Y 100X Y

Note : In both addition and subtaction maximumpossible error = sum of absolute errors

Problem : 2.3I f L = 2.06 cm 0.02 cmB = 1.11 cm 0.03cmWhat are (L + B) and (L B) equal to ?

Sol. L + B = 3.17 cm 0.05 cmL B = 0.95 cm 0.05 cmPlease note that actual values i.e. 2.06 cm and 1.11 cmare added in case of (L + B) and subtracted in case of(L B), but absolute errors are added in both cases.

*Problem : 2.4Two objects A and B are of lengths 5 cm and 7cmdetermined with errors 0.1 cm and 0.2 cm respetively.What is the error in determining (a) the total lenth and(b) the difference in their lengths ?

Sol . a = 5cm, a = 0.1cmb = 7 cm, b = 0.2cmIf x = a + b is the total length, then x = a + b = 0.1 + 0.2 = 0.3 cm and x = ( 5+7) 0.3= (12 0.3) cm.If 'x' is the difference between the lengths,then x = a + b = 0.3cm and

x 5 7 0.3 2 0.3 cm Note : I f a ' constant' is multipl ied with an obser ved/

measured value, the result will have its absolute errorequal to 'constant' times the absolute er ror in theobser ved / measur ed value. Thi s means, thepercentage error in the result will remain unchanged.

Problem : 2.5

I f L = 2.01 m 0.01m, what is 3L ?Sol. Here, we should equate 3L with

(L + L + L) to find out the absolute error in 3L.

Thus 3L = (3 2.01) m 3 0.01 m

= 6.03 m 0.03m Ans.

Problem : 2.6

If L1 = 2.02 m 0.01 m, L2 = 1.02m 0.01 m, determineL1 + 2L2

Sol. L1 + 2L2 = L1 + L2 + L2 = (2.02m0.1m) + (1.02m0.01m) + (1.02 m 0 .01m)

= 4.06 m 0.03 m

i i i ) Combinat ion of er r or s in case ofmultiplication

Suppose a physical quantity Z is the product ofthe physical quantities X and Y

Z = XYLet X and Y be the respective absolute

errors in X and Y. then the values of X and Y will beX X and Y Y

Let the error in Z be ZZ = XYlogZ = logX +logYdifferentiating both sidesdZ dX dYZ X Y

Maximum relative errorZ X Y

Z X Y

In multiplication (Z=XY), Maximum relative error Z X YZ X Y

iv) Combination of errors in case of divisionSuppose X,Y, Z are physical quantitie such that

XZY

Let X and Y be the respective absoluteerrors in X and Y. then the values of X and Y will beX X and Y Y

Let the error in Z be ZXZY

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logZ = logX logYdifferentiating both sidesdZ dX dYZ X Y

Maximum relative errorZ X Y

Z X Y

In division Maximum relative error Z X YZ X Y XZ

Y

Note : Both in multiplication and division Maximumrelative error in the result is the sum of the relativeerrors of the quantities being multiplied or divided.

Similarly Both in multiplication and divisionMaximum percentage error in the result is the sumof the percentage errors of the quantities beingmultiplied or divided.

*Problem : 2.7The length and breadth of a rectangular object are25.2 cm and 16.8cm respectively and have beenmeasured to an accuracy of 0.1cm. Find the relativeerror and percentage error in the area of the object.

Sol. Area A = l bl = 25.2 cm and l = 0.1cmb = 16.8 cm and b = 0.1cm

Relative error in area A bA b

0.1 0.125.2 16.8

= 0.004 + 0.006 = 0.01A 100 0.01 100

A

Percentage error ( a = 1%.

* Problem : 2.8In an experiment to determine the value of accelerationdue to gravity g using a simple pendulum, the measuredvalue of length of the pendulum is 31.4cm known to 1mm accuracy and the time period for 100 oscillationsof pendulum is 112.0s known to 0.01s accuracy. Findthe accuracy in determining the value of g.

Sol. (Accuracy is to be taken as the error involved)l = 31.3 cm and l = 1mm = 0.1cm

T = 112.0100 = 1.12s and T = 0.01s

Formula for g is 2 2g 4 T

g T2g T

=0.1 0.012

31.4 1.12

= 0.003 + 0.02 = 0.023

Relative error in determining g is 0.023 and percentageerror = 0.023 100 = 2.3%

Problem : 2.9 I f L = 20.04 m 0.01 m B = 2.52 m 0.02 mWhat are the values of (L B) and (L / B)?

Sol. Given errors are absolute errors, while the rule says thatpercentage errors are to be added. Hence, the first stepwill be to convert the given absolute errors intopercentage errors.

L = 20.04 m 0.01

20.04 100%

= 20.04 m 0.05%

B = 2.52 m 0.022.52

100%

= 2.52 m 0.79%

L B = (20.04 2.52)m2 (0.05 + 0.79)% = 50.50 m2 0.84%This is the result. However, since the data given in thequestion was in terms of absolute errors, so we shouldgive our result also in absolute errors.

L B = 50.50 m2 0 .8 4100 50.50m

2

= 50.50 m2 0.42 m2 Ans.Note : The result should be rounded to which place of decimal

is explained later in this chapter.Similarly,LB

= 20.04 m2.52 m

(0.05 + 0.79)%

= 7.95 0.84% = 7.95 0 .8 41 0 0 7.95

= 7.95 0.067.

v) Combination of errors in case of a quantityr ised to different powers Suppose the physical quantity z depends onthe quantity x as z=kxn where k is a constant and nis a real numberLet x and z be the respective absolute errorsin x and z.

Let nz kx

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log z = log k + n log x Differentiating both sides

z xn

z x

If Maximum relative error Z XnZ X nZ kX

Percentages error in z = n times percentage errorin x

Similar ly Let Z =p q

r

A .BC

Where A, B, C and Z are variables, and p, qand r are constants.

log Z = p log A + q log B r log CDifferentiating both sidesdZZ = p.

dAA + q.

dBB r..

dCC

Z A B Cp q r

Z A B C

Maximum fractional error in Z isZ A B Cp q r

Z A B C

Note : All quantities are positiveMaximum %error in Z =

p (%error in A) + q(%error in B)+ r(%error in C)

Problem : 2.10One side of a cube is measured as a = 4.03 1 %.What is its volume ?

Sol. Volume = a3 = (4.03 m 1%)3

= 4.033 m3 (3 1)% = 65.45 m3 3 %

* Problem : 2.11We know that the density of a cube can be measuredby measuring its mass and length of its side. I f themaximum errors in the measurement of mass andlength are 3% and 2% respectively, what is themaximum error in the measurement of the density ofthe cube ?

Sol. M LHere 100 = 3%, 100 = 2%

M L

Density 3M Mass(M)d = = V Length(L)

3Md = L

Maximum error in the measurement of densityd M L 100 = 100 +3 100

d M L

= 3 + (3 2) = 3+ 6 = 9%Maximum error in the measurement of density = 9%

* Problem : 2.12

Error in the measurement of radius of a sphere is 1%.Find the error in the measurement of volume.

Sol.r 100

r

=1%

Volume of sphere 33V = r4

Error in the measurement of volumeV r 100 = 3 100

V r

= 3 1 =3%

* Problem : 2.13An experiment measures quantities a, b, c and then x is

calculated as2

3ab

xc

. I f the percentage error in a, b,c are 1% , 3% and 2% respectively, find themaximum percentage error in x.

Sol. Herea b c 100 = 1%, 100 = 3%, 100 = 2%

a b c

2

3abx = c

Maximum percentage error in x isx a b c

100 = 100 + 2 100 + 3 100x a b c

= 1 + (2 3) + (3 2) = 13%Maximum percentage error in x is 13%

* Problem : 2.14The percentage error in the mass and speed are 2%and 3% respectively. How much will be the maximumerror in kinetic energy calculated using mass and ve-locity ?

Sol.m 100 = 2% , 100 = 3%

m

Kinetic energy. 21E = m2

Maximum error in kinetic energyE m 100 = 100 + 2 100

E m

Maximum error in kinetic energy= 2% + 2 3%= 2% + 6% =8%

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*Problem : 2.15 I n the measurement of a physi cal quanti ty

2

133

A BX

C D

.The percentage errors introduced in the

measurement of the quantities A, B, C and D are 2%,2%, 4% and 5% respectively. Then the minimumamount of percentage of error in the measurement ofX is contributed by which quantity ?

Sol. 2

133

A BX =

C D

A B C D% 2%, % 2%, % 4%, % 5%

A B C D

X A B 1 C 1 D = 2 + + + X A B 3 C 3 D

Percentage error in XX = 100

X

A B 1 C= 2 100 + 100 + A B 3 C

D + 3

D

1 2 2 + 2 + 4 3 5 22.34%3

The minimum amount of error is contributed by C.

*Problem : 2.16

Dimensional formula for a physical quantity X isM1L3T2. The error in measuring the quantities M, Land T respectively are 2% , 3% and 4%. What is themaximum percentage error that occurs in measuringthe quantity X ?

Sol. X = M1 L3T-2dM dL dT 100 = 2%, 100 = 3%, 100 = 4%M L T

Maximum percentage error inM L TX = 100 + 3 100 + 2 100

M L T

2 3 3 (2 4) 2 9 8 Maximum error in X 19%

*Problem : 2.17In Poiseuilles method of determination of viscosity,what is the physical quantity that requires greateraccuracy in measurement ?

Sol. Formula 4Pr

= 8Vl

the inner radius of the capillary tube appears with a 4thpower in the equation. So if there is an error it becomes4 times. Hence inner radius must be measured with highaccuracy.

*Problem : 2.18In an experiment of simple pendulum the error inmeasurement of length of the pendulum L and timeperiod T are 3% and 2% respectively. What is the

maximum percentage error in the value of 2L

T ?

Sol. l T 100 = 3%, 100 = 2%

L T

Maximum percentage error in the value of

2L L T = 100 + 2 100

L TT

= 3 + (2 2)

= 3 + 4Maximum percentage error in the value of 2

LT

= 7%

*Problem : 2.19The measured mass and volume of a body are 2.42 gand 4.7 cm3 respectively with possible errors 0.01g,and 0.1 cc. Find the maximum error in density.

Sol. Density, d = Mass(m)

Volume(V)Here m = 2.42 g, m = 0.01 g

v = 4.7 cm3 V = 0.1 ccMaximum error in density

d m V 100 = 100 + 100d m V

0.01 0.1= 100 + 1002.42 4.7

= 0.413% + 2.127%

Maximum error in density = 2.54%

* Problem : 2.20A rectangular metal slab of mass 33.333g has its length8.0cm, breadth 5.0cm and thickness 1mm. The mass ismeasured with accuracy up to 1mg with a sensitivebalance. The length and breadth are measured with avernier calipers having a least count of 0.01cm. Thethickness is measured with a screw gauge of least count0.01mm. Calculate the percentage accuracy in den-sity caluclulated from above measurements.

Sol. The precentage error itself gives us the percentage ac-curacy.

mass(M) Md=

length( ) breadth(b) height(h) bh

The relative error is given by

d M b hd M b h

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Given l = 8.0 cm and = 0.01cm;b = 5.0 cm and b = 0.01cmh = 1mm and h = 0.01mm andM = 33.333g and M = 1mg = 0.001g

Hence,M 0.001

M 33.333

0.01 b 0.01,8.0 b 5

and

h 0.01h

The percentage error is given byM b h 100

m b h

0.001 0.01 0.01 0.01 10033.333 8.0 5 1

= 0.003+0.125+0.2+1.0 = 1.328.

1.328 1.3The percentage error is 1.3%

* Problem : 2.21The error in the measurement of the length of asimple pendulum is 0.1% and the error in the timeperiod is 2%. What is the possible percentage of errorin the physical quantity having the dimensional formulaLT2.

Sol. Percentage error in LT2 isL Ta 2. 100

L T

2 20.1 100 4.1%100 100

*Problem : 2.22

The heat generated in a circuit is dependent on theresistance, current and time of flow of electrical current.If the errors measured in the above are 1%, 2% and 1%respectively, what is the maximum error in heat ?

Sol. Heat 2i RTH =

jMaximum error in measuring the heat

H i R T100 = 2 100 + 100+ 100

H i R T

Here J is a constant.Giventhat

R i T100 = 1%, 100 = 2% 100 = 1%R i T

H 100 = 2 2 + 1 +1H

= 4 + 1 + 1

Maximum error in heat = 6%

2.12. SIGNIFICANT FIGURESThe digits of a number r epr esenting a

measurement that are definitely known, plus onemore digit added at the end which is estimatedare called significant digits or significant figures.Example : Suppose least count of a scale used tomeasure length of a rod is 0.1 cm.

Suppose one end of the rod is coinciding withzero of the scale and the other end is in between twomarks reading 14.5 cm and 14.6 cm.So length of the rod is more than 14.5cm but lessthan 14.6cm

One observer may think that the correctmeasurement is perhaps 14.52 cmHere 1,4 and 5 are reliable and last digit 2 is guess-work and thus not reliable.

So in 14.52 cm there are 4 significant figures.The last significant figure is called the 'first

doubtful figure' or the 'least significant figure'. It isalso called 'uncertain digit' . It is 2 in 14.52 cm.Rules to count Significant Figures :

1) All non-zero digits in a given number aresignificant without any regard to the location of thedecimal point if any.

6,482 has four significant digits648.2 or 64.82 or 6.482 all have the same

number (four) of significant digits.2) All zeros occurring between two non-zero

digits are significant without any regard to thelocation of decimal point if any.

206008 has six significant figures.206.008 or 2.06008 has also got six significant

digits.3) If the number is less than one, all the zeros to

the right of the decimal point but to the left of thefirst non-zero digit are not significant.

In the number 0.000608 the three zeros to theright of decimal point and upto 6 are not significant.The zero that is conventionally placed before decimalpoint is obviously not significant. The given numberhas three significant figures. The zero between 6 and8 is significant from rule 2.

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4(a) All zeros to the right of a decimal point aresignificant if the number is more than one. 60.00contains four significant digits.

4(b) All zeros to the right of the last non-zerodigit after the decimal point are significant.

0.007800 has four significant figures. The zerobetween decimal point and 5 is not significant fromrule-3. But the zeroes after eight are significant.Otherwise it would have been reported as 0.0078,only with two significant digits.

5(a) All zeros to the right of the last non-zerodigit in a number having no decimal point are notsignificant.

4060 contains three significant figures as thezero to the right of six is not significant.

40,6000 has also got only three significantfigures.

5(b) However, there is an exception to rule 5(a).All zeros to the right of the last non-zero digit in anumber having no decimal point will be significantif they come from an actual measurement.

Suppose a distance measured in metres isreported as 4.500 m having four significant figures.Expressed in cms it will be 450 cm. It has only twosignificant figures. To have four significant figures,we should express it as 450.0 cm. Similarly, if wewant to express the value in km, we have to writethe result as 0.004500 km and not as 0.0045 km.When expressed in mm, the value becomes 4500mm. It has only two significant figures from rule5(a). Now arises a problem. We cannot write 4500mm in any way as we have done with cm and km tohave four significant figures.

To avoid such ambiguity, we write the numberin scientific notation as powers of 10. Now 4500mm is written as 4.500103 mm and by rule 4(a) thenumber has 4 significant figures. Similarly the resultis expressed as 4.500 102cm and 4.500 103 km.Note : Very large and very small quantities of physicalmeasurement are usually expressed in scientificnotation to denote accuracy of measurements.

* Problem : 2.23Write down the number of significant figures in thefollowing.(i) 0.007 (ii) 2.64 1024 (iii) 0.2370(iv) 6.320 (v) 6.032 (vi) 0.0006032

Sol. (i) 0.007 has one significant figure.(ii) 2. 64 1024 has three significant figures.(iii) 0.2370 has four significant figures.(iv) 6.320 has four significant figures,(v) 6.032 has four significant figures.(vi) 0.0006032 has four significant figures.

2.13. ROUNDING OFF THE NUMBERThe process of omitting the non-significant digits

and retaining only the desired number of significantdigits, incorportating the required modifications tothe last significant digit is called 'rounding off thenumber'.Rules for Rounding off Numbers :i) The preceding digit is raised by 1 if theimmediate insignificant digit to be dropped is morethan 5.Ex : 5728 is to be rounded off to three significantfigures. Here 8 is the nonsignificant digit to bedropped and is more than 5. Hence 5728 = 5730.ii) The preceding digit is to be left unchanged ifthe immediate insignificant digit to be dropped is lessthan 5.Ex : 5728 is to be rounded off to two siginificantfigures. Here the immediate or first nonsignificantdigit to be dropped is 2(along with 8 also) and is lessthan 5. Hence 5728 = 5700. In the above two example1 and 2, zeros are to be placed in the places of digitsdropped. Otherwise the values will change drasticaly.iii) If the immediate insignificant digit to be droppedis 5 then there will be two different cases (a) If thepreceding digit is even, it is to be unchanged and 5is dropped.Ex : 6.7258 is to be rounded off to two decimal places(or three significant figures). The digit to be droppedhere is 5(along with 8) and the preceding digit 2 iseven and hence to be retained as two only.

6.7258 = 6.72

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(b) If the preceding digit is odd, it is to be raisedby 1.

6.7158 is to be rounded off to two decimalplaces (or three significant figures). As the precedingdigit 1 is odd, it is to be raised by 1 as 2.

6.7158 = 6.72In the above two example 3(a) and 3(b) zeros

should not be placed in places of decimal digitsdropped. Otherwise the number of significant digitswill drastically change.

* Problem : 2.24

Round off to 3 significant figures:(i) 20.96 (ii) 0.0003125

Sol. (i) 20.96 has four significant figures. The fourth signifi-cant figure is more than 5 and hence on rounding off tothree significant figures, the given measurement willbecome 20.9 + 0.1 i.e., 21.0.

(ii) 0.0003125 has four significant figures. The fourthsignificant figure is 5 and hence on rounding off tothree significant figures, the given measurement willbecome 0.000312 or 3.12104. This is because 2 be-fore 5 is an even number.

2.14. RULES FOR ARITHMETIC OPER- ATIONS WITH SIGNIFICANT FIGURESBecause of the errors involved in measurements,

the measured values of physical quantities havelimited number of significant figures. To obtain manyphysical quantities we have to make arithmeticoperations (addition, subtraction, multiplication,division etc.) of the measured quantities. To obtainlinear momentum, we have to multiply mass withvelocity. To obtain velocity we have to dividedistance travelled with time taken. Now, any reusltof such operation should also indicate the errorsinvolved in the original measured values. The resultcan never be more accurate than the originalindividual measured values.

So the final result of ar ithmetic operationsshould never have more significant figures thanthe least number of signigicant figures in theoriginal components.

This results in the following rules.

Rule 1: In addition or subtraction the final resultshould retain only that many decimal places asare there in the number with the least decimalplaces.Note : In addition and subtraction, the rule is in termsof decimal places (and not in terms of significantfigures)Rule 2: In multiplication or division, the finalresult should retain only that many significantfigures as are there in the or iginal number withthe least number of significant figures.2.14.1. EXPLANATION OF THE RULESi) Addition and subtraction

In addition or subtraction of numbers, beforeperforming the operation we must identify thenumber with the least number of digits afterdecimal point (say n ) and round off other numbersto one extra digit in the decimal part (say n + 1). Thefinal result is again rounded off , such that it containssame number of decimal places as that of the numberwith least number of decimal places (say n) amongall other numbers. * Problem : 2.25

Find the value of 2.2 + 4.08 + 3.125 + 6.3755.Sol. Out of all the four numbers 2.2 has got the least number

of decimal places one.Hence we should retain only two decimal places in theremaining numbers. Hence 4.08 remains as it is 3.125 isto be rounded off as 3 : 12 (as 2 before 5 is even)6.3755 is to be rounded off as 6.38 (as 7 before 5 isodd). Now adding 2.2 + 4.08 + 3.12 + 6.38 = 15.78.Finally we should have only one decimal place andhence 15.78 is to be rounded off as 15.8.

* Problem : 2.26A stick has a length of 12.132 cm and another stickhas a length of 12.4 cm.(a) I f the two sticks are placed end to end, what is

their total length ?(b) I f the two sticks are placed side by side, what is

the difference in their lengths ?Sol . (a) Let lengths of the sticks are named as

1 2l = 12.132cm, l = 12.4 cmHere 2 l has one decimal place and 1 l has to be roundedoff to have only two decimal placesl1 + l2 12.13 + 12.4 = 24.53.

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This is to be rounded off to have one decimal placeonly.The total length is 24.5cm(b) l1 = 12.132, l2 = 12.4.

l1 - l2 = 12.4 - 12.132Here 12.4 has only decimal place and hence 12.132should have only two decimal places.

l1 - l2 = 12.4 - 12.13 = 0.27This should be rounded off to have only one decimalplace.

l1 - l2 = 0.3Hence difference of their lenghts is 0.3cm

* Problem : 2.27Find the value of 44.8 21.235.

Sol. Only one decimal place is there in 44.8. Hence the othernumber is to be rounded off to have two decimal places.21.235 = 21.24 (as 3 before 5 is odd).Now 44.8 21.24 = 23.56.Finally this is to be rounded off to one decimal place

44.8 21.235 = 23.6.

* Problem : 2.28

Solve with due regard to significant figures.(i) 46.7 - 10.4 = (ii) (3.0 108) + (4.5 106) =

Sol. (i) 46.7 - 10.04Here 46.7has one decimal place, and 10.04 hastwo deci-mal places.

46.7 - 10.04 = 36.66.The result should have only one decimal place.The result is 36.7.

ii) 8 63.0 10 + 4.5 10 6 6= 0.03 10 + 4.5 10 = 4.53 10-6

Here 4.5 10-6 has only one decimal place and 0.03 10-6can have two decimal places.This result should be rounded off to one decimal place. (3.0 108) + (4.5 106) = 4.5 10-6

ii) Multiplication and divisionIn multiplication or division of numbers,

before performing the operation we must identify thenumber with the least number of significant figures(say n) and round off other numbers to one extradigit (say n + 1). The final result is again roundedoff, such that it contains same number of significant

figures (say n)as that of the number with leastnumber of significant figures among all othernumbers.Note : The final result of arithmetic operations shouldnever have more significant figures than the leastnumber of signigicant figures in the originalcomponents.iii) In square root value of the given number,there can be same number of significant figuresthat are present in the given number.

Example : 58.97 7.679iv) In case of Transcendental Functions:Transcendental functions have the same number ofsignificant figures as their arguments.

Non-algebraic functions like sine, sin1exponential and logarithmic functions are calledtranscendental functions.

Arguments are quantities on which someoperation is performed e.g. in case of sin 30, 30 isargument; in case of log x, x is argument, etc.Example :

sin (60) = 0.89sin (60.0) = 0.886sin (60.00) = 0.8660sin (60.000) = 0.86603

* Problem : 2.29

Find the product of 1.2, 2.54 and 3.257 with due regardto significant figures.

Sol. Out of the three numbers 1.2 has got the least number ofsignificant figures two. We should round off the othernumbers to 2 + 1 = 3 significant figures and carry themultiplication. 2.54 has three significant figures andhence needs no rounding off. 3.257 is to be rounded offto 3.26.1.2 2.54 3.26 = 9.93648. But the result should belimited to the least number of significant digits that istwo digits only.The final answer should be written as 9.9 after rounding off.

* Problem : 2.30Find the value of /53.2 with due regard to significantfigures.

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Sol. Out of the two numbers 53.2 has three significant dig-its. So p should be written with 3 + 1 = 4 significantentfigures. = 3.1415 = 3.142 (as 1 is odd it is raised by one).

Now 3.142 0.059060153.2

This is to be rounded off to three significant figures0.059060 = 0.0591

* Problem : 2.312.31 Find out the results of the following operations.(i) 117.3 0.0024 (ii) 9.27 41(iii) 42 0.041 (iv) 124.2 + 52.487(v) 124.2 52.487 (vi) 58.97(vii) (17.5)2

Sol. (i) 0.0024 has 2 significant figures. Hence 117.3 isrounded off to have 2 + 1 = 3 significant figures. Itbecomes 117 only.Now 117 0.0024 = 0.2808This is to be rounded off to have two significant figuresonly. The result is 0.28.(ii) 41 has only 2 significant digits. Hence 9.27 canhave 2 +1 = 3 significant digits. It has 3 significantdigits only. No need for rounding off.

9.27 0.226097541

This is to be rounded off to two

significant digits. The result is 0.23.

(iii) Both number have two significant digits.42 0.041 = 1.722. This is to be rounded off to 2 sig-nificant digits. The result is 1.7.(iv) As this is a sum, we have to consider decimalplaces. 124.2 has only one decimal place. Hence 52.487is to be rounded off to 1 + 1 = 2 decimal places.It becomes as 52.487 = 52.49Now 124.2 + 52.49 = 176.69. This is to be rounded offto one decimal place. The result is as 176.69 = 176.7.(v) Here again, we have124.2 52.487 = 124.52.49 = 71.71. This is to berounded off to one decimal placeThe result is 71.7.(vi) 58.97 = 7.679. This has got the same numberof significant digits (4) as 58.97. But for square rootsits is customary to have the number of significant fig-ures one less than the number that is, 7.68.(vii) (17.5)2 = 306.25. The original number 17.5 hasonly 3 significant digits. And hence the result will be306 (Here we have to consider the significant figures).

* Problem : 2.32The mass of 1.2 cm3of a certain substance is 5.74g.calculate its density with due regard to significant fig-ures.

Sol. Mass, m = 5.74 g, volume, v = 1.2 cm3

density,3

m 5.74 gd = = v 1.2 cm

5.74 has significant digits and 1.2 has two significantdigits.

no need for rounding off.5.74 d = = 4.783333.1.2

This is to be rounded off to two significant digits. The density of the given substance is 4.783 = 4.8 g cm3

* Problem : 2.33I f a circular piece of tin has a measured radius of 2.6cm, then what is its circumference ?

Sol. r = 2.6 cmCircumference of circular disc = 2 r = 2 3.1428 2.6

Here 2.6 has onley 2 significant digits. Hence in theabovemultiplication value shoud be written with2 +1 = 3 significant figures.

= 3.1428 = 3.14Circumference = 2 3.14 2.6 = 16.328This is to be rounded off to 2 significant digits. Cir-cumference is 16cm.

* Problem : 2.34The diameter of a sphere is 4.24 m. Calculate its surfacearea with due regard to significant figures.

Sol. Diameter d = 4. 24 m

Radius d 4.24r = = = 2.122 2

surface area of sphere = 4 r2

= 4 3.1428 2.12 2.12In the above multiplication 2.12 has 3 significantfigures.Hence 3.1428 is rounded off to have 3 + 1 = 4 significantfigures. It becomes 3.143.Surface area = 4 3.143 2.12 2.12 = 56.5056.50 this to be rounded off to have 3 significant figures. Area is 56.5m2

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* Problem : 2.35 Each side of a cube is measured to be 7.203m. What is(i) the total surface area and (ii) the volume of thecube to appropriate significant figures ?

Sol. Length of a side = 7.203 m(i) Total surface area = 6a2

= 6 7.203 7.203 = 311.29

this to be rounded off to 4 significant figures as 7.203as 4 significant figures.

Total surface area is 2311.3 m(ii) Volume of the cube = a3

= 7.203 7.203 7.203

= 373.71

This is to be rounded off to 4 significant figures as7.203 as 4 significant figures.

The volume = 373.7 m3.

* Problem :2.36The length of a rod is 2.5 cm and diameter is 2.5 mm.Find the volume of the rod with due consideration tosignificant figures.

Sol. l = 2.5 cm,diameter = 2.5 mm = 2.5 101 cm = 0.25 cmRadius, r = 0.125 cm

Volume of the rod = 2r l= 3.1428 0.125 0.125 2.5

Here 2.5 has two significant figures. Round off the re-maining numbers to 2 + 1 = 3 significant figures.

Volume of the rod = 3.14 0.125 0.125 2.5

= 0.1226. This has to be rounded off to two significant figures.Volume of the rod0.12 cm3

Short Answer Questions1) How do the random errors differ from systematic

errors?

2) What is rounding off a number and what arethe rules to be followed in it?

3) What are the rules for arithmetic operations withsignificant figures?

4) What are random errors? Distinguish betweenradom errors and systematic errors

5) What is meant by significant figures? How arethese counted?

6) State the different types of errors present in ameasurement.

7) Define the terms (i) mean absolute error(ii) relative error and (iii) percentage error. Howare they calculated?

8) Round off to 3 significant figures giving therules followed (i) 25.87 (ii) 0.05134.

Very Short Answer Questions1) What is an error? What are constant errors?2) Mention different kinds of errors.3) Explain briefly what are systematic errors.4) What are the causes for environmental errors?5) What type of errors are met with in a

calorimetric experiment?6) What are gross errors? Give an example.7) What are random errors? Give an example8) Distiguish between accuracy and precision.9) What is mean absolute error? State its formula.10) What is relative error? State its formula.11) What is absolute error? State its formula12) Are all the significant figures reliable? Give an

example. 13) What is rounding off a number?14) What are significant figures ? Give an example.

Asseses Yourself1. By taking precautions, can we minimize the

random errors ?Ans. No2. Why is that for both very small as well as very

large (in addition to ordinary) quantities, thephysical measurements are usually in scientificnotation, with powers of ten ?

Ans.To denote precesion and accuracy of themeasurement.