Question bank physics

8/11/2019 Question bank physics

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Physics Std. 11th, Maharashtra Board, English Medium


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Std. XI Sci.

Perfect Physics

Prof. ChandrabhushanChoudhary

(M.Sc., Magadh University)

Prof. Umakant N. Kondapure(M.Sc., B.Ed., Solapur)

Salient Features:

Exhaustive coverage of syllabus in Question Answer Format.

Covers answers to all Textual Questions.

Textual Questions are represented by * mark.

Covers relevant NCERT questions.

Simple and Lucid language.

Neat, labelled and authentic diagrams.

Solved & Practice Numericals duly classified

Multiple Choice Questions for effective preparation.

Target PUBLICATIONS PVT. LTD.Mumbai, Maharashtra

Tel: 022 6551 6551

Website:www.targetpublications.in

www.targetpublications.org

email :[email protected]

Written according to the New Text book (2012-2013) published by the Maharashtra State

Board of Secondary and Higher Secondary Education, Pune.


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Std. XI Sci.Perfect Physics

Target Publications Pvt Ltd.

Second Edition : July 2012

Price : 170/-

Printed at:Uday OffsetLower Parel,Mumbai

Published by

Target PUBLICATIONS PVT. LTD.Shiv Mandir Sabhagriha,Mhatre Nagar, Near LIC Colony,Mithagar Road,Mulund (E),Mumbai - 400 081Off.Tel: 022 6551 6551email: [email protected]


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PREFACE

In the case of good books, the point is not how many of them you can get through, but rather how many can

get through to you.

Physics is a science that deals with matter and energy and their actions upon each other in the fields ofheat, mechanics, aeronautics, sound, light and electricity. It covers a wide range of phenomena, from the

smallest subatomic particles to the largest galaxies. It is closely related to other natural sciences and in a

sense encompasses them. In order to study such a vast science and to master it, one needs to understand and

grasp each and every concept thoroughly.

Hence to ease these tasks we bring to you Std. XI Sci. : PERFECT PHYSICS a complete and

thorough guide critically analysed and extensively drafted to boost the students confidence. Topic wise

classified question and answer format of this book helps the student to understand each and every concept

thoroughly. The book provides answers to all textual questions marked with *. Important definitions,

statements and laws are specified with Italic representation. Formulae are provided in every chapter which

are main tools to tackle difficult problems. Solved problems are provided to understand the application of

different concepts and formulae. Practice problems and multiple choice questions help the students, to test

their range of preparation and the amount of knowledge of each topic.

And lastly we would like to thank the publisher for helping us to take this exclusive guide to all

students. There is always room for improvement and hence we welcome all suggestions and regret any

errors that may have occurred in the making of this book.

A book affects eternity; one can never tell where its influence stops.

Best of luck to all the aspirants!

Yours faithfully

Publisher


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Contents

No. Topic Name Page No.

1 Measurements 1

2 Scalars and Vectors 22

3 Projectile Motion 43

4 Force 68

5 Friction in Solids and Liquids 102

6 Sound Waves 130

7 Thermal Expansion 149

8 Refraction of Light 175

9 Ray Optics 200

10 Electrostatics 228

11 Current Electricity 261

12 Magnetic Effect of Electric Current 283

13 Magnetism 305

14 Electromagnetic Waves 326

Note: All the Textual questions are represented by * mark

All the Intext questions are represented by # mark


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TARGET Publications Std. XI Sci.: Perfect Physics

1Measurement

01 MEASUREMENT1.0 Introduction

Q.1. What is physics?

Ans: i. Physics is the branch of science whichdeals with the study of nature andnatural phenomena. The word Physicsis derived from the greek word fusismeaning nature. Fusis was firstintroduced by the ancient scientistAristotle.

ii. In physics, various physicalphenomenon are explained in terms offew concepts and laws. The effort is tovisualize the physical world asmanifestation of some universal laws in

different domains.Q.2. Explain the domains which explain scope of

physics.

Ans: Basically, there are two domains in the scope ofphysics: (i) macroscopic domain and (ii)microscopic domain.i. Macroscopic domain:

The macroscopic domain includesphenomena at the laboratory, terrestrialand astronomical scales.The macroscopic phenomena dealsmainly with the branch of classical

mechanics which includes subjects likemechanics, electrodynamics, optics andthermodynamics.

ii. Microscopic domain:The microscopic domain includes atomic,molecular and nuclear phenomena.The microscopic domain of physicsdeals with the constitution and structureof matter at the minute scales of atomsand other elementary particles.

So we can see that the scope of physics is trulyvast. It covers a tremendous range of physical

quantities.Q.3. What are physical quantities?

Ans: Those quantities, which can be measuredi.e. subjected equally to all three elements ofscientific study, namely detailed analysis,precise measurement and mathematicaltreatment, are called physical quantities.Example: Mass, length, time, volume,pressure, force, etc.

1.1 Need for Measurement

*Q.4. What is the need for measurement of aphysical quantity?

Ans: i. Experiments and measurements form thebasis of physics.

ii. To study various phenomenon inphysics scientist have performed someexperiments.

iii. These experiments require measurementof physical quantities such as mass,length, time, volume, etc.

iv. Based on the observations of physicalphenomena, scientists have developedvarious laws and theories.

v. For the experiment verification ofvarious theories, each physical quantityshould be measured precisely.

vi. Therefore, accurate measurement ofphysical quantities with appropriateinstruments is necessary.

vii. For example, if I say, I am waiting for along time. In the given statement, thephysical quantity time is not definedprecisely. A numerical value for time,

which is measured on a watch isnecessary. Therefore, measurement isnecessary in physics.

1.2 Unit for Measurement

Q.5. How can magnitude of physical quantity be

expressed?

Ans: i. The magnitude of a physical quantity xis expressed in the following way:

Magnitude of physical quantity =Numerical value of physical quantity Size of its unit

x= nu

Where, n = number of times the unit istaken.

u = size of unit of physical quantity.

ii. For example, If the length of a clothpiece is 5 metre, it means that the clothpiece is 5 times as long as the standardunit of length (i.e. metre).


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TARGET Publicationstd. XI c .: Per ect Phys cs

Measurement2

*Q.6. What is meant by unit of a physical quantity?

Ans: i. The reference standard used for themeasurement of a physical quantity is

called the unit of that physical quantity.ii. Example:

Physical

Quantity

Standard (unit)

Length metre, centimeter, inch, feet,etc.

Mass kilogram, gram, pound etc.

Q.7. State the essential characteristics of a good

unit.

Ans: A good unit should have the following

characteristics:

i. It should be well defined.

ii. It should be easily available and

reproducible at all places.iii. It should not be perishable.

iv. It should be invariable.

v. It should be universally accepted.

vi. It should be comparable to the size ofthe measured physical quantity.

vii. It must be easy to form multiples or submultiples of the unit.

Q.8. Why do we choose different scale for

measurement of same physical quantity?

Ans: i. Choice of unit depends upon itssuitability for measuring the magnitudeof a physical quantity underconsideration. Hence we choosedifferent scale for same physicalquantity.

ii. For example: To measure the diameterof a rod we use a centimeter, formeasuring the height of a building weuse a metre, for measuring the distancebetween two towns we use a kilometreand for measuring the distance of stars

from earth we use a light year (thedistance covered by light in one year).

iii. One convenient way of measuring big aswell as small quantities is to usemultiples and submultiples of a unit.

Note:

Following table is used for expressing larger as wellas smaller unit of same physical quantity in a choiceof unit.

Prefix Symbol Power

of 10

Prefix Symbol Power

of 10

Exa E 1018 deci d 101Peta P 1015 centi c 102Tera T 1012 milli m 103

Giga G 109 micro 106Mega M 106 nano n 109

Kilo k 10

3

angstrom 1010

Hecto h 102 pico p 1012Deca da 101 femto f 1015

atto a 1018

1.3 System of Units

Q.9. What is system of unit?

Ans: i. Units are classified into two groupsfundamental units and derived units. Aset of fundamental and derived units iscalled a system of units.

ii. In 1832, Gauss suggested to select anythree physical quantities as fundamental

quantities.*Q.10. State different types of system of units.

OR

Explain briefly the various system of units.

Ans: The whole set of units, i.e., all the basic andderived units taken together, forms a system ofunits. System of units are classified mainly intofour types:i. C.G.S. system:

It stands for Centimeter- Gram-Secondsystem. In this system fundamentalquantities i.e. length, mass and time are

measured in centimeter, gram andsecond, respectively. It is French metricsystem of unit.

ii. M.K.S. system:It stands for Metre-Kilogram-Secondsystem. In this system, fundamentalquantities i.e. length, mass and time aremeasured in metre, kilogram andsecond, respectively. It is a large Frenchmetric system of unit.

iii. F.P.S. system:It stands for Foot-Pound-Second system.

In this system, length, mass and time aremeasured in foot, pound and secondrespectively. It is a British imperialsystem.

iv. S.I. system:It stands for Standard Internationalsystem. This system has replaced allother systems mentioned above. It hasbeen internationally accepted & is beingused all over world.


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3Measurement


#Q.11.Can you call a physical quantity large or

small without specifying a standard for

comparison?

Ans: No, we cannot call a physical quantity large orsmall without specifying a standard forcomparison.

1.4 S.I. units

*Q.12.What is S.I. system of units? Explain its

need.

OR

Write a short note on S.I. units.

Ans: S.I. system of units:

i. Use of different systems of units becamevery inconvenient for exchangingscientific information between differentparts of the world.

ii. To overcome this difficulty, it became

necessary to develop a common systemof units.

iii. In October 1960, at the EleventhInternational General Conference ofweights and measures in Paris, acommon system of units was accepted.This system of units is called SystemeInternationale dUnits.It is abbreviatedas S.I. units.

iv. S.I. units consists of seven fundamentalunits, two supplementary units and large

number of derived units.v. Nowadays, S.I. system has replaced allthe other system of units and is greatlyused to exchange scientific data betweendifferent parts of the world.

1.5 Fundamental and derived units

*Q.13.What are fundamental quantities? State

two examples of fundamental quantities.

Write their S.I. and C.G.S. units.

Ans: Fundamental quantities:

The physical quantities which do not dependon any other physical quantities for their

measurements are called fundamental

quantities.

Examples:mass, length etc.

Fundamental

quantities

S.I. unit C.G.S. unit

Mass kilogram (kg) gram (g)

Length metre (m) centimeter (cm)

*Q.14.What are fundamental units? State the S.I.

units of seven basic fundamental quantities.

Ans: Fundamental units:

i. The units used to measure fundamentalquantities are called fundamental units.

ii. There are seven fundamental quantities

accepted in S.I. system.Fundamental quantities with their correspondingunits are given in following table.

Fundamental quantity S. I. unit Symbol

Length metre m

Mass kilogram kg

Time second s

Electric current ampere A

Temperature kelvin K

Luminous intensity candela cd

Amount of substance mole mol

Supplementary Units

Plane angle radian rad

Solid angle steradian sr

*Q.15.What are derived quantities and derived

units? State with two examples and their

corresponding S.I. and C.G.S. units.

Ans: i. Derived quantities:Physical quantities other than

fundamental quantities which depend on

one or more fundamental quantities for

their measurements are called derived

quantities.

Examples: speed, acceleration, force,etc.

ii. Derived unit:

The units of derived quantities which

depend on fundamental units for their

measurements are called derived units.

Examples:

Derived quantity S.I. unit C.G.S. unit

Speed m/s cm/s

Force N dyne

Density kg/m3 g/cm3

Acceleration m/s2 cm/s2


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Measurement4

Q.16. What are the rules for writing the S.I. units

of physical quantities?

OR

What conventions should be followed while

writing S.I. units of physical quantities?

Ans: Following convention should be followedwhile writing S.I. units of physical quantities:

i. For a unit named after a person, thesymbol has a capital initial letter.For example, N for newton, J for joule,W for watt, Hz for hertz. Symbols of theother units are not written with capitalinitial letter.

ii. Full name of a unit, even when it isnamed after a person, is written withsmall letter. E.g. unit of force is writtenas newton and not as Newton; unit ofpower is written as watt and not as Watt.

iii. The symbol for other units are written in

small letters. For example, symbol formetre is m, for second is s, forkilogram is kg and so on.

iv. Symbols of units are not to be expressedin plural form. For example, 10 metresis written as 10 m and not as 10 ms. Thisis because, 10 ms, means 10 metre second.

v. Full stop and any other punctuationmark should not be written after thesymbol. e.g. kg and not kg., or N andnot N.

Q.17. Explain the advantages of S.I. system of units.

Ans: The advantages of S.I. system of unit are asfollows:

i. It is comprehensive, i.e. its small set ofseven fundamental units cover the needsof all other physical quantities.

ii. It is coherent, i.e., its units are mutuallyrelated by rules of multiplication anddivision with no numerical factor otherthan 1.For example, 1 ohm = 1 volt / 1 ampere.

iii. S.I. system being a decimal or metricsystem, writing of very large or verysmall numerical value is simplified byusing prefixes to denote decimalmultiples and submultiples of the S.I.units.For example, 1 m (micrometer) =106 m.

iv. The joule is the unit of all forms ofenergy. Hence, the joule provides a linkbetween mechanical and electrical units.

Q.18. Classify the following quantities into

fundamental and derived quantities:

Length, Velocity, Area, Electric current,

Acceleration, Time, Force, Momentum,

Energy, Temperature, Mass, Pressure,

Magnetic induction, Density.

Ans:

Fundamentalquantities

Derivedquantities

TimeTemperatureMassLengthElectric current

Velocity, Area,Acceleration,Force,Momentum,Energy, Pressure,Magneticinduction, Density

Q.19. Classify the following units into

fundamental, supplementary and derived

units:

newton, metre, candela, radian, hertz,square metre, tesla, ampere, kelvin, volt,

mol, coulomb, farad, steradian.

Ans:

Fundamental

units

Supplementary

unitsDerived

units

metre,candela,ampere,kelvin, mol

radiansteradian

newton,hertz,square metre,tesla,volt,coulomb,farad

Q.20. State the special names of the following

derived quantities in S.I. units:

i. Force ii. Energy

iii. Power iv. Pressure

v. Frequency vi. Electric charge

vii. Electric potential

viii. Electric resistance

ix. Magnetic induction

Mention their symbols.

Ans:

Sr

No.

Physical quantity S.I. unit Special name

i. Force kg. m/s2 newton (N)ii. Energy, work kg . m2/s2 joule (J)iii. Power kg . m2/s3 watt (W)iv. Pressure kg / s2m pascal (Pa)v. Frequency s1 hertz (Hz)vi. Electric charge A . s coulomb (C)vii. Electric potential kg . m2/s3A volt (V)viii. Electric resistance kg . m2/s3A2 ohm ()ix. Magnetic induction kg / s2A tesla (T)


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5Measurement


*Q.21.Explain the methods to measure length.

Ans: Methods for measurement of length

There are broadly two methods formeasurement of length.

i. Direct methods:

a. A metre scale is used for

measurement of length from 103

m to 102 m using differentinstruments.

b. A vernier callipers is used for themeasurement of length upto theaccuracy of 104m.

c. Further accuracy can be attainedby using spherometer and screwgauge. The accuracy by usingthese instruments is upto 105 mfor smaller lengths.

ii. Indirect methods:It includes measurement of largedistances, such as distance between twoplanets, diameter of sun, distance ofstars from the earth etc.

*Q.22.Explain the method to measure mass.

Ans: Method for measurement of mass:

i. Mass is fundamental property of matter.It does not depend on the temperature,pressure or location of the object inspace. The S.I. unit of mass is kilogram

(kg).ii. Mass of commonly available object can

be determined by a common balance.

iii. For atomic levels, instead of kilogramatomic unified mass unit is used.

1 unified atomic mass unit = 1 u =0.8333 101of the mass of an atom ofcarbon-12 in kg

Isotope, 126 C including mass of

electrons = 1.66 1027 kg.

iv. For measurement of atomic orsubatomic particles, mass spectrographyis used. This method uses the propertythat mass of the charged particle isproportional to radius of the trajectorywhen particle is moving in uniformelectric and magnetic field.

v. Large masses in the universe likeplanets, stars etc. can be measured byusing Newtons law of gravitation.

*Q.23.Explain the method for measurement of time.

Ans: Method for measurement of time:

i. To measure any time interval, a clock isneeded. Atomic standard of time is nowused for the measurement of time.

ii. In atomic standard of time periodicvibrations of cesium atom is used.

iii. One second is time required for9, 192, 631, 770 vibrations of cesiumatomic clock. This corresponds totransition between two hyperfine energystates of cesium-133 atom.

iv. The cesium atomic clocks are veryaccurate.

v. The national standard of time intervalsecond as well as the frequency ismaintained through four cesium atomicclocks.

Q.24. What is parallax angle (Parallactic angle)?

Ans: Angle between the two directions along whicha star or planet is viewed at the two points ofobservation is called parallax angle(Parallactic angle).

It is given by =b

DWhere, b = separation

between two points of observations.D = Distance of source from any point ofobservation.

#Q.25.How to determine the distance of different

stars from the Earth?

Ans: i. Parallax method is used to determinedistance of different stars from the earth.ii. To measure the distance D of a far

distant star or planet S, select twodifferent observatories (E1and E2).

iii. The star or planet should be able toobserve from E1 and E2 observatoriessimultaneously i.e. at the same time.

iv. E1 and E2 are separated by distance bas shown in figure. E1E2= b S

Measurement of length between two stars

D D

E1 E2bEarth


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Measurement6

v. The angle between the two directionsalong which the star or planet is viewed,can be measured. It is and is calledparallax angle or parallactic angle.

E1SE2= vi. The star or planet is far away from the

(earth) observers, hence

b


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7Measurement


Electricpotential V =

W

q

[L2M1T3A1]

Electricpermittivity 0=

2

2

q

4 r F

[L3M1T4A2]

Electriccapacity C =

q

V

[L2M1T4A2]

Magnetic flux = B.A [L2M1T2A1]Pole strength

m =F

B

[L1M0T0A1]

Magneticpermeability =

B

H

[L1M1T2A2]

Note:

Following table helps to write S.I. unit and

dimensions of various derived quantities:

Derived

quantity

Formula S.I.

unit

Dimensions

Area A = L2 m2 [L2M0T0]Volume V = L3 m3 [L3M0T0]Density = M/V kg/m3 [L3M1T0]Velocity orspeed

v = s/t m/s [L1M0T1]

Acceleration a = v/t m/s2 [L1M0T2]Momentum P = mv kg

m/s[L1M1T1]

Force F = ma N [L1M1T2]Impulse J = F.t Ns [L1M1T1]Work W = F.s J [L2M1T2]

Kinetic energy K.E.= 12

mv2 J [L2M1T2]

PotentialEnergy

P.E.= mgh J [L2M1T2]

PowerP =

W

t

J/s orW

[L2M1T3]

PressureP =

F

A

N/m2 [L1M1T2]

Note:

Students can write K for temperature, I for current,C for luminous intensity and mol for mole.

Q.27. State principle of homogeneity. Use this

principle and find conversion factor

between the units of the same physical

quantity in two different system of units.

Ans: Principle of homogeneity:

i. The dimensions of all the terms on thetwo sides of a physical equation must besame. This is called the principle ofhomogeneity of dimensions.

ii. This principle is based on the fact thattwo physical quantities can be addedtogether or subtracted from one anotheronly if they have same dimensions.

iii. Consider dimensions of a physicalquantity in two system of units:

a b c a b c1 1 1 2 2 2L M T and L M T .

iv. If n is the conversion factor in twosystems, then applying principle ofhomogeneity, we have,

a b c a b c1 1 1 2 2 2L M T n L M T =

n =a b c

1 1 1

2 2 2

L M T

L M T

*Q.28.State the uses of dimensional analysis.

Ans: Uses of dimensional analysis:i. To check the correctness of a physical

equation.

ii. To derive the relationship between differentphysical quantities.iii. To find the conversion factor between the

units of the same physical quantity in twodifferent systems of units.

*Q.29. Explain the use of dimensional analysis to

check the correctness of a physical equation.

Ans: Correctness of a physical equation by

dimensional analysis:

A physical equation is correct, only if thedimensions of all the terms on both sides ofthat equations are the same.

For example, consider the equation of motion.v = u + at .(1)Writing the dimensional equation of everyterm, we getv = [L1M0T1]u = [L1M0T1]a = [L1M0T2]t = [L0M0T1]at = [L1M0T2] [L0M0T1] = [L1M0T1]As dimensions of both side of equation issame, physical equation is dimensionallycorrect.

*Q.30.Explain the use of dimensional analysis tofind the conversion factor between units of

same physical quantity into different

system of units.Ans: Conversion factor between units of same

physical quantity:i. Suppose we have to find the conversion

factor between the units of force i.e.newton in S.I. system to dyne in C.G.S.system.


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Measurement8

ii. Let n be the conversion factor betweenthe units of force.

iii. Dimension of force in S.I. system is

[L 11 M11 T1

2] and in CGS system is

[L 12 M12 T2

2]

iv. Suppose, 1 newton = n dyne ....(1)

1[L1

1 M1

1 T12

] = n [L1

2 M1

2 T22

]

n =1 1 21 1 1

1 1 22 2 2

L M T

L M T

=

1 1 2

1 1 1

2 2 2

L M T

L M T

....(2)

By expressing L, M and T into itscorresponding unit.We have,

n =11 2

m kg second

cm g second

.(3)

Since, 1 m = 100 cm and 1 kg = 1000g,we have,

n =100cm

cm

2100g

(1)g

n = 1021031 = 105Hence, the conversion factor, n = 105Therefore, from equation (1), we have, 1 newton = 105dyne.

*Q.31.Time period of a simple pendulum depends

upon the length of pendulum (l) and

acceleration due to gravity (g). Usingdimensional analysis obtain an expression

for time period of simple pendulum.

Ans: Expression for time period of a simple

pendulum by dimensional analysis:

i. Time period (T) of a simple pendulumdepends upon length (l) and accelerationdue to gravity (g) as follows:T lagbi.e. T = k lagb .(1)

Where k = proportional constant whichis dimensionless.

ii. The dimensions of T = [L0

M0

T1

]The dimensions of l= [L1M0T0]The dimensions of g = [L1M0T2]Taking dimensions on both sides ofequation (1),

[L0M0T1] = [L1M0T0]a [L1M0T2]b[L0M0T1] = [La + bM0T2b]

iii. Equating corresponding power of L, Mand T on both sides, we geta + b = 0 .(2)

and 2b = 1

b =1

2

Substituting b in equation (2), we get

a =1

2

iv. Putting values of a and b in equation (1),we have,

T = k1 12 2g

l

T = k

1 12 2

1

2

= k = k g g

g

l l l

v. Experimentally, it is found that k = 2

T = 2g

l

This is the required expression fortime period

of simple pendulum.Q.32. Point out the drawbacks of dimensional

analysis.

Ans: Drawbacks of dimensional analysis:i. While deriving a formula the

proportionality constant cannot be found.ii. The formula for a physical quantity

depending on more than three otherphysical quantities cannot be derived. Itcan be checked only.

iii. The equations of the type v = u + atcannot be derived. They can be checkedonly.

iv. The equations containing trigonometricalfunctions (sin , cos , etc), logarithmicfunctions (log x, log x3, etc) and

exponential functions (ex,2xe etc) can

neither be derived nor be checkedbecause they are independent of L , Mand T.

v. To derive the formula for a physicalquantity, we must know all the physicalquantities on which it depends.

#Q.33. a. If two quantities have samedimensions, do they represent the

same physical content?

b. A dimensionally correct equation

need not actually be a correct

equation but dimensionally incorrect

equation is necessarily wrong. Justify.

Ans: a. When dimensions of two quantities aresame, they do not represent the samephysical content.


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9Measurement


Example:Modulus of rigidity, pressure, Youngs

modulus and longitudinal stress.

b. i. To justify the statement, let ustake an example of a simplependulum, having one bobattached to a string. It oscillates

under the action of gravity. Theperiod of oscillation of simplependulum depends upon its length(l), mass of the bob (m) andacceleration due to gravity (g).

ii. An expression for its time periodby the method of dimension canbe found out as follows:T = k lxgymz .(1)Where, k is dimensionlessconstant and x, y, z are exponents.

iii. By taking dimensions on both

sides of equation (1), we have,[L0M0T1] = [L1M0T0]x[L1M0T2]y [L0M1T0]z

[L0M0T1] = [Lx + yMz T2y] .(2)iv. Equating dimensions of equation

(2) on both sides we havex + y = 0 , z = 0, 2y = 1

x = y y = 1

2

x =1

2 z = 0

Then equation (1) becomesT =

1

2kl g1

2

= kg

l

T = kg

l .(3)

v. The value of constant k cannot beobtained by the method ofdimensions. It does not matter ifsome number multiplies the righthand side of formula given byequation (3) because that does not

affect dimensions.

T = kg

lis not correct formula

unless we put value of k = 2 Hence, dimensionally correctequation need not actually becorrect equation.

vi. But dimensionally incorrectequation is necessarily wrong.

Let us consider the formula1

2mv = mgh .(4)

We have to check whether it iscorrect or not.

vii. For that write the dimensions ofL.H.S. and R.H.S.

L.H.S. = [L1M1T1]R.H.S. = [L2M1T2]Since the dimensions of R.H.S.and L.H.S. are not correct theformula given by equation (4) is

incorrect. Here,1

2 which is

dimensionless constant does notplay any role.

#Q.34.Whether all constants are dimensionless

or unitless?

Ans:All constants need not be dimensionless orunitless.Example Non dimensional constants arepure numbers, i.e. , e etc., and alltrigonometric functions. Whereas quantitieslike plancks constant, gravitational constantetc., possess dimensions and also have aconstant value. They are dimensional constant.

1.7 Order of magnitude and significant figures

*Q.35.Explain with example the term order of

magnitude of a physical quantity.

OR

What is order of magnitude? Explain withsuitable examples.

Ans: i. The order of magnitude of a physicalquantity is defined as the value of its

magnitude rounded off to the nearest

integral power of 10.

ii. To find the order of magnitude of aphysical quantity, it is first expressed inthe form: P 10Qwhere, P is a numberbetween one and ten and Q is an integer(positive or negative).

iii. If P is 5 or less than 5, the power of10 (i.e. 10Q) gives the order ofmagnitude.

iv. If P is greater than 5, add 1 to the powerof to get the order of magnitude.

Examples:a. Speed of light in air = 3 108m/s

Order of magnitude = 108m/s (3 < 5)

b. Mass of an electron = 9.1 1031kg Order of magnitude = 1030kg (9.1 > 5)


15/24

11Measurement


Q.39. Explain the rules for rounding off the

numbers of the significant figures with

examples.

Ans: Rules for rounding off the numbers:

While rounding off numbers in measurement,following rules are applied.i. If the digit to be dropped is smaller than

5, then the preceding digit should be leftunchanged. eg. 7.34 is rounded off to 7.3

ii. If the digit to be dropped is greater than5, then the preceding digit should beraised by 1.eg. 17.26 is rounded off to 17.3

iii. If the digit to be dropped is 5 followedby digits other than zero, then thepreceding digit should be raised by 1.eg. 7.351, on being rounded off to firstdecimal, becomes 7.4

iv. If the digit to be dropped is 5 or 5

followed by zero, then the precedingdigit is not changed if it is even.eg. 3.45, on being rounded off, becomes3.4

v. If the digit to be dropped is 5 or 5followed by zeros, then the precedingdigit is raised by 1 if it is odd. eg. 3.35,on being rounded off, becomes 3.4

1.8 Accuracy and errors in measurements

*Q.40.Explain in brief, accuracy and errors in

measurements.Ans: i. Accuracy is the closeness of the

measurement to the true or known value.ii. Accuracy of the measurement depends

upon the accuracy of the instrumentused for measurement.

iii. Defect in measurement of physicalquantities can lead to errors andmistakes.

iv. Lesser the errors, more is the accuracy inthe measurement of a physical quantity.

v. For example, when we measure volume

of a bar, the length is measured with ametre scale whose least count is 1 mm.The breadth is measured with a verniercalliper whose least count is 0.1 mm.Thickness of the bar can be measuredwith a micrometer screw gauge whoseleast count is 0.01 mm.

vi. Thus, the smaller the magnitude of aquantity, the greater is the need formeasuring it accurately.

Q.41. What is error? Classify it into differentcategories.

Ans: i. The difference between measured valueand true value of a physical quantity is

called error.

ii. It is the uncertainty in measurement of aphysical quantity.

Error = Measured value True valueiii. Errors are classified into following four

groups:

a. Instrumental error (Constant error)

b. Systematic error (Persistent errors)

c. Personal error (Human error)

d. Random error (accidental error)

*Q.42.What is constant (instrumental) error?

Explain about its cause and remedies.

Ans: i. Constant (instrumental) error:

If the same error is repeated every time

in a series of observations, the error is

said to be constant error.

ii. Constant error is caused due to faultyconstruction of measuring instruments.

iii. Example: If a thermometer is notgraduated properly, i.e. one degree onthe thermometer actually corresponds to0.99, the temperature measured by sucha thermometer will differ from its valueby a constant amount.

iv. In order to minimise constant error,measurements are made with differentaccurate instruments.

*Q.43.What is systematic error? Write down the

cause and minimisation of systematic error.

Ans: i. Systematic errors:

Those errors which occur due to

defective setting of an instrument is

called systematic error.

ii. These errors are due to known reasons,i.e. fault in instrument, improperattention, change in condition, etc.

For example, if the pointer of anammeter is not pivoted exactly at thezero of the scale, it will not point to zerowhen no current is passing through it.

iii. These errors can be minimized bydetecting the sources of errors.


16/24


Measurement12

*Q.44.What is personal error? Give its remedies.

Ans: i. Personal error (Human error):

The errors introduced due to fault of an

observer taking readings are called

personal errors.

Forexample, Error due to non removalof parallax between pointer and itsimage in case of a magnetic compassneedle, errors made in counting numberof oscillations while measuring theperiod of simple pendulum.

ii. They vary from person to person.

iii. These errors can be reduced to someextent by asking different observers totake the measurement.

*Q.45.What is random error (accidental)? How it

can be minimized?

Ans: i. The errors which are caused due tominute change in experimental conditions

like temperature, pressure or fluctuation

in voltage while the experiment is being

performed are called random errors.

ii. Random error cannot be eliminatedcompletely but can be minimized tolarge extent.

Q.46. State general methods to minimise effect of

errors.

Ans: Methods to minimise effect of errors:i. Taking a large magnitude of the quantity

to be measured.

ii. Taking large number of readings andcalculating their mean value.

iii. Using an instrument whose least countis as small as possible.

Q.47. What are mistakes? Explain causes of

mistakes and their remedies.

Ans: i. The faults caused by the carelessness of

an untrained experimenter are calledmistakes.

ii. Causes: Mistake are commited due to

a. lack of skill

b. faulty observation

c. wrong readings

iii. Mistakes can be avoided by careful and

properly trained experimenter.

Q.48. Distinguish between Mistakes and Errors in

measurements.

Ans:

Mistakes Errors

i. Mistake is a fault onthe part of theobserver.

Error is a fault dueto other reasons,such as, limitationsof human senses,instruments etc.

ii. Mistakes can betotally avoided bytaking proper care.

Errors cannot beeliminated, they canbe reduced.

iii. Mistakes are causeddue to lack of taking

precautions,carelessness in takingand recordingreadings, carelessness

in calculations etc.

Errors are causeddue to limitations ofinstruments used formeasurement,personal error due tolimitations of senses

etc.Q.49. Define and explain the terms:

*i. Most probable value (mean value)

*ii. Absolute error

iii. Mean absolute error

*iv. Relative error

*v. Percentage error

Ans: i. Most probable value (mean value):

a. For the accurate measurement of aphysical quantity, we take large

number of readings and find themean or average value of thequantity. This value is called themost probable value or meanvalue and can be considered to betrue value of the quantity.

b. If a1, a2, a3,an are nnumber of readings taken formeasurement of a quantity, thentheir mean value is given by,

amean =n

a........aa n21 +++

ma =1

n

n

ii 1

a=

ii. Absolute error:

The magnitude of the difference between

mean value and each individual value is

called absolute error (a).absolute error = |mean valuemeasured

value|


17/24

13Measurement


| a1| = m 1a a

Similarly |a2 | = m 2a a

|an | = m na a iii. Mean absolute error:

The arithmetic mean of all the absolute

errors is called mean absolute error inthe measurement of the physical quantity.

| ma | = 1 2 na a ........ a

n

+ + +

=n

ii 1

1a

n =

iv. Relative error:

The ratio of the mean absolute error in

the measurement of a physical quantity

to its most probable value is called

relative error.

Relative error=m

m

a

a

v. Percentage error:

The relative error multiplied by 100 is

called the percentage error.

Percentage error=m

m

a

a

100%

Percentage error in different cases :

a. If the error in a is a, then the

percentage error isa

a

100

b. If the error in a is a, then thepercentage error in

an= ana

100%

c. If the error in measurement of a isa and the error in measurementof b is b then, the percentageerror in ab is given by,

ab = a ba b + 100

d. If the error in measurement of ais a and the error inmeasurement of b is b then the

percentage error in,a

bis

a

b=

a b

a b

+

100

Formulae

1. Measure of physical quantity = Numericalvalue size of unit. i.e. M = nu

2. Relation between numerical value and size ofunit: n1u1= n2u2

3. Conversion factor of a unit in two system of

units

n =a

1

2

L

L

1

2

M

M

c1

2

T

T

4. Average value or mean value

ma = 1 2 3 na a a .. a

n

+ + + +=

1

n i 1

n

ia=

5. If x = x1x2, then maximum errorx = x1+ x2

6. If x = m1x n2x , then error in measurement

x

x

=

1

1

m x

x

+

2

2

n x

x

7. Absolute error = Average value Measured

value| an| = | aman|

8. Mean absolute error

| ma | = 1 2 na a ... a

n

+ + +

=n

1 n

i 1ia

=

9. Relative (fractional) error

=m

m

a

a

10. Percentage error

=m

m

a100

a

%

Some practical units in term of S.I. unit

Practical units Abbreviation S.I. unit

1 Angstrom 1010m

1 Micronm 10

6

m1 Nanometer nm 109m

1 Light year ly 9.46 1015m

1 Astronomical unit AU 1.496 1011m

1 Atomic mass unit amu 1.66 1027kg

1 Torr T 1 mm of Hg


18/24


Measurement14

Type I : Problems based on dimensional analysis

*Example 1

Find the dimensions of the following

i. Power ii. Force

iii. ElectricPermittivity

Solution:

i. PowerWork

Time [L2M1T3]

ii. Force Mass Acceleration [L1M1T2]

iii.Electric

permittivity 0=2

2

q

4 r F [L3MT4A2]

Example 2

Check the correctness of formula v2= u2+ 2as by

using dimensional analysis.Solution:

Dimensions of v2 = [L1M0T1]2= [L2M0T2]

Dimensions of u2 = [L1M0T1]2

= [L2M0T2]Dimensions of a = [L1M0T2]Dimensions of s = [L1M0T0]2 is dimensionless quantity.Substitute dimensions in L.H.S. ofequation

L. H. S. = v2= [L2M0T2] .(1)Substitute dimensions in R.H.S. equationR.H.S. = u2+ 2 as=[L2M0T2] + {[L1M0T2] [L1M0T0]}

R.H.S.= [L2M0T2] + [L2M0T2] .(2)From the equations (1) and (2) andoriginal one each term in the formula,have same dimensions, hence theformula is dimensionally correct.

*Example 3

If length L, force F and time T are taken as

fundamental quantities. What would thedimensional equation of mass and density?

Solution:

i. Force = Mass Acceleration

Mass =Force

Acceleration

Dimensional equation of mass1

1 2

FDimension of force

Dimension of acceleration L T

=

= [F1L1T2]

Dimensional equation of mass= [F1L1T2]

ii. Density =Mass

Volume

Dimensional equation of density

= Dimension of massDimension of volume

=1 1 2

3

F L T

L

= [F1L4T2]

*Example 4

Derive an expression of kinetic energy of a body

of mass m and moving with velocity v, using

dimensional analysis.

Solution:

Kinetic energy of a body depends upon mass (m)and velocity (v) of the body.

Let K.E. mxvy K.E. = kmxvy .(1)where k = dimensionless constant of proportionality.Taking dimensions in both sides of equation (1)

[L2M1T2] = [L0M1T0]x[L1M0T1]y

= [L0MxT0] [LyM0Ty]

= [L0 + yMx + 0T0y]

[L2M1T2] = [LyMxTy] .(2)Equating dimensions of L, M, T on both sides ofequation (2)

y = 2 and x = 1 also y = 2 y = 2Substituting x, y in equation (1), we haveK.E. = kmv2

Example 5

A calorie is a unit of heat or energy and it equal

about 4.2 J where 1 J = 1 kg m2s2. Suppose, we

employ a system of units in which unit of mass

equals kg, the unit of length equals m, and the

unit of time is s. Show that a calorie has a

magnitude of 4.2 212in terms of the new

units. (NCERT)Solution: 1 cal. = 4.2 kg m2s2

S.I. system New system

L1= 1 m L2= mM1= 1 kg M2= kgT1= 1 second T2= second

Solved Examples


19/24

15Measurement


Dimensional formula of energy is [L2M1T2]According to the question,

4.2[ 21L1

1M2

1T ] = n [ 22L

1

2M2

2T ]

Hence, magnitude of calorie in the new system isgiven by conversion factor, n.

n = 4.22 1 2

1 1 1

2 2 2

L M T

L M T

= 4.2

2 1 21 m 1 kg 1 s

m kg s

n = 4.2 212

Example 6

Density of oil is 0.8 g cm3in C.G.S. unit. Find its

value in S.I. units.

Solution:C.G.S unit S.I unitL2= 1 cm L1= 1 m = 100 cm .(1)M2= 1 g M1= 1 kg = 1000 gT2= 1 s T1= 1 s

Dimensions of density is [L3M1T0]0.8 g cm3= conversion factor (n) kg m3 .(2)0.8 [L 32

M 12 T20] = n [L 31

M 11 T10]

n =3 1 0

2 2 23 1 0

1 1 1

0.8[L M T ]

[L M T ]

n = 0.8

3 1 0

2 2 1

1 1 2

L M T

L M T

= 0.8

3 1 0

1 cm 1 g 1s100 cm 1000g 1s

= 0.8 [102]3[103]1= 0.8 [10]6[103]

n = 0.8 103Putting the value of n in equation (2), weget, 0.8 g cm3= 0.8 103kg m3.Density of oil in S.I unit = 0.8 103kg m3

Example 7

Assume that the speed (v) of sound in air depends

upon the pressure (P) and density (

) of air, thenuse dimensional analysis to obtain an expression

for the speed of sound.

Solution:

It is given that speed (v) of sound in air dependsupon the pressure (P) and density () of the air.So we can write, v = k Pab (1)where k is a dimensionless constant and a and b arepowers to be determined.Dimensions of v = [L1M0T1]

Dimensions of P = [L1M1T2]Dimensions of = [L3M1T0]Substitute the dimensions of the quantities on bothsides of equation (1)

[L1M0T1] = [L1M1T2]a[L3M1T0]b

[L1M0T1] = [LaMaT2a] [L3bMbT0][L1M0T1] = [La 3bMa + bT2a]

Comparing the powers of L, M and T on both sides,we get

2a = 1 a = 12

Also, a + b = 0

12

+ b = 0 b = 12

Substituting these values in equation (1)v = k P1/2. 1/2

v = kP

*Example8The hydrostatic pressure P of a liquid column

depends upon the density , height h of liquid

column and also an acceleration g due to

gravity. Using dimensional analysis, derive a

formula for pressure P.

Solution:

Let P hxygz

P = k hxygz . (1)Where k is the constant of proportionality.

Dimensionally,[L1M1T2] = [L1M0T0]x [L3M1T0]y [L1M0T2]z

[L1M1T2]=[Lx 3y + zMyT2z]Comparing the power of L, M and T on the bothsides, we gety = 1, x 3y + z = 1 and 2z = 2Solving we get, x = 1, y = 1 and z = 1From equation (1), P = k hgAssuming k = 1, we get

P = hg

Example 9The value of G in C.G.S system is

6.67 108dyne cm2g2. Calculate its value in S.I.

system.

Solution:

C.G.S systemL1= 1 cmM1= 1 gT1= 1 s

S.I. systemL2= 1 m = 100 cm

M2= 1 kg = 1000 gT2= 1 s


20/24


Measurement16

Dimensional formula of gravitational constant =[L3M1T2]

6.67 108dyne cm2g2= Conversion factor (n) Nm2kg2 ....(1)

6.67 108[L 31 M1

1 T 21

] = n [L 32 M1

2 T 22

]

n = 6.67 108

3 1 2

1 1 1

2 2 2

L M TL M T

n = 6.67 1083 1 2

1 cm 1 g 1 s

100 cm 1000 g 1 s

n = 6.67 108103106

n = 6.67 1011From equation (1),

6.67 108dyne cm2g2= 6.67 1011Nm2kg2

Value of G in S.I. system= 6.67 1011Nm2kg2

*Example 10

Using the method of dimension, show that

1 joule = 107erg.

Solution:

Dimension of work = [L2M1T2] .(1)

S.I. system C.G.S system

Dimension=[ 21L M11 T1

2] Dimension=[L 22 M12 T2

2]

L1= 1 m = 100 cm

M1= 1 kg = 1000 g

L2= 1 cm

M2= 1 g

T1= 1 s T2= 1 s

To show that: 1 J = 107erg

Let 1 J = n erg ....(2)Conversion factor, n, is given by,

1 [L 21 M11 T

21 ] = n [L 22 M

12 T

22 ]

n =2 1 2

1 1 1

2 2 2

L M T

L M T

=

21 21000 g 100 cm 1 s

g cm 1 s

= 103104n = 107

Putting the value of n in equation (2), we have,

1 J = 107erg

Type II: Problems based on order of magnitude

and significant figures

Example 11Add 7.21, 12.141 and 0.0028 and express the

result to an appropriate number of significant

figures.

Solution:

7. 2112.141

0.0028

Sum 19.3538

++=

In the given problem minimum number of digitsafter decimal is 2. Result will rounded off upto two places of

decimal. Corrected rounded off sum = 19.35

*Example 12

State the order of magnitude of the following:i. Acceleration due to gravity g = 9.81 m/s2

ii. The gravitation constant

G = 6.67 1011Nm2/kg2

iii. The period of rotation of the earth about its

own axis.

Solution:

i. Acceleration due to gravity g = 9.81 ms2= 9.81 100m/s2

9.81 > 5 Order of magnitude of acceleration due to

gravity = 10

0+1

= 10

1

ms

2

ii. The gravitational constantG = 6.67 1011Nm2/ kg2

6.67 > 5 Order of magnitude gravitational constant

= 1011 + 1= 1010Nm2/kg2iii. The period of rotation of the earth about its

own axisT = 8.64 104s

8.64 > 5 Order of magnitude period of rotation = 104 + 1

= 105s

*Example 13

Determine the number of significant figures in

the following measurements.

i. 0.05718 ii. 93.26

iii. 2.35 1019 iv. 1.3725 109

Solution:

i. 0.05718 significant figures are 5, 7, 1, 8 (Zeroof left side of number is not significant)

Number of significant figures = 4


21/24

17Measurement


ii. 93.26Significant figures = 9, 3, 2, 6

Number of significant figures = 4iii. 2.35 1019

Significant figures = 2, 3, 5 Number of significant figures = 3iv. 1.3725 109

Significant figures = 1, 3, 7, 2, 5 Number of significant figures = 5

Example 14

The length, breadth and thickness of a

rectangular sheet of metal are 4.234 m, 1.005 m

and 2.01 cm respectively. Give the area and

volume of the sheet to correct significant figures.

(NCERT)

Solution:

Given: l= 4.234 m, b = 1.005 m,t = 2.01 cm = 2.01 102m

To find: A = ?, V = ?Formula: i. A = lb ii. V = lb tCalculation:

From formula (i),A = 4.234 1.005 = 4.255As the rectangular sheet has two surfaces, wemultiply the above answer by 2. total area of rectangular sheet = 4.255 2

= 8.510 m2In correct significant figure, A = 8.510 m2

From formula (ii)

V = 4.234 1.005 2.01 102

= 0.0855289 m3In correct significant figure, V = 0.085 m3

Example 15

The mass of a box measured by a grocers

balance is 2.3 kg. Two gold pieces of masses

20.15 g and 20.17 g are added to the box. What is

(i) the total mass of the box? (ii) the difference in

the masses of the pieces to correct significant

figures? (NCERT)

Solution:

i. Total mass of the box= (2.3 + 0.02017 + 0.02015) kg= 2.34032 kg

Since, the last number of significant figure is 2,therefore, the total mass of the box = 2.3 kgii. Difference of mass = (20.17 20.15)

= 0.02 gSince, there are two significant figures so thedifference in masses to the correct significant figuresis 0.02 g.

Example 16Find the order of magnitude of force exerted by

the sun on the earth. Mass of the sun = 1.99 1030

kg, Mass of earth = 5.97 1024 kg, Distance

between Earth and the Sun = 1.49 1011 m.

G = 6.67 1011Nm2/kg2

Solution:

Given: MS = 1.99 1030kg, ME= 5.97 1024kgR = 1.49 1011m, G = 6.67 1011Nm2/kg2

To find: Order of magnitude of force?

Formula: F = G. S E2

M .M

R

Calculation:

From formula,

F =( )

11 30 24

211

6.67 10 1.99 10 5.97 10

1.49 10

= 35.69 1021N

= 3.569 1022

N 3.569 < 5

Order of magnitude = 1022N

Type III: Problems based on errors in

measurement

Example 17In an experiment to find the density of a solid, the

mass and volume of the solid were found to be

400.3 0.02 g and 75.6 0.01 cm3 respectively.

Find the percentage error in the determination of

its density.Solution:

Given: M = 400.3 g, M = 0.02 g,V = 75.6 cm3, V= 0.01 cm3

To find: Percentage error in = ?

Formula:

100%

Calculation:

From formula,

M

M=

0.02

400.3

= 0.00005

andV

V=

6.75

01.0= 0.00013

Relation between mass, volume and density is= M/V

=M

M+

V

V(neglecting sign)

= 0.00005 + 0.00013 = 0.00018


22/24


Measurement18

Hence, the percentage error in the determination ofthe density of the solid is

100% = (0.00018 100)% = 0.018%

Example 18

Find the percentage error in energy E =

1

2 mv2

,

where m = (52.4 0.2) kg and v = (25.6 0.1) m/s.

Solution:

Given: m = 52.4 kg, v = 25.6 m/s, m = 0.2 kg,v = 0.1 m/s

To find: Percentage error in E = ?Formula:

Percentage error in E =m v

2m v

+

100%

Calculation:

From formula,

Percentage error in E = 0.2 0.1252.4 25.6

+

100%

= 1.16%

Type IV: Miscellaneous

Example 19

In an experiment, width of given object was

found to be 1.55 cm, 1.54 cm, 1.53 cm, 1.52 cm,

1.49 cm in successive trials. Calculate mean

absolute error, relative error and percentage

error.

Solution:Given: a1= 1.55 cm, a2= 1.54 cm, a3= 1.53 cm,a4= 1.52 cm, a5= 1.49 cm, n = 5

To find: i. Mean absolute error = ?ii. Relative error = ?

iii. Percentage error = ?Formula:

i. Mean absolute error = 1 2 na a ..... a

n

+ + +

ii. Relative error =m

m

a

a

iii. Percentage error =m

m

a

a 100%

Calculation:

Mean of readings, 1 2 3 4 5ma a a a a

a5

+ + + +=

5

49.152.153.154.155.1 ++++=

ma 526.15

63.7== cm

1a 1.526 1.55 0.024 0.024 = = =

2a 1.526 1.54 0.014 0.014 = = =

3a 1.526 1.53 0.004 0.004 = = =

4a 1.526 1.52 0.006 0.006 = = =

5a 1.526 1.49 0.036 0.036 = = = From formula (i),

1 2 3 4 5m

| a | | a | | a | | a | | a |a

5

+ + + + =

=0.024 0.014 0.004 0.006 0.036

5

+ + + +

=0.084

5= 0.0168

ma = 0.0168 cm

From formula (ii),

Relative error =0.0168

1.526= 0.011

Relative error = 0.011cmFrom formula (iii),Percentage error = 0.011 100 = 1.1Percentage error = 1.1%

*Example 20

An object was weighed by a physical balance and

following readings were obtained:

5.04 g, 5.06 g, 4.97 g, 5.00 g, 4.93 g.

Find i. Mean value

ii. Mean absolute error

iii. Percentage error

Solution:Given: a1= 5.04 g, a2= 5.06 g, a3= 4.97 g

a4= 5.00 g, a5= 4.93 g, n = 5To find: i. Mean value = ?

ii. Mean absolute error = ?iii. Percentage error = ?

Formula:

i. Mean value = 1 2 3 4 5ma a a a a

a5

+ + + +=

ii. Mean absolute error = 1 2 na a ..... an

+ + +

iii. Percentage error =m

m

a

a

100%

Calculation:

i. From formula (i),

ma = 5.04 5.06 4.97 5.00 4.935

+ + + +

= 25.005

= 5.00 g

ma =5.00 g


23/24

19Measurement


ii. Absolute errors:|a1| = |ama1| = |5.00 5.04| = |0.04| = 0.04|a2| = |ama2| = |5.00 5.06| = |0.06| = 0.06|a3| = |ama3| = |5.00 4.97| = |0.03| = 0.03|a4| = |ama4| = |5.00 5.00| = 0.00|a5| = |ama5| = |5.00 4.93| = |0.07| = 0.07From formula (ii),

ma = 1 2 3 4 5a a a a a5

+ + + +

=0.04 0.06 0.03 0.00 0.07

5

+ + + +

=0.20

5= 0.04

ma = 0.04 g

iii. From formula (iii),

Percentage error =0.04

100%5.00

=4

5= 0.8

Percentage error = 0.8%

Type I : Problem based on Dimensional analysis

1. The acceleration due to gravity of a place is9.8 ms2. Find its value in km h2.

2. If the value of atmospheric pressure is106dyne cm2in CGS system, find its value in

S.I. system.3. Check the correctness of the relation = I,

where is the torque acting on a body, I is themoment of inertia and is angularacceleration.

4. Check the correctness of T = 2g

1

5. Assuming that the mass M of the largest stone

that can be moved by a flowing river depends

upon v the velocity, the density of water

and on g, the acceleration due to gravity.Show that M varies with sixth power of the

velocity of flow.

6. Assuming that the critical velocity vc of aviscous liquid flowing through a capillary tubedepends only upon the radius r of the tube,density and the coefficient of viscosity ofthe liquid, find the expression for criticalvelocity.

Type II: Order of magnitude and significant

figures

7. Find the order of magnitude of following data.i. height of a tower 4325 mii. weight of a car 789 kgiii. Charge on electron 1.6 1019C

8. Round off the following numbers as indicatedi. 15.654 upto 3 digitsii. 1426 upto 5 digitsiii. 5.996 105upto 3 digits

9. Add 3.8 106and 4.2 105with due regardto significant figures.

10. The diameter of a sphere is 2.78 m. Calculateits volume with due regard to significantfigures.

Type III: Errors in measurement

11. Two different masses are determined as(23.7 0.5) g and (17.6 0.3) g. What is thesum of their masses?

12. The lengths of two rods are recorded asl1= (25.2 0.1) cm and l2= (16.8 0.1) cm.Find their combined length.

Type IV : Miscellaneous

13. The length of a rod as measured in anexperiment was found to be 2.48 m, 2.46 m,2.49 m, 2.50 m and 2.48 m. Find the meanabsolute error, relative error and percentageerror.

1. Which of the following is the fundamental unit?(A) Length, force, time(B) Length, mass, time(C) Mass, volume, height

(D) Mass, velocity, pressure2. S.I. unit of energy is joule and it is equivalent to

(A) 106erg (B) 107erg(C) 107erg (D) 105erg

3. Which of the following pairs are similar?(A) work and power(B) momentum and energy(C) force and power(D) work and energy

Practice Problems

Multiple choice questions


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27. The diameter of the paper pin is measuredaccurately by using(A) Vernier callipers(B) Micrometer screw gauge(C) Metre scale(D) A measuring tape

28. For measurement of mass of subatomic

particles following method is suitable(A) Gravitational method(B) Mass spectro graph(C) Method based on Newtons law of

gravitation(D) Lever balance

29. An atomic clock makes use of(A) cesium-133 atom (B) cesium-132 atom(C) cesium-123 atom (D) cesium-131 atom

30. Dimensional equation for temperature gradientis(A) [L0M1T1K0] (B) [L1M0T0K1](C) [L1M0T0K1] (D) [L1M0T0K1]

31. The two stars S1 and S2 are located atdistances d1and d2respectively. Also if d1 >d2then following statement is true.(A) The parallax of S1and S2are same.(B) The parallax of S1 is twice as that of S2(C) The parallax of S1 is greater than

parallax of S2(D) The parallax of S2 is greater than

parallax of S1

32. 0.00849 contains ______ significant figures.(A) 6 (B) 5(C) 3 (D) 2

33. 50.000 contains ______ significant figures.(A) 5 (B) 3(C) 2 (D) 1

34. 3.310 102has ______ significant figures.(A) 6 (B) 4(C) 2 (D) 1

35. Which of the following is NOT a fundamental

quantity?(A) temperature (B) electric charge(C) mass (D) electric current

36. Which of the following is NOT a unit of time?(A) hour (B) nano second(C) microsecond (D) light year

37. Identify the pair, whose dimensions are equal(A) torque and work (B) stress and energy(C) force and stress (D) force and work

38. The dimensional formula [L2M1T2] represents.(A) moment of force (B) force(C) acceleration (D) momentum

39. Order of magnitude of (106 + 103) is(A) 1018 (B) 109(C) 106 (D) 103

40. Nano size of the gold has ______ colour.(A) Pink (B) yellow(C) golden (D) red

1. 127008 km h2 2. 105Nm2

6. cK

vr

=

7. i. 103m ii. 103kgiii. 1019C

8. i 15.7 ii. 14260iii. 6.00 105

9. 4.6 105 10. 11.25 m311. 41.3 0.8 g 12. 42.0 0.2 cm13. 0.01 m, 0.004 m, 0.4%

1. (B) 2. (C) 3. (D) 4. (B)

5. (B) 6. (C) 7. (A) 8. (D)9. (A) 10. (D) 11. (D) 12. (B)13. (C) 14. (C) 15. (B) 16. (A)17. (C) 18. (D) 19. (A) 20. (C)21. (A) 22. (B) 23. (B) 24. (A)25. (A) 26. (D) 27. (B) 28. (B)29. (A) 30. (D) 31. (D) 32. (C)33. (A) 34. (B) 35. (B) 36. (D)37. (A) 38. (A) 39. (C) 40. (A)

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