Ramgarhia Polytechnic - UNITS AND DIMENSIONS PHYSICS... · 2020. 4. 10. · 1 UNITS AND DIMENSIONS...

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UNITS AND DIMENSIONS

PHYSICAL QUANTITIES

The quantities which can be

measured directly or indirectly

are called physical quantities.

Examples: Length, mass,

time, temperature, electric

current, speed, density, area

TYPES OF PHYSICAL

QUANTITIES

Physical quantities are of two

types:

1. Fundamental or Basic

quantities

2. Derived quantities

FUNDAMENTAL OR BASIC

QUANTITIES

Physical quantities which are

not related to each other or to

other quantities are called

fundamental or basic

quantities.

BOiqk rwSIAW

rwSIAW, ijnWH nUM is`Dy jW Ais`Dy qOr 'qy mwipAw jw skdw hY, BOiqk rwSIAW kihMdy hn[

audwhrxW: lMbweI, puMj, smW, qwpmwn, ibjleI krMt, cwl, Gxqw, KyqrPl

BOiqk rwSIAW dIAW iksmW:

BOiqk rwSIAW do qrWH dIAW huMdIAW hn:

1. mùFlIAW rwSIAW 2. ivauNqpMn rwSIAW

mu`FlIAW rwSIAW

BOiqk rwSIAW, ijnWH dw Awps iv`c jW dUjIAW rwSIAW nwl koeI sMbMD nhIN huMdw,nUM mu`FlIAW rwSIAW ikhw jWdw hY[

audwhrxW: lMbweI, puMj, smW, qwpmwn, ibjleI krMt

2

DERIVED QUANTITES

Physical quantities which can

be obtained from fundamental

or basic quantities by

multiplication or division are

called derived quantities.

Examples:, area, volume,

density, weight, force, power,

energy, pressure, speed

UNIT

To measure a physical

quantity, it is compared with a

standard of same kind. This

chosen standard is called unit.

TYPES OF UNITS

Units are of two types:

1. Fundamental or Basic

units

2. Derived units

FUNDAMENTAL OR BASIC

UNITS

Units which are not related to

each other or to other units

are called fundamental or

basic units.

ivauNqpMn rwSIAW

BOiqk rwSIAW, ijnWH nUM mùFlIAW rwSIAW qoN guxW jW Bwg kr ky pRwpq kIqw jw skdw hY, nUM ivauNqpMn rwSIAW ikhw jWdw hY[

audwhrxW: KyqrPl, GxPl dbwA, cwl, , Gxqw, Bwr, bl, SkqI, aUrjw

iekweI

iksy BOiqk rwSI nUM mwpx leI ies dI ausy iksm dy stYNfrf nwl qulnw kIqI jWdI hY[ies cuxy hoey stYNfrf nUM iekweI kihMdy hn[

iekweIAW dIAW iksmW

iekweIAW dIAW do iksmW huMdIAW hn:

1. mùFlIAW iekweIAW 2. ivauNqpMn iekweIAW

mu`FlIAW iekweIAW

iekweIAW, ijnWH dw Awps iv`c jW dUjIAW iekweIAW nwl koeI sMbMD

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Examples: units of Length,

mass, time, temperature,

electric current,

DERIVED UNITS

Units which can be obtained

from fundamental or basic

units by multiplication or

division are called derived

units.

Examples: units of, area,

volume, density, weight, force,

power, energy, pressure,

speed

Area = length x breadth

= mx m = m2

m2 is derived unit of area

Volume = length x breadth x

height

= m x m x m = m3

m3 is derived unit of volume

Speed = distance / time

= m/s =ms-1

ms-1 is derived unit of speed

nhIN huMdw nUM mu`FlIAW iekweIAW ikhw jWdw hY[audwhrxW: lMbweI, puMj, smW, qwpmwn, ibjleI krMt dIAW iekweIAW

ivauNqpMn iekweIAW

iekweIAW, ijnWH nUM mùFlIAW iekweIAW qoN guxW jW Bwg kr ky pRwpq kIqw jw skdw hY, nUM ivauNqpMn iekweIAW ikhw jWdw hY[

audwhrxW: KyqrPl, GxPl ,dbwA, cwl, , Gxqw, Bwr, bl, SkqI, aUrjw dIAW iekweIAW

KyqrPl = lMbweI x cOVweI

= mItr x mItr = mItr2

mItr2 KyqrPl dI ivauNqpMn iekweI hY[

GxPl = lMbweI x cOVweI x aucweI = mItr x mItr x mItr = mItr3

mItr3 GxPl dI ivauNqpMn iekweI hY[

cwl = dUrI / smW

= mItr/ sYikMf

mItr/ sYikMf cwl dI ivauNqpMn iekweI hY

4

MEASUREMENT OF

PHYSICAL QUANTITY

Measurement of physical

quantity (Q) involves two

steps:

a) Selection of unit (u)

b) Numerical value (n)

Q = n u

In system I, Q= n1 u1

In system II, Q = n2 u2

Hence, n1 u1 = n2 u2

Numerical value of a

quantity is inversely

proportional to its unit.

n α 1/u

CHARACTERISTICS OF

UNITS

1. It should be well-defined.

2. It should be of suitable

size.

3. It should be easily

reproducible.

4. It should be easily

accessible.

5. It should not change with

time.

6. It should not change with

place.

[BOiqk rwSI dw mwp

BOiqk rwSI (Q) dw mwp iv`c do stYp huMdy hn: a) iekweI dI cox (u) b) numYrIkl mùl (n)

Q = n u

pRxwlI 1 iv`c, Q= n1 u1

pRxwlI 2 iv`c, Q = n2 u2

ies leI, n1 u1 = n2 u2

iksy BOiqk rwSI dw numYrIkl mùl ies dI iekweI dy ault AnupwqI huMdw hY[ n α 1/u

iekweIAW dIAW ivSySqwvW 1. ieh cMgI qrWH pRIBwiSq hoxw

cwhIdw hY[ 2. ieh Xog Awkwr dw hoxw

cwhIdw hY[ 3. ies nUM sOiKAW hI

punrauqpwdnXog hoxw cwhIdw hY[

4. ieh sOiKAW hI phuMcXog hoxw cwhIdw hY[

5. ieh smyN nwl qbdIl nhIN hoxw cwhIdw[

5

7. It should not be

perishable.

8. It should not be affected

by physical conditions

like temperature,

pressure, humidity etc.

9. It should be accepted all

over the world.

SYSTEMS OF UNITS

There are four systems of

units-

1. C.G.S. System: In this

system, the unit of

length is centimeter, the

unit of mass is gram and

the unit of time is

second.

2. F.P.S. System: In this

system, the unit of

length is foot, the unit of

mass is pound and the

unit of time is second.

3. M.K.S. System: In this

system, the unit of

length is meter, the unit

of mass is kilogram and

the unit of time is

second.

6. ieh jgwH nwl qbdIl nhIN hoxw cwhIdw[

7. ieh ^qmXog nhIN hoxw cwhIdw[

8. ies 'qy BOiqk hwlqW ijvyN qwpmwn, dbwA, is`l Awid dw pRBwv nhIN hoxw cwhIdw[

9. ieh pUrI dunIAW iv`c svIkwrq hoxw cwhIdw hY[

iekweIAW dIAW pRxwlIAW

iekweIAW dIAW cwr pRxwlIAW huMdIAW hn-

1.sI.jI.AYs.pRxwlI:ies pRxwlI iv`c lMbweI dI iekweI sYNtImItr, puMj dI iekweI grwm Aqy smyN dI iekweI sYikMf huMdI hY[

2. AYP.pI.AYs.pRxwlI:ies pRxwlI iv`c lMbweI dI iekweI Pu`t, puMj dI iekweI pONf Aqy smyN dI iekweI sYikMf huMdI hY[

3. AYm.ky.AYs.pRxwlI:ies pRxwlI iv`c lMbweI dI iekweI mItr, puMj dI iekweI iklogrwm Aqy smyN dI iekweI sYikMf huMdI hY[

6

3. S.I.(System

International de Units):

In this system, there are

seven basic units and two

supplementary units.

Basic Units:

1. metre for length

2. kilogram for mass

3. second for time

4. kelvin for temperature

5. ampere for electric

current

6. candela for luminous

intensity

7. mole for amount of

substance

Supplementary Units:

1. radian for plane

angle.

2. steradian for solid

angle

4. AYs. AweI.:ies pRxwlI iv`c s`q byisk Aqy do pUrk iekweIAW huMdIAW hn[

byisk iekweIAW: 1. lMbweI leI mItr 2. puMj leI iklogrwm 3. smyN leI sYikMf 4. qwpmwn leI

kYlivn 5. ibjleI kurMt leI

AYmpIAr 6. pRkwSmwn qIbrqw

leI kYNfylw 7. pdwrQ dI mwqrw

leI mol

splImYNtrI iekweIAW:

1. smql kox leI ryfIAn

2. Tos kox leI strfIAn

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ADVANTAGES OF SI

1. It is a rational System

of units: In this system

only one unit is used for

one physical quantity.

For example, newton for

all types of forces, joule

for all types of energies.

2. It is coherent system of

units: In this system all

the derived units are

obtained by multiplying

or dividing basic units,

without involving any

numerical factor.

3. It is closely related to

C.G.S. system of units.

4. It is metric system of

units. In this system

power of 10 is used.

AYs AweI dy Pwiedy

1. ieh iekweIAW dI qrksMgq pRxwlI hY[ies pRxwlI iv`c ie`k rwSI leI ie`ko iekweI dI vrqoN huMdI hY[audwrhx dy qOr 'qy ,bl dIAW swrIAW iksmW leI inaUtn, swrIAW aUrjw dIAW iksmW leI jUL iekweI [

2. ieh iekweIAW dI susMgq(AnukUl) pRxwlI hY[ies pRxwlI iv`c swrIAW ivauNqpMn iekweIAW byisk iekweIAW nUM guxW jW Bwg kr ky pRwpq kIqw jWdw hY, ibnWH koeI numYrkIl PYktr Swiml kIqy[

3. ieh pRxwlI iekweIAW dI sI.jI.AYs. pRxwlI nwl nyVlw sMbMD r`KdI hY[

4. ieh iekweIAW dI mIitRk pRxwlI hY[ies pRxwlI iv`c 10 dI pwvr dI vrqoN huMdI hY[

8

TABLE 1

Basic Quantity

C.G.S. SYSTEM

M.K.S. SYSTEM

F.P.S. SYSTEM

Length centimetre metre Foot

Mass gram kilogram Pound

Time second second second

TABLE 2

TABLE 3

S.No.

Quantity Unit Symbol

1 Plane Angle radian rad

2 Solid angle steradian sr

S.No.

Quantity Unit Symbol

1 Length metre m

2 Mass kilogram kg

3 Time second s

4 Temperature kelvin K

5 Electric Current ampere A

6 Luminous Intensity

candela cd

7 Amount of substance

mole mol

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MULTIPLES SUBMULTIPLES

PREFIX SYMBOL VALUE

deci d 10-1

centi c 10-2

milli m 10-3

micro µ 10-6

nano n 10-9

pico p 10-12

femto f 10-15

atto a 10-18

SOME OTHER UNITS

1. Micron (micrometre)= 10-6 m

2. Angstrom (Å) = 10-10 m ( It is unit of length convenient on

the atomic scale

3. Fermi (F) = 10-15 m

4. Astronomical unit (AU) = 1.496 x 1011m

5. Light year = 9.46 x 1015m

6. Parsec = 3.08 x 1016m

PREFIX SYMBOL VALUE

Deca da 101

Hector h 102

Kilo k 103

Mega M 106

Giga G 109

Tera T 1012

Peeta P 1015

Exa E 1018

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IMPORTANT CONVERSIONS

1kg = 1000 g

1m = 100 cm

1 cm = 10 mm

1m = 1000 mm

1 inch = 2.54 cm

1 foot =12 inch = 30.48 cm

1 pound = 453 g

1 day = 24 h = 86400 s

1 year = 365 days= 31536000 s

1 quintal = 100 kg

1 ton = 10 quintal = 1000 kg

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DIMENSIONS OF A

PHYSICAL QUANTITY

Dimensions of a physical

quantity are the powers to

which the basic units must be

raised to obtain the derived

units of that quantity.

If [M], [L] and [T] denote

mass, length, time, then

Area = length x breadth

= [L] x[L]

= [L2]

= [M0L2T0]

Area has zero dimensions in

mass, 2 dimensions in length,

0 dimensions in time.

Volume = length x breadth x

height

=[L] x[L]x [L]

= [L3]

= [M0L3T0]

BOiqk rwSIAW dy AwXwm

iksy BOiqk rwSI dy AwXwm auh Gwq hn, ijnWH 'qy mùFlIAW iekweIAW nUM auBwirAw jwvy qW jo aus rwSI dIAW ivauNqpMn iekweIAW pRwpq ho skx[

jy [M], [L] and [T] puMj, lMbweI Aqy smyN nUM drswaux qW

KyqrPl = lMbweI x cOVweI

= [L] x[L]

= [L2]

= [M0L2T0]

KyqrPl iv`c puMj dy 0, lMbweI dy 2, smyN dy 0 AwXwm huMdy hn[

GxPl = lMbweI x cOVweI x aucweI

=[L] x[L]x [L]

= [L3]

=[M0L3T0]

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Volume has zero dimensions

in mass, 3 dimensions in

length, 0 dimensions in time.

Speed = Distance / Time

= [L]/ [T]

= [LT-1]

Speed has 1 dimension in

length, -1 dimension in time.

DIMENSIONAL FORMULA

Dimensional formula of a

physical quantity is defined as

the expression which shows

which basic unit enters into

the derived units of that

quantity and with what

powers.

Dimensional formula for area

is [M0L2T0]. This shows that

length enters in the derived

units of area with power of 2.

DIMENSIONL EQUATION

Dimensional equation of a

physical quantity is defined as

the equation obtained by

GxPl iv`c puMj dy 0 ,lMbweI dy 3, smyN dy 0 AwXwm huMdy hn[

cwl = dUrI / smW

= [L]/ [T]

= [LT-1]=[M0L1T-1]

cwl iv`c puMj dy 0 , lMbweI dy 1, smyN dy -1 AwXwm huMdy hn[

AwXwm sUqr

iksy BOiqk rwSI dw AwXwm sUqr auh hY jo drswauuNdw hY ik aus rwSI dIAW ivauNqpMn iekweIAW iv`c ikhVI mu`FlI iekweI AwauNdI hY Aqy iks Gwq nwl[

KyqrPl dw AwXwm sUqr hY [M0L2T0]. ieh drswauNdw hY ik KyqrPl dIAW ivauNqpMn iekweIAW iv`c lMbweI 2 dI Gwq nwl AwauNdI hY[

AwXwm smIkrn

iksy BOiqk rwSI dI smIkrn auh smIkrn hY, jo rwSI nUM ies dy AwXwm sUqr dy brwbr krn nwl pRwpq huMdI hY, nUM AwXwm smIkrn

13

equating quantity with its

dimensional formula .

Dimensional equation for area

is

Area = [M0L2T0]

ikhw jWdw hY[

KyqrPl dI AwXwm smIkrn hY

KyqrPl = [M0L2T0]

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LIST OF DIMENSIONAL

FORMULAE

1. Area = length x breadth

= [L] x[L]

= [L2]

= [M0L2T0]

2. Volume=length x breadth

x height

=[L] x[L]x [L]

= [L3]

=[M0L3T0]

3. Density = mass/ volume

=[M]/ [L3]

=[ML-3]

=[M1L-3T0]

4. Speed = Distance / Time

= [L]/ [T]

= [LT-1]

= [M0 L1T-1]

5. Velocity=displacement/ti

me

= [L]/ [T]

AwXwm sUqrW dI sUcI

1. KyqrPl = lMbweI x cOVweI = [L] x[L]

= [L2]

= [M0L2T0]

2. GxPl = lMbweI x cOVweI x aucweI = [L] x[L]x [L]

= [L3]

=[M0L3T0]

3. Gxqw = puMj / GxPl

=[M]/ [L3]

=[ML-3]

=[M1L-3T0]

4. cwl = dUrI /smW = [L]/ [T]

= [LT-1]

= [M0 L1T-1]

5. vyg = ivsQwpn / smW = [L]/ [T]

= [LT-1]

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= [M0 L1T-1]

6. Acceleration=velocity/time

= [LT-1]/ [T]

= [LT-2]

= [M0 L1T-2]

7. Momentum=mass x

velocity

= [M] X [LT-1]

=[M LT-1]

=[M1 L1T-1]

8. Force = mass x

acceleration

= [M] X [LT-2]

=[M LT-2]

=[M1 L1T-2]

9. Work = force x distance

= [M1 L1T-2] x[L]

= [M1 L2T-2]

10. Power = work / time

= [M1 L2T-2]/ [T]

= [M1 L2T-3]

11. Energy = work

= [M1 L2T-2]

12. Pressure = Force / Area

= [M1 L1T-2]/[L2]

= [M0 L1T-1]

6. pRvyg = vyg / smW = [LT-1]/ [T]

= [LT-2]

= [M0 L1T-2]

7. sMvyg = puMj x vyg

= [M] X [LT-1]

=[M LT-1]

=[M1 L1T-1]

8. bl = puMj x pRvyg

= [M] X [LT-2]

=[M LT-2]

=[M1 L1T-2]

9. kMm = bl x dUrI = [M1 L1T-2] x[L]

= [M1 L2T-2]

10. SkqI = kMm /smW = [M1 L2T-2]/ [T]

= [M1 L2T-3]

11. aUrjw = kMm = [M1 L2T-2]

12. dbwA = bl / KyqrPl

= [M1 L1T-2]/[L2]

=[M1 L-1T-2]

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=[M1 L-1T-2]

13. Surface Tension =

Force / length

= [M1 L1T-2]/[L]

=[M1 L0T-2]

14. Gravitational

Constant(G)

F = G m1 xm2 / d2

G= F d2 / m1 x m2

= [M1L1T-2][L2]/ [M]x[M]

= [M-1 L3T-2]

15. Coefficient of

viscosity (η)

F = η Av /d

η = F d / A v

= [M1L1T-2]x[L]/ [L2] x [LT-1]

= [M1L-1T-1]

16. Angle = arc /radius

= [L] /[L]

= [L0]

= [M0L0T0]

17. Angular

Displacement = angle

covered = [M0L0T0]

13. sqHw qxwA = bl / lMbweI = [M1 L1T-2]/[L]

= [M1 L0T-2]

14. grUqvI siQr AMk (G) F = G m1 xm2 / d2

G= F d2 / m1 x m2

=[M1L1T-2][L2]/

[M]x[M]

= [M-1 L3T-2]

15. icpicpwiht dw guxWk (η)

F = η Av /d

η = F d / A v

=[M1L1T-2]x[L]/ [L2] x [LT-1]

= [M1L-1T-1]

16. kox = cwp / ArD ivAws = [L] /[L]

= [L0]

= [M0L0T0]

17. koxI ivsQwpn = kox= [M0L0T0]

17

18. Angular velocity =

angular displacement /

time

= [L0] /[T]

= [L0T-1]

= [M0L0T-1]

19. Angular

Acceleration = angular

velocity / time

= [L0T-1] /[T]

= [L0T-2]

= [M0L0T-2]

20. Torque = Force x ┴

distance

= [M1 L1T-2] x[L]

= [M1 L2T-2]

21. Angular momentum

= momentum x ┴

distance

= [M1 L1T-1] x [L]

= [M1 L2T-1]

22. Time Period = time

for one vibration

= [T]

= [M0L0T1]

18. koxI vyg = koxI ivsQwpn / smW = [L0] /[T]

= [L0T-1]

= [M0L0T-1]

19. koxI pRvyg = koxI vyg / smW = [L0T-1] /[T]

= [L0T-2]

= [M0L0T-2]

20. tOrk = bl x ┴ dUrI = [M1 L1T-2] x[L]

= [M1 L2T-2]

21. koxI sMvyg = sMvyg x ┴ dUrI = [M1 L1T-1] x [L]

= [M1 L2T-1]

22. AwivRqI kwl = [T]

= [M0L0T1]

18

23. Frequency = 1 /

Time period

= 1/ [T]

= [T-1]

= [M0L0T-1]

24. Planck's Constant

(h)

Energy, E = h ν

h = E / v

= [M1 L2T-2] / [T-1]

=[M1 L2T-1]

25. Impulse = Force x

time

= [M1 L1T-2] x [T]

= [M1 L1T-1]

26. Stress = Force /

area = [M1 L1T-2]/[L2]

=[M1 L-1T-2]

27. Strain = Change in

dimensions (length,

volume) / original

dimensions

= [L]/[L] = [L0]

= [M0L0T0]

23. AwivRqI = 1 / AwivRqI kwl

= 1/ [T]

= [T-1]

= [M0L0T-1]

24. plYNk siQr AMk(h) aUrjw , E = h ν

h = E / v

= [M1 L2T-2] / [T-1]

=[M1 L2T-1]

25. Awvyg = bl x smW

= [M1 L1T-2] x [T]

= [M1 L1T-1]

26. qxwA = bl /KyqrPl

= [M1 L1T-2]/[L2]

=[M1 L-1T-2]

27. stryn = AwXwmW iv`c qbdIlI/ Asl AwXwm = [L]/[L] = [L0]

= [M0L0T0]

19

28. Modulus of

Elasticity

= Stress /Strain =[M1 L-1T-2]/ [M0L0T0]

=[M1 L-1T-2]

29. Moment of Inertia =

mass x (┴ distance )2

= [M] x [L]

= [ML2]

= [ML2T0]

28. Lckqw dw mwfUls = strYs / stryn =[M1 L-1T-2]/ [M0L0T0]

=[M1 L-1T-2]

29. jVHqw dw mumYNt = puMj x (┴ dUrI )2 = [M] x [L]

= [ML2]

= [ML2T0]

20

DIMENSIONAL QUANTITIES

The quantities which have

certain dimensions are called

dimensional quantities.

Examples: Length, mass,

time, area, volume, density,

speed, force, power

DIMENSIONLESS

QUANTITIES

The quantities which have no

dimensions are called

dimensionless quantities.

Examples: Angle, strain,

DIMENSIONAL CONTANTS

The constants which have

certain dimensions are called

dimensional constants.

Examples: Gravitational

constant, Planck's constant,

Coefficient of viscosity

DIMENSIONLESS

CONSTANTS

The constants which have no

dimensions are called

dimensionless constants.

Examples: 1,2,3...... π, e,

AwXwmI rwSIAW

rwSIAW ijnHW dy koeI nw koeI AwXwm huMdy hn, nUM AwXwmI rwSIAW ikhw jWdw hY[ audwhrxW: lMbweI, puMj, smW, KyqrPl, GxPl, Gxqw, cwl, bl, SkqI AwXwmhIx rwSIAW rwSIAW ijnHW dy koeI AwXwm nhIN huMdy, nUM AwXwmhIx rwSIAW ikhw jWdw hY[ audwhrxW: kox, stryn AwXwmI siQr AMk siQr AMk ijnHW dy koeI nw koeI AwXwm huMdy hn, nUM AwXwmI siQr AMk ikhw jWdw hY[ audwhrxW: gurUqvI siQr AMk, plYNk siQr AMk, AwXwmhIx siQr AMk

siQr AMk ijnHW dy koeI AwXwm nhIN huMdy, nUM AwXwmhIx siQr AMk ikhw jWdw hY[

audwhrxW:

1,2,3...... π, e,

21

PAIRS OF PHYSICAL

QUANTITIES HAVING SAME

DIMENSIONS

1. Speed and Velocity

2. Frequency and Angular

velocity

3. Work, Energy, Torque

4. Pressure , Stress,

Modulus of Elasticity

5. Impulse and Momentum

6. Angular Momentum and

Planck's constant

BOiqk rwSIAW dy joVy ijnWH dy AwXwm ie`ko ijhy hn

1. cwl Aqy vyg 2. AwivRqI Aqy koxI vyg 3. kMm, aUrjw , tOrk 4. dbwA, strY`s, lckqw

dw mwfUls 5. Awvyg Aqy sMvyg 6. koxI sMvyg Aqy plYNk

siQr AMk

22

PRINCIPLE OF

HOMOGENEITY OF

DIMENSIONS

It states that dimensions of

each term in a physical

relation must be the same.

Explanation:

v = u + a t

Dimensions of v = [M0 L1T-1]

Dimensions of u = [M0 L1T-1]

Dimensions of a = [M0 L1T-2]

Dimensions of t = [M0 L0T1]

Dimensions of a t = [M0 L1T-2]

x [M0 L0T1] = [M0 L1T-1]

Hence, the dimensions of

each term in the physical

relation are the same.

AwXwmW dI iekswrqw dw isDWq

iksy vI BOiqk smIkrn iv`c hryk trm dy AwXwm ie`ko ijhy hoxy cwhIdy hn[ ivAwiKAw: v = u + a t

v dy AwXwm = [M0 L1T-1]

u dy AwXwm = [M0 L1T-1]

a dy AwXwm = [M0 L1T-2]

t dy AwXwm = [M0 L0T1]

a t dy AwXwm = [M0 L1T-2] x [M0 L0T1] = [M0 L1T-1]

ies qrWH , BOiqk smIkrn iv`c hryk trm dy AwXwm ie`ko ijhy hn[

23

Q. If x = at +bt2 + c, where x

is distance and t is time,

what are dimensions of

constants a, b and c.

Ans. Given, x = at +bt2 + c

According to principle of

homogeneity of dimensions,

dimensions of each term must

be the same.

Dimensions of x = Dimensions

of at

[x] = [at]

[a] = [x] /[t] = [L] /[T] = [LT-1]


of bt2

[x] = [bt2]

[b] = [x] /[t2] = [L] /[T2] = [LT-2]


of c

[x] = [c]

[c = [x] = [L]

Q. If v = at +bt2 , where v is

velocity and t is time , what

are dimensions of

constants a and b.

pR. jy x = at +bt2 + c, ij`Qy x dUrI Aqy t smW hY, siQr AMkW a, b Aqy c dy AwXwm kI hn?

h`l . id`qw hY, x = at +bt2 + c

AwXwm dI iekswrqw dy inXm Anuswr hryk trm dy AwXwm ie`ko ijhy hoxy cwhIdy hn[

x dy AwXwm = at dy AwXwm

[x] = [at]

[a] = [x] /[t] = [L] /[T] = [LT-1]

x dy AwXwm = bt2 dy AwXwm

[x] = [bt2]

[b] = [x] /[t2] = [L] /[T2] = [LT-2]

x dy AwXwm = c dy AwXwm

[x] = [c]

[c = [x] = [L]

pR. jy v = at +bt2, ij`Qy v vyg Aqy t smW hY, siQr AMkW a Aqy b dy AwXwm kI hn?

24

Ans. Given, v = at +bt2




be the same.

Dimensions of v = Dimensions

of at

[v] = [at]

[a] = [v] /[t] = [LT-1] /[T] = [LT-2]

Dimensions of v = Dimensions

of bt2

[v] = [bt2]

[b] = [v] /[T2] = [LT-1] /[T2] =

[LT-3]

Q. The Vander Waal gas equation

is (P+ a/V2)(V-b) = RT. Find the

dimensions of 'a' and 'b' by using

dimensional analysis

Sol. Given equation is

(P+ a/V2)(V-b) = RT




be the same.

Dimensions of P =

Dimensions of a/V2

h`l . id`qw hY, v = at +bt2


v dy AwXwm = at dy AwXwm

[v] = [at]

[a] = [v] /[t] = [LT-1] /[T] = [LT-2]

v dy AwXwm = bt2 dy AwXwm

[v] = [bt2]

[b] = [v] /[T2] = [LT-1]/ [T2]

= [LT-3]

pR. (P+ a/V2)(V-b) = RT vWfr vwl gYs smIkrn hY[ AwXwmI ivSlySx dI vrqoN nwl 'a" Aqy 'b' dy AwXwm pqw kro [

h`l : id`qI smIkrn hY :

(P+ a/V2)(V-b) = RT


P dy AwXwm = a/V2 dy AwXwm

25

[a/V2] = [P]

[a] = [P V2] = [ M1L-1T-2][L3]2

= [ M1L5T-2]

Dimensions of b = Dimensions of

V

[b] =[V] = [ L3]

[a/V2] = [P]

[a] = [P V2] = [ M1L-1T-2][L3]2

= [M1L5T-2]

b dy AwXwm = V dy AwXwm

[b] =[V] = [ L3]

26

APPLICATIONS OF

DIMENSIONAL EQUATIONS

1. Checking the

correctness of a physical

relation

2. Conversion of units from

one system of units into

another system

3. Derivation of equations

1. Checking the

correctness of a

physical relation

The method of dimensions

can be used to check the

correctness of a physical

relation or formula. For this

principle of homogeneity of

dimensions is used. If the

dimensions of each term in

the given relation are the

same, relation is correct. If not

same, the relation is not

correct.

AwXwmI smIkrnW dy aupXog

1. iksy BOiqk rwSI dI shIkrqw jWc krnw

2. iekweIAW dI ie`k pRxwlI qoN dUjI pRxwlI 'c qbdIl krnw

3. smIkrnW dw ivauNqpMn krnw

1. iksy BOiqk rwSI dI shIkrqw jWc krnw

AwXwmW dy FMg nUM iksy BOiqk smIkrn jW PwrmUly dI shI hox dI jWc leI vriqAw jw skdw hY[ies mksd leI AwXwmW dI iekswrqw dy inXm dI vrqoN kIqI jWdI hY[jy kr id`qI hoeI smIkrn jW PwrmUly iv`c hryk trm dy AwXwm ie`ko ijhy hox qW smIkrn shI hY[jy kr ie`ko ijhy nhIN hn qW smIkrn shI nhIN hovygI[

28

Example1. Check the

correctness of the relation v

= u + a t.

Dimensions of v = [M0 L1T-1]

Dimensions of u = [M0 L1T-1]

Dimensions of a = [M0 L1T-2]

Dimensions of t = [M0 L0T1]

Dimensions of a t = [M0 L1T-2]

x [M0 L0T1] = [M0 L1T-1]

Since the dimensions of each

term in the physical relation

are the same, the equation is

correct.

Example2. Check the

correctness of the relation s

= u t + ½ a t2.

Dimensions of s = [M0 L1T0]

Dimensions of ut = [M0 L1T-1]x

[M0 L0T1]= [M0 L1T0]

Dimensions of ½ a t2 =[M0

L1T-2]x [M0 L0T1]2 = [M0 L1T0]

Since the dimensions of each

term in the physical relation

are the same, the equation is

correct.

audwhrx 1. smIkrn v = u + a t dI shIhox dI jWc kro[

v dy AwXwm = [M0 L1T-1]

u dy AwXwm = [M0 L1T-1]

a dy AwXwm = [M0 L1T-2]

t dy AwXwm = [M0 L0T1]

a t dy AwXwm = [M0 L1T-2] x [M0 L0T1] = [M0 L1T-1]

ikauN ik BOiqk smIkrn iv`c hryk trm dy AwXwm ie`ko ijhy hn ies leI smIkrn shI hY[

audwhrx 2. smIkrn s = u t + ½ a t2 dI shIhox dI jWc kro[

s dy AwXwm = [M0 L1T0]

ut dy AwXwm = [M0 L1T-1] x [M0 L0T1]= [M0 L1T0]

½ a t2 dy AwXwm = [M0 L1T-2] x [M0 L0T1]2 = [M0 L1T0]

ikauN ik BOiqk smIkrn iv`c hryk trm dy AwXwm ie`ko ijhy hn ies leI smIkrn shI hY[

29

2. Conversion of units from

one system to another

Dimensional equations can be

used to convert units of a

physical quantity from one

system of units to another.

We know, a physical quantity

can be expressed as

Q = n U

In system of units I,

Q = n1 U1

In system of units II

Q = n2 U2

Hence, n2 U2 = n1 U1

Or n2 = n1 U1 / U2 ....(1)

Now, U1 = [M1xL1y T1z]

U2 = [M2xL2y T2z]

From (1)

n2 = n1 [M1xL1y T1z] / [M2xL2y

T2z]

= n1[M1xL1y T1z / M2xL2y T2z]

n2=n1[M1/M2]x[L1/L2]y[T1/T]z

3. ie`k pRxwlI qoN dUjI pRxwlI iv`c iekweIAW qbdIl krnw

AwXwmI smIkrnW nUM iksy BOiqk rwSI dIAW iekweIAW nUM ie`k pRxwlI qoN dUjI pRxwlI iv`c qbdIl krn leI vriqAw jw skdw hY[

AsIN jwxdy hW ik iksy BOiqk rwSI nUM ies qrWH iliKAw jw skdw hY-

Q = n U

iekweIAW dI pRxwlI 1 iv`c,

Q = n1 U1

iekweIAW dI pRxwlI 2 iv`c

Q = n2 U2

ies leI, n2 U2 = n1 U1

Or n2 = n1 U1 / U2 ....(1)

hux, U1 = [M1xL1y T1z]

U2 = [M2xL2y T2z]

(1) qoN

n2=n1 [M1xL1y T1z] / [M2xL2y T2z]

= n1[M1xL1y T1z / M2xL2y T2z]

n2=n1[M1/M2]x[L1/L2]y[T1/T]z

30

Example: Convert 1 newton

into dyne using method of

dimensions.

Solution: newton is SI unit of

force. dyne is unit of force in

CGS system.

n1 = 1

n2 =?

Dimensional formula of force

= [M1L1T-2]

Hence, x =1 , y= 1 , z = -2

SI CGS System

M1 = 1 kg M2 = 1 g

L1 = 1 m L2 = 1 cm

T1 = 1 s T2 = 1 s

n2= n1[M1/M2]x[L1/L2]y[T1/T]z

=1[1kg/1g]1[1m/1 cm]1[1s/1s]-2

=1[1000g/1g]1[100cm/1

cm]1[1s/1s]-2

= 1 x 1000 x 100 x1

= 100000 = 105

Hence, 1 newton = 105 dyne

audwhrx :AwXwmW dy qrIky dI vrqoN nwl 1 inaUtn nUM fweIn iv`c bdlo[

h`l: inaUtn bl dI AYs. AweI. iekweI hY[fweIn sI.jI.AYs. pRxwlI iv`c bl dI iekweI hY[

n1 = 1

n2 =?

bl dw AwXwmI sUqr hY

= [M1L1T-2]

ies leI, x =1 , y= 1 , z = -2

AYs AweI sIjIAYspRxwlI

M1 = 1 kg M2 = 1 g

L1 = 1 m L2 = 1 cm

T1 = 1 s T2 = 1 s

n2= n1[M1/M2]x[L1/L2]y[T1/T]z

=1[1kg/1g]1[1m/1 cm]1[1s/1s]-2

=1[1000g/1g]1[100cm/1

cm]1[1s/1s]-2

= 1 x 1000 x 100 x1

= 100000 = 105

ies leI, 1 inaUtn= 105 fweIn

31

Example: Convert 1 joule

into erg using method of

dimensions.

Solution: joule is SI unit of

work. erg is unit of work in

CGS system.

n1 = 1

n2 =?

Dimensional formula of work=

[M1L2T-2]

Hence, x =1 , y= 2 , z = -2

SI CGS System

M1 = 1 kg M2 = 1 g

L1 = 1 m L2 = 1 cm

T1 = 1 s T2 = 1 s

n2= n1[M1/M2]x[L1/L2]y[T1/T]z

=1[1kg/1g]1[1m/1 cm]2[1s/1s]-2

=1[1000g/1g]1[100cm/1

cm]2[1s/1s]-2

= 1 x 1000 x (100)2 x1

= 10000000 = 107

Hence, 1 joule = 107 erg

audwhrx :AwXwmW dy qrIky dI vrqoN nwl 1 inaUtn nUM fweIn iv`c bdlo[

h`l: jUl kMm dI AYs. AweI. iekweI hY[Arg sI.jI.AYs. pRxwlI iv`c kMm dI iekweI hY[

n1 = 1

n2 =?

kMm dw AwXwmI sUqr hY

= [M1L2T-2]

ies leI, x =1 , y= 2 , z = -2

AYs AweI sIjIAYspRxwlI

M1 = 1 kg M2 = 1 g

L1 = 1 m L2 = 1 cm

T1 = 1 s T2 = 1 s

n2= n1[M1/M2]x[L1/L2]y[T1/T]z

=1[1kg/1g]1[1m/1 cm]2[1s/1s]-2

=1[1000g/1g]1[100cm/1

cm]2[1s/1s]-2

= 1 x 1000 x (100)2 x1

= 10000000 = 107

ies leI, 1 jUl= 107 Arg

32

Example : Convert density

of 13.6 g/cm3 into kg/m3

using method of

dimensions.

Solution: g/cm3 is unit of

density in CGS system and

kg/m3 is SI unit of density.

n1 = 13.6

n2 =?

Dimensional formula of

density = [M1L-3T0]

Hence, x =1 , y= -3 , z = 0

CGSSystem SI

M1 = 1 g M2 = 1 kg

L1 = 1 cm L2 = 1 m

T1 = 1 s T2 = 1 s

n2= n1[M1/M2]x[L1/L2]y[T1/T]z

=13.6[1g/1kg]1[1cm/1m]-

3[1s/1s]0

=13.6[1g/1000g]1[1cm/100

cm]2[1s/1s]0

= 13.6 x (1/1000)1 x (1/100)-3

x1

audwhrx: AwXwmW dy qrIky dI vrqoN nwl 13.6 g/cm3 dI Gxqw nMU kg/m3 iv`c bdlo[

h`l : g/cm3 Gxqw dw sI.jI.AYs pRxwlI iv`c and kg/m3 AYs AweI iv`c iekweI hY[

n1 = 13.6

n2 =?

Gxqw dw AwXwm sUqr= [M1L-3T0]

ies leI, x =1 , y= -3 , z = 0

CGSSystem SI

M1 = 1 g M2 = 1 kg

L1 = 1 cm L2 = 1 m

T1 = 1 s T2 = 1 s

n2= n1[M1/M2]x[L1/L2]y[T1/T]z

=13.6[1g/1kg]1[1cm/1m]-

3[1s/1s]0

=13.6[1g/1000g]1[1cm/100

cm]2[1s/1s]0

= 13.6 x (1/1000)1 x (1/100)-3

x1

33

=13.6 (1/1000) x (100)3x1

= 13.6 x1/1000 x 1000000

=13.6 x1000 = 13600

Hence, 13.6 g/cm3 = 13600

kg/m3

=13.6 (1/1000) x (100)3x1

= 13.6 x1/1000 x 1000000

=13.6 x1000 = 13600

ies leI, 13.6 g/cm3 = 13600 kg/m3

34

III. Derivation of relations

The method of dimensions

can be used to derive the

relation for a quantity.

Example: Weight (W) of a

body depends on its mass

(m) and acceleration due to

gravity (g). Derive relation

for W.

Solution:

Suppose W α mx gy

W = k mx gy ……(1)

Dimensions of W =[M1L1T-2]

Dimensions of m = [M1L0T0]

Dimensions of g = [M0L1T-2]

Put these values in eq. (1),

[M1L1T-2] =[M1L0T0]x[M0L1T-2]y

=[MxL0T0][M0LyT-2y]

III. smIkrnW ivauNqpMn krnw

AwXwm dy qrIky nwl iksy rwSI leI smIkrn ivauNqpMn kIqI jw skdI hY[

audwhrx: iksy vsqU dw Bwr (W ) ies dy puMj (m) Aqy gurUqvI pRvyg (g) qy inrBr krdw hY[W leI sMbMD ivauNqpMn kro[

h`l :

mMn lE, W α mx gy

W = k mx gy ……(1)

W dy AwXwm=[M1L1T-2]

m dy AwXwm = [M1L0T0]

g dy AwXwm = [M0L1T-2]

ieh kImqW (1) iv`c Bro

[M1L1T-2] =[M1L0T0]x[M0L1T-2]y

=[MxL0T0][M0LyT-2y]

35

[M1L1T-2] =[MxLyT-2y] …(2)

Compare powers of M,L and

T on both sides of (2)

x = 1 ………(3)

y = 1 ………(4)

-2y = -2 …….(5)

Put values of x and y in eq.

(1),

W = k m1 g1

= kmg

Take k=1,

W = mg

This is the required equation.

Example2. Time period (T)

of a simple pendulum may

depend on mass (m) of its

bob, length (l) of the

pendulum and acceleration

due to gravity (g).Derive

relation for T.

Solution:

[M1L1T-2] =[MxLyT-2y] …(2)

(2) dIAW dono pwsy M,L Aqy T dIAW pwvrW dI qulnw kro

x = 1 ………(3)

y = 1 ………(4)

-2y = -2 …….(5)

(1) iv`c x Aqy y dIAW kImqW Bro

W = k m1 g1

= kmg

k=1 lE

W = mg

ieh loVINdI smIkrn hY.

audwhrx 2. ie`k swDwrn pYNfUlm dw foln kwl (T) ies dy bOb dy puMj (m), pYNfUlm dI lMbweI (l) Aqy gurUqvI pRvyg (g) 'qy inrBr kr skdw hY[T vwsqy sMbMD ivauNqpMn kro[

h`l :

36

Suppose mMn lE

T α mx ly gz T α mx ly gz

Or T = k mx ly gz ….(1) Or T = k mx ly gz ….(1)

Dimensions of T = [M0L0T1] T dy AwXwm = [M0L0T1]

Dimensions of m = [M1L0T0] m dy AwXwm = [M1L0T0]

Dimensions of l = [M0L1T0] l dy AwXwm = [M0L1T0]

Dimensions of g = [M0L1T-2] g dy AwXwm = [M0L1T-2]

Put these values in eq.(1) ieh kImqW smIkrn (1) iv`c Bro

[M0L0T1]=[M1L0T0]x[M0L1T0]y[M0L1T-2]z [M0L0T1]=[M1L0T0]x [M0L1T0]y [M0L1T-2]z

=[MxL0T0] [M0LyT0] [M0LzT-2z] =[MxL0T0] [M0LyT0] [M0LzT-2z]

[M0L0T1] =[MxLy+zT-2z] ……(2) [M0L0T1] =[MxLy+zT-2z] ……(2)

Compare powers of M, L and T on both sides M , L Aqy T dIAW pwvrW dI qulnw kro

x =0 ……..(3) x =0 ……..(3)

y + z = 0 ……(4) y + z = 0 ……(4)

-2z = 1 …….(5) -2z = 1 …….(5)

From (5), 2z =-1 or z = -1/2 (5) qoN 2z =-1 or z = -1/2

From (4) y -1/2 = 0 or y = ½ (4) qoN y -1/2 = 0 or y = ½

Put these values in eq. (1) ieh kImqW smIkrn (1) iv`c Bro

T = k m0 l1/2 g-1/2 T = k m0 l1/2 g-1/2

= k x 1 l1/2 / g1/2 = k x 1 l1/2 / g1/2

37

= k (l/g)1/2 = k (l/g)1/2

= k√l/g = k√l/g

Take k= 2 π Take k= 2 π

T = 2 π√l/g T = 2 π√l/g

38

LIMITATIONS OF

DIMENSIONAL ANALYSIS

1. The method of

dimensions does not

give the value of

dimensionless constant

(k). It has to be

calculated by some other

method.

2. The method of

dimensions cannot be

used to derive relations if

the factors on which

quantity depends are not

known.

3. The method of


used to derive relations if

the quantity depends on

more than three factors.

4. The method of


used to derive relations

which involve

dimensionless factors

like sin , cos, log etc.

AwXwmI ivSlySx dIAW KwmIAW

1. AwXwm dw qrIkw AwXwmhIx siQr AMk dy mùl bwry nhIN d`sdw[ies nUM iksy hor FMg nwl pqw krnw pYNdw hY[

2. AwXwm dw qrIkw PwrmUly ivauNqpMn krn leI nhIN vriqAw jw skdw jy kwrk ijnHW 'qy rwSI inrBr krdI hY, pqw nw hox[

3. AwXwm dw qrIkw PwrmUly ivauNqpMn krn leI nhIN vriqAw jw skdw jy rwSI iqMn qoN v`D kwrkW 'qy inrBr krdI hY[

4. AwXwm dw qrIkw PwrmUly ivauNqpMn krn leI nhIN vriqAw jw skdw ijnHW iv`c AwXwmhIx kwrk ijvyN sin , cos, log Awid hox[

39

ERRORS qrùtIAW

The difference between actual

value and measured value of

a physical quantity is called

error.

iksy BOiqk rwSI dI Asl mùl Aqy mwpy mu`l 'c Prk nUM qrùtI ikhw jWdw hY[

TYPES OF ERRORS

Errors are of three types:

1. Systematic Errors

2. Random Errors

3. Gross Errors

1. Systematic Errors:

Errors which are due to

known reasons (causes) and

which are always of same

sign or type are called

systematic errors.

(a) Instrumental Errors:

Errors which are due to some

defect in the measuring

instrument are called

instrumental errors.

(b) Environmental

Errors

qru`tIAW dIAW iksmW

qru`tIAW iqMn iksmW dIAW huMdIAW hn:

1. kRmbD qru`tIAW 2. byqrqIb qru`tIAW 3. grOs qru`tIAW

1. kRmb`D qru`tIAW

qru`tIAW ijhVIAW jwxU kwrnW kr ky Aqy hmySW ie`ko icMn /iksm dIAW huMdIAW hn, nUM kRmb`̀D qru`tIAW ikhw jWdw hY[

(a) XMqirk qru`tIAW: qru`tIAW jo mwpx vwly XMqr iv`c iksy nuks kwrn huMdIAW hn , nUM XMqirk qru`tIAW ikhw jWdw hY[

(A) vwqwvrnI qru`tIAW:

40

Errors which arise due to

changes in atmospheric

conditions like temperature,

pressure are called

environmental errors.

(c) Errors Due to

imperfection

Minimization of Systematic

Errors:

Because reasons of

systematic errors are known,

these errors can be minimized

by applying proper correction.

Random Errors:

The errors which are due to

unknown reasons and which

are not always of the same

sign or type are called random

errors.

Minimization of Random

Errors:

These errors can be

minimized by repeating the

experiment a number of times

and then taking arithmetic

mean of all the observations.

qru`tIAW jo vwqwvrnI hwlqW ijvyN qwpmwn, dbwA Awid kwrn hoNd iv`c AwauNdIAW hn ,nUM vwqwvrnI qru`tIAW ikhw jWdw hY[

(d) gYrsMpUrnqw kwrn qru`tIAW

kRmbD qru`tIAW nUM G`t krnw:

ikauN ik kRmb`D qru`tIAW dy kwrn pqw huMdy hn, ies leI iehnW qr`utIAW nUM Xog soD lgw ky G`t kIqw jw skdw hY[

byqrqIb qru`tIAW

qru`tIAW ijhVIAW nw jwxU kwrnW kr ky Aqy hmySW ie`ko icMn /iksm dIAW nhIN huMdIAW hn, nUM byqrqIb qru`tIAW ikhw jWdw hY[

byqrqIb qru`tIAW nUM G`t krnw

iehnW qru`tIAW nUM pRXog nUM keI vwr duhrw ky Aqy swrIAW pRyKxW dw gixqk AOsq lY ky G`t kIqw jw skdw hY[

41

Gross Errors

The errors which arise due to

human mistakes are called

gross errors.

Minimization of Gross

Errors:

These errors can be

minimized by doing the

experiment very carefully.

ABSOLUTE ERROR

The magnitude of the

difference between the true

value and measured value of

a physical quantity is called

absolute error.

This is denoted by | Δa |.

RELATIVE ERROR

The error obtained by dividing

absolute error by actual value

is called relative error.

Relative error = Δamean/ amean

PERCENTAGE ERROR

The error obtained by taking

percentage of relative error is

called percentage error.

grOs qru`tIAW

ieh qru`tIAW mnu`KI glqIAW kwrn hoNd iv`c AwauNdIAW hn[

grOs qru`tIAW nUM G`t krnw

ieh qru`tIAW pRXog nUM iDAwn nwl krn nwl G`t kIqIAW jw skdw hY[

inrpyK qru`tI

iksy BOiqk rwSI dI Asl mùl Aqy mwpy gey mu`l iv`c Prk dI mwqrw nUM inrpyK qru`tI ikhw jWdw hY[

ies nUM | Δa | nwl drswieAw jWdw hY[

swpyKI qru`tI

qru`tI jo inrpyK qru`tI nUM Asl mùl nwl Bwg krn 'qy pRwpq huMdI hY nUM swpyKI qru`tI ikhw jWdw hY[

swpyKI qru`tI = Δamean/ amean

pRqISq qru`tI

qru`tI jo swpyKI qru`tI dI pRqISqqw lYx nwl pRwpq huMdI hY nUM pRqISq qru`tI ikhw jWdw hY[

pRqISq qru`tI δa = (Δamean/amean) × 100%

42

Percentage error, δa = (Δamean/amean)×100%

SIGNIFICANT FIGURES

The digits which are used to express the result with proper

accuracy are called significant figures.

Rules for Significant Figures:

1. All non-zero digits are significant.

Example: The number of significant figures in 463298 is 6.

2. All zeros between nonzero digits are also significant.

Example: The number of significant figures in 6032 is 4.

3. The zeros after the decimal point and before non-zero digit

are not significant.

Example: The number of significant figures in 0.007 is 1.

4. The zeros to the right of decimal point and also right of

nonzero digits are significant.

Example: The number of significant figures in 0.2370 is 4.

5. Zeros after non zero digit are not significant.

The number of significant figures in 100000000 is 1.

43

mh`qvpUrx AMk

AMk, jo iksy irjlt nUM Xog Su`Dqw nwl pRgt krn leI vrqy jWdy hn, nUM mh`qvpUrx AMk ikhw jWdw hY[

mh`qvpUrx AMkW leI inXm

1. swry gYr-isPr AMk mh`qvpUrx huMdy hn[ audwhrx: 463298 iv`c 6 mh`qvpUrx AMk hn[

2. gYr-isPr AMkW ivckwr AwauNdIAW swrIAW isPrW vI mh`qvpUrx huMdIAW hn[ audwhrx: 6032 iv`c 4 mh`qvpUrx AMk hn[

3. dSmlv ibMdU dy s`jy pwsy pr gYr-isPr AMk dy K`by pwsy isPrW mh`qvpUrx nhIN huMdIAW hn[ audwhrx: 0.007 iv`c 1 mh`qvpUrx AMk hY[

4. dSmlv ibMdU dy s`jy pwsy Aqy gYr-isPr AMk dy s`jy pwsy isPrW mh`qvpUrx huMdIAW hn[ audwhrx: 0.2370 iv`c 4 mh`qvpUrx AMk hn[

5. gYr-isPr AMkW dy s`jy pwsy isPrW mh`qvpUrx nhIN huMdIAW[ audwhrx: 100000 iv`c 1 mh`qvpUrx AMk hY[

44

QUESTIONS FROM PREVIOUS

EXAMINATIONS

SECTION A

Fill in the blanks:

1. Dimensional formula of potential energy is......

[May'19]

2. S.I. unit of Planck's constant is .............

[May'19]

3. Dimensional formula of pressure is........... [Nov'18]

4. Dimensional formula of force is ............

[Nov'18]

5. The dimensional formula of energy is ....... [May'18]

6. Number of fundamental and base units in S.I. system is ....

[May'18]

7. Dimensional formula of Gravitation constant is ........

[May'17]

8. Energy and power have ......................dimensions.

[Nov'16]

9. SI unit of temperature is..............

[May'16]

10. 1 nanometer is equal to ...................

[May'15]

11. The dimensional formula of compressibility is .........

[Nov'14]

12. Number of significant figures in 0.345200 are ..........

[Nov'14]

13. A physical quantity can be completely measured if we know the

numerical value and its ...............

[Dec'14]

14. The dimensional formula of torque is ...........

[May'13]

15. The dimensional formula of kinetic energy is ..........

[Dec'12]

45

16. The dimensional formula of pressure is ...............

[May'11]

17. Plane angle and solid angle are the two ........... units in SI

system. [Dec'10]

18. If zero of vernier scale and main scale do not coincide the type

of error is called..........

[Dec'10]

State True or Force:

1. One Fermi is equal to 10-15 m. [May'18][May '19][May'13]

2. Arithmetic mean of all the absolute errors is called mean absolute

error. [Nov'18]

3. Dimensional formula of work and energy is not same. [Nov'18]

4. The equation v2 + u2 = 2as is dimensionally correct.

[Nov'16]

5. SI unit of pressure is dyne/cm2. [Nov'14]

6. One nanometre is equal to 10-9 m.

[Dec'14]

7. Young's modulus has the same dimension as pressure.

[May'12]

8. SI unit of compressibility is Nm-1.

[Dec'12]

9. SI is coherent systems of units. [May,11]

10. Dimensions of work and torque are same.

[Dec'10]

Multiple Choice questions:

1. The significant figures in 0.09 are

a) 1 b) 2 c) 3 d) 4 [May'19]

2. Dimensional formula of angular velocity is same as that of

a) Linear velocity b) acceleration c) frequency d) speed

[Nov '18]

3. The significant figures in 96.000 are

a) 3 b)4 c) 5 d) 6 [May'18]

46

4. How many significant figures are there in 40.00?

a) 1 b) 2 c) 3 d) 4 [Nov'16]

5. The significant figures in 0.07805 is

a) 3 b) 4 c) 5 d) 6 [May'16]

6. Which of the following is dimensionless constant/quantity?

a) Young's modulus b) Poisson ratio c) Bulk modulus d) Modulus

of rigidity [Dec'14]

7. The significant figures in 0.09806 is

a) 3 b) 4 c) 5 d)6 [May'13]

8. The formula P= (x2-b)/at relates power (P) distance (d) and time

(t) . The dimensional formula of 'a' is

a) [ML1T2] b) [M1L1T-2] c) [M1L0T2] d) [M1L2T-2] [May '12]

9. How many significant figures are there in 30.00?

a) 1 b) 2 c) 3 d) 4 [May'11]

10. The wavelength associated with a particle of mass 'm' and

moving with velocity v is given by λ = h/mv where h is Planck's

constant. The dimensional formula of 'h' is

a) ML2T b) M1L-1T-1 c) ML1T-1 d) ML2T-1

[Dec 11]

11. Dimensional formula for density is

a) [ML3T] b) [ML-3T0] C) [M3LT3] d) [MLT]

[Dec'10]

SECTION B

1. Convert 1 newton of force into dyne using dimensional

analysis. [May'19][May'18][Dec'10]

2. What is meant by error? Define systematic and random

error.

[May'19]

3. The Vander Waal gas equation is (P+ a/V2)(V-b) = RT. Find

the dimensions of 'a' and 'b' by using dimensional analysis.

[May'19]

4. Check the accuracy of the relation v2-u2=2as

[Nov'18]

5. Check the accuracy of the relation s= ut +1/2 at2.

[Nov'18]

47

6. What are significant figures? Give rules for finding significant

figures. [May'17]

7. Check the correctness of relation t = 2 π √l/g where l is

length and g is acceleration due to gravity.

[Nov'16]

8. State and prove principle of homogeneity of Dimensions.

[May'16]

9. Derive the expression for centripetal force (F) acting on a

body of mass (m) moving with velocity (v) in a circle of radius

(r) using dimensional analysis.

[Nov'14]

10. What is meant by error? Define systematic and random

errors. [Nov'14]

11. Write six pairs of physical quantities which have same

dimensions. [Dec'14]

12. Using method of dimensions, convert 1 joule into ergs.

[May'12]

13. What is a unit? Give different systems of units.

[Dec'12]

14. Check the correctness of the relation, v = u + at by the

method of dimensions.

[May'11]

15. What are limitations of dimensional analysis.

[Dec'10]

SECTION C

1. Check the correctness of relation t = 2 π√l/g where l is length

and g is acceleration due to gravity.

[May'19]

2. A body of mass m is moving with uniform speed v in a circle of

radius r. Find an expression for centripetal force F by method of

dimensions. [Nov'18]

3. a) Check the correctness of the relation t = 2 π √l/g where t is

time period of pendulum, l is length and g is acceleration due to

gravity.

b) Explain different systems of units. [May'18]

48

4. The density of mercury is 13.6 gm/cm3. Convert its value into

kg/m3. [Nov'14] [Dec '12]

5. a) The wavelength λ associated with a moving particles

depends upon its mass m, velocity v and Planck's constant h.

Show dimensionally that λ α h/mv .

c) The maximum error in the measurement of mass and length

are 3% and 2% respectively. Find the maximum error in the

measurement of density.

[Nov'16]

6. The wavelength of a moving electron depends upon its mass

m, its velocity v and Planck's constant h. Prove that λ α h/mv.

[May'16][Dec'10]

7. Define fundamental and derived units. [Dec'14]

8. What are significant figures? Give rules for finding significant

figures with examples.

[May'13]

9. a) Write five pairs of physical quantities which have same

dimensions.

b) Deduce the dimensions of gravitational constant. [Dec'13]

10. Define

(a) Instrumental error

(b) Error due to imperfect theory or formulae

(c) Random error. Give example of each type. How can the

above error be removed or minimised.

[May'12]

11. Prove that 1 joule = 107 ergs using dimensional equation.

[May'11]

12. The frequency of vibration (n) of a stretched string depends

upon the load applied T, length l, and its mass per unit length u.

Find dimensionally the formula of frequency. [Dec '11]

13. The Pressure P exerted by a liquid column depends on its

height h, density d and acceleration due to gravity g. Derive an

expression for P with the help of dimensions.

[Dec'10]

Ramgarhia Polytechnic - UNITS AND DIMENSIONS PHYSICS... · 2020. 4. 10. · 1 UNITS AND DIMENSIONS...

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Transcript of Ramgarhia Polytechnic - UNITS AND DIMENSIONS PHYSICS... · 2020. 4. 10. · 1 UNITS AND DIMENSIONS...