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UNIT & DIMENSION

Preface

IIT - JEE Syllabus : Unit & DimensionIIT - JEE Syllabus : Unit & DimensionIIT - JEE Syllabus : Unit & Dimension IIT - JEE Syllabus : Unit & DimensionIIT - JEE Syllabus : Unit & Dimension Unit & Dimensions, Dimensional analysis, Least count, Significant figure, Methods of measurement and Error analysis for physical quantities.

Fundamental concepts of the Physics start from this chapter. Basically the terms & concepts which are illustrated in this topic will be used in so many ways because all Physical quantities have units. It is must to measure all Physical quantities so that we can use them. In this chapter we will have an over view of different units of different Physical quantities. We will learn the dimension and dependence of the unit of any Physical quantity on fundamental quantities or unit. Entire topic is illustrated very systematically with respective examples so that the students can understand the fundamentals very easily & quickly. Students are advised to read every point of supplementary very carefully which is given at the end of the topic. Generally, students are not able to find out the Dimension of unseen or new quantity as their basic concepts are not clear & then they read the dimensions like a parrot. It should be avoided & they should develope themselves, so that they can find out the dimensions of any given quantity.

Total number of Questions in Units & Dimension are :

In chapter Examples .......................................... 27

UNIT & DIMENSION

2.1.2.6 Mole :

The amount of a substance that contains as many elementary entities (Molecules or atoms if the substance is monoatomic) as there are number of atoms in .012 kg of carbon - 12 is called a mole. This number (number of atoms in 0.012 kg of carbon-12) is called Avogadro constant and its best value available is 6.022045 x 10 23.

2.1.2.7 Candela:

The S.I. unit of luminous intensity is 1cd which is the luminous intensity of a blackbody of

surface area

600 000, m^

(^2) placed at the

temperature of freezing platinum and at a pressure of 101,325 N/m 2 , in the direction perpendicular to its surface.

Examples based on Definition of fundamental Units

Ex.2 A man seeing a lighting starts counting seconds until he hears thunder. He then claims to have found an approximate but simple rule that if the count of second is divided by an integer, the result directly gives in km, the distance of the lighting source. What is the integer if the velocity of sound is 330 m/s Sol. If n is the integer then according to the

problem

t in s n

= dist in km.

t in s n = (v) t

n =

v

330 x 10 −^3 = 3

Ex.3 In defining the standard of length we have to specify the temperature at which the measurement should be made. Are we justified in calling length a fundamental quantity if another physical. quantity, temperature, has to be specified in choosing a standard. Sol. Yes, length is a fundamental quantity. One metre is the distance that contains 1650 763.73 wavelength of orange-red light of Kr - 86. Hence, the standard metre is

independent of temperature. But the length of object varies with temperature and is given by the relation. L (^) t = L 0 (1 + αt) ∴ We usually specify the temperature at which measurement is made. Ex.4 Which of the following sets cannot enter into the list of fundamental quantities in any system of units (A) length ; mass ; velocity (B) length ; time ; velocity (C) mass ; time; velocity (D) length ; time, mass

Sol.[B] Since velocity = length time i.e. in this set a quantity is dependent on the other two quantities Where as fundamental quantities are independent. 2.2 Derived units : Units of derived quantities are called derived units. Physical quantity units ΙΙ ΙΙΙ llustration^ Volume = (length)^3 m^3 Speed = length/time m/s 2.3 Supplementary units : The units defined for the supplementary quantities namely plane angle and solid angle are called the supplementary units. The unit for plane angle is rad and the unit for the solid angle is steradian. Note : The supplemental quantities have only units but no dimensions (will be discussed later)

3. PRINCIPAL SYSTEM OF UNITS 3.1 C.G.S. system [centimetre (cm) ; gram (g) and second (s)] 3.2 F.P.S system [foot ; pound ; second] 3.3 M.K.S. system [meter ; kilogram ; second] 3.4 S.I. (system of international) In 1971 the international Bureau of weight and measures held its meeting and decided a system of units. Which is known as the international system of units.

Examples based on Units

Ex.5 The acceleration due to gravity is 9.80 m/s 2. What is its value in ft/s 2? Sol. Because 1 m = 3.28 ft, therefore

9.80 m/s 2 = 9.80 × 3.28 ft/s 2 = 32.14 ft/s 2 Ex.6 A cheap wrist watch loses time at the rate of 8.5 second a day. How much time will the watch be off at the end of a month?

Sol. Time delay = 8.5 s/day

= 8.5 × 30 s/ (30 day) = 255 s/month = 4.25 min/month.

5. DIMENSIONAL FORMULA

It is an expression which shows how and which of the fundamental units are required to represent the unit of physical quantity. Different quantities with units. symbol and dimensional formula,

Quantity Symbol Formula S.I. Unit D.F. Displacement s — Metre or m M^0 LT^0 Area A l × b (Metre) 2 or m 2 M^0 L^2 T^0 Volume V l × b × h (Metre) 3 or m 3 M^0 L^3 T^0

Velocity v v = (^) t

s ∆

m/s M^0 LT–

Momentum p p = mv kgm/s MLT–

Acceleration a a = t

v ∆

m/s 2 M^0 LT–

Force F F = ma Newton or N MLT– Impulse – F × t N.sec MLT– Work W F. d N.m ML^2 T–

Energy KE or U K.E. =

mv 2 Joule or J ML^2 T–

P.E. = mgh

Power P P = (^) t

W

watt or W ML^2 T–

Density d d = mass/volume kg/m 3 ML–3T^0

4. DIMENSIONS

Dimensions of a physical quantity are the powers to which the fundamental quantities must be raised to represent the given physical quantity. ΙΙ ΙΙΙ llustration : Force (Quantity) = mass × acceleration

= mass × velocity time = mass ×

length ( time) 2 = mass × length × (time) – So dimensions of force : 1 in mass 1 in length –2 in time and Dimensional formula : [MLT –2^ ]

Universal gas constant R PV = nRT (^) mol.K

Joule ML^2 T–2θ– Mechanical equivalent J W = JH — M^0 L^0 T^0 of heat

Charge Q or q I = t

Q

Coulomb or C M^0 L^0 TA

Current I — Ampere or A M^0 L^0 T^0 A

Electric permittivity ε 0 F =

4 πε 0.^2

12 r

q q ( .) .

coul N m

2 2 or^2

2 N m

C

M–1^ L–3A^2 T^4

Electric Potential V V = (^) q

∆W

Joule/coul ML^2 T–3^ A–

Intensity of electric field E E =

F

q

N/coul. MLT–3A–

Capacitance C Q = CV Farad M–1^ L–2T^4 A^2

or relative permittivity εr εr =

ε ε 0 —^ M

0 L 0 T 0

Resistance R V = IR Ohm ML^2 T–3^ A–

Conductance S S =

R Mho^ M

–1 L –2 T –3 A 2

Specific resistance ρ ρ =

RA

l Ohm × meter ML^3 T–3^ A–

or resistivity

Conductivity or σ σ =

ρ Mho/meter^ M

–1 L–3T 3 A 2

specific conductance

Magnetic induction B F = qvBsinθ Tesla or weber/m 2 MT–2^ A– or F = BIL

Magnetic flux φ e =

d dt

φ Weber ML 2 T–2A– Magnetic intensity H B = μ H A/m M^0 L–1T^0 A

Magnetic permeability of free space or medium μ 0 B =

μ π

0 4

r^2

I dlsin θ N amp 2 MLT

–2A–

Coefficient of self or Mutual inductance L e = L. (^) dt

dI Henery ML^2 T–2^ A–

Electric dipole moment p p = q × 2l C.m. M^0 LTA

Magnetic dipole moment M M = NIA amp.m^2 M^0 L^2 AT^0

Dielectric constant

Examples based on Dimensions

Ex.7 (a) Can there be a physical quantity which has no unit and dimensions (b) Can a physical quantity have unit without having dimensions Sol. (a) Yes, strain (b) Yes, angle with units radians Ex.8 Fill in the blanks (i) Three physical quantities which have same dimensions are ............................ (ii) Mention a scalar and a vector physical quantities having same dimensions ........................... Sol. (i) Work, energy, torque (ii) Work, torque Ex.9 Choose the correct statement (s) (A)all quantities may be represented dimensionally in terms of the base quantities (B) all base quantity cannot be represented dimensionally in terms of the rest of the base quantities (C) the dimension of a base quantity in other base quantities is always zero. (D) the dimension of a derived quantity is never zero in any base quantity. Sol. [A,B,C] (B) all the fundamental base quantities are independent of any other quantity (C) same as above

Ex.10 If velocity (V), time (T) and force (F) were chosen as basic quantities, find the dimensions of mass. Sol. Dimensionally : Force = mass × acceleration

Force = mass ×

veloc time

Mass = (^) velocity

Force ×time

mass = FTV – Ex.11 In a particular system, the unit of length, mass and time are chosen to be 10cm, 10gm and 0.1s respectively. The unit of force in this system will be equivalent to

(A)

N (B) 1N

(C) 10N (D) 100 N

Sol. Dimensionally F = MLT – In C.G.S system 1 dyne = 1g 1 cm (1s)– In new system 1x = (10g) (10 cm) (0.1s) –

1 x

1 dyne = (^10) g

1 g × (^10) cm

1 cm 2 1 s

10 s (^) − 

1 dyne =

10 000, × 1x 10 4 dyne = 1x 10 x = 10 5 dyne = 1 N

x =

10 N

Ex.12 If the units of length and force are increased four times, then the unit of energy will (A) increase 8 times (B) increase 16 times (C) decreases 16 times (D) increase 4 times Sol. Dimensionally E = ML 2 T – E = (MLT –2^ ) (L) E' = (4) (MLT –2^ ) (4L) E' = 16 (ML 2 T –2^ ) Note : 5.1 Two physical quantities having same dimensions can be added or subtracted but there is no such restriction in division and multiplication. (Principle of homogeneity) ΙΙ ΙΙΙ llustration : Using the theory of dimensions, determine the dimensions of constants ‘a’ and ‘b’

in Vander Wall’s equation. (^)  

V

P a (V – b) = RT

Sol.

a V 2 must have the same dimensions as that of

P (because it is added to P) Dimension of b must be same as that of V.

L^6

[a ] = ML –1^ T – [a] = ML 5 T – [b] = L 3

The dimensional formula for work is (M 1 L 2 T –2^ )

n 2 = n (^1)

M

M

1 2

1

L

N

M

O

Q

P

L

L

1 2

2

L

N

M

O

Q

P

T

T

1 2

2

L

N

M

O

Q

P

−

n 1 = 5 J, n 2 = 1 Substituting values we get

M

M

1 2

L^1

N

M

O

Q

P

L

L

1 2

L^2

N

M

O

Q

P

T

T

1 2

L^2

N

M

O

Q

P

− ...(2)

Similarly the dimensional formula for velocity is (Mº L 1 T –1^ ). Hence, conversion formula for velocity is

n 2 = n (^1)

M

M

1 2

0

L

N

M

O

Q

P

L

L

1 2

1

L

N

M

O

Q

P

T

T

1 2

1

L

N

M

O

Q

P

−

Here n 1 = 0.5 ms–1^ , n 2 = 1, Substituting values we get

1= 0.

L

L

1 2

L^1

N

M

O

Q

P

T

T

1 2

L^1

N

M

O

Q

P

− ...(3)

Dividing (2) by (1),1 =

L

L

1 2

L

N

M

O

Q

P ,

L 2 =

L 1

m = 0.5 m

Substituting value of

L

L

1 2

L

N

M

O

Q

P in (3), we get

1= 0.5 × 2

T

T

1 2

1

L

N

M

O

Q

P

− ,

T

T

1 2 = 1, T 2 = 1s

Substituting value of

L

L

1 2

L

N

M

O

Q

P and

T

T

1 2

L

N

M

O

Q

P in (1)

M

M

1 2

L

N

M

O

Q

P × 2 × 1,

M

M

1 2

L

N

M

O

Q

P ,

M 2 = 20 M 1 as M 1 = 1kg, M 2 = 20 kg. Hence units of mass, length and time are 20 kg, 0.5 m and 1 sec respectively 6.2.2 Conversion of a quantity from a given system to new hypothetical system

Examples Ex.18 The density of a substance is 8 g/cm 3. Now we have a new system in which unit of length is 5cm and unit of mass 20g. Find the density in this new system Sol. In the new system ; Let the symbol of unit of length be La and mass be Ma.

Since 5cm = 1 La ⇒ 1cm =

La

20g = 1Ma ⇒ 1g =

Ma

D = 8 g/cm 3 = (^3)

5

1 La

Ma 20

 ×

D = 50 Ma/(La) 3 = 50 units in the new system 6.3 To check the dimensional correctness of a given relation Examples based on Dimensional correctness Ex.19 Find the correct relation

F =

mv r

2 2

or mv r

2

Sol. Checking the dimensionally correctness of relation

F =

mv r

2 2 L.H.S. = MLT –

R.H.S. =

M LT

L

2 = ML^

0 T –2 ; LHS ≠ RHS

F =

Mv r

2

LHS = MLT –

RHS = M LT

L

( −1 2) = MLT –2 ; LHS = RHS

Hence dimensionally second relation is correct Limitation : It is not necessary that every dimensionally correct relation, physically may be correct 6.4 As a research tool to derive new relation

Examples based on Deriving new relation Ex.20 To derive the Einstein mass - energy relation Sol. E = f ( m , c) E = k Mx^ C y ML 2 T –2^ = M x^ (LT –1^ ) y ML 2 T –2^ = Mx^ L y^ T –y Comparing the coefficients x = 1 ; y = + 2 Through experiments ; k = 1 ∴ E = mc 2 Ex.21 When a small sphere moves at low speed through a fluid, the viscous force F opposing the motion, is found experimentaly to depend on the radius ‘r’, the velocity v of the sphere and the viscosity η of the fluid. Find the force F (Stoke’s law) Sol. F = f (η ; r ; v) F = k. η. r. v MLT –2^ = (ML –1^ T –1^ ) x^ (L) y^ (LT –1^ ) z MLT –2^ = M x^ L –x + y + z^ T –x–z comparing coefficients x = 1 , –x + y + z = 1 ; – x – z = – 2 x = y = z = 1 F = kηvr F = 6πηvr As through experiments : k = 6π

Ex.22 A gas bubble from an explosion under water, oscillates with a period T proportional to p a^ d b^ E c^ where p is the static pressure, d is the density of water and E is the total energy of explosion. Find the values of a,b, and c.

Sol. a = –

; b =

and c =

7. LIMITATIONS OF THE APPLICATION OF

DIMENSIONAL ANALYSIS

7.1 If the dimensions are given, then the physical quantity may not be unique as many physical quantities can have same dimensions. Examples based on Limitation of dimensional analysis Ex.23 If there is a physical quantity whose dimensional formula is [ML 2 T –2^ ]. Determine the physical quantity. Sol. It may be torque, work or energy.

7.2 Since numerical constant have no dimensions.

Such as

, 1, 6π etc, hence these can’t be deduced by the methods of dimensions. 7.3 The method of dimensions cannot be used to derive relations other than product of power functions.

ΙΙ ΙΙΙ llustration : S = ut +

2 at^

(^2) , y = asinωt

Note : However the dimensional correctness of these can be checked 7.4 The method of dimensions cannot be applied to derive formula if in mechanics a physical quantity depends on more than three physical quantities. As then there will be less number (=3) of equations than the unknowns. However the dimensional correctness of the equation can be checked

ΙΙ ΙΙΙ llustration : T = 2π (^) mgl

I

cannot be derived by theory of dimensions. 7.5 Even if a physical quantity depends on three physical quantities, out of which two have same dimensions, the formula cannot be derived by theory of dimensions. ΙΙ ΙΙΙ llustration : Formula of the frequency of a

tunning fork f =

d L^2

 (^) v

Note : However the dimensional correctness can be checked.

8. SIGNIFICANT DIGITS 8.1 Normally decimal is used after first digit using powers of ten,

ΙΙ ΙΙΙ llustration : 3750 m will be written as 3.750 x10^3 m

8.2 The order of a physical quantity is expressed in power of 10 and is taken to be 1 if ≤ (10)1/2^ = 3. and 10 if > 3. ΙΙ ΙΙΙ llustration : speed of light = 3 x 10^8 , order = 10 8 Mass of electron = 9.1 x 10 –31^ , order = 10 – 8.3 Significant digits : In a multiplication or division of two or more quantities, the number of significant digits in the answer is equal to the number of significant digits in the quantity which has the minimum number of significant digit

9.3 Propagation of error (Addition and Subtraction) : Let error in x is ± ∆x, and error in y is ± ∆y, then the error in x + y or x – y is ± (∆x + ∆y). The errors add. 9.4 Multiplication and Division : Let errors in x, y, z are respectively ± ∆x, ± ∆y and ± ∆z. Then error in a quantity f (defined as)

f =

c

a b z

x y

is obtained from the relation

f

∆f = | a |

x

∆x

• | b | (^) y

∆y

• | c | (^) z

∆z

. The

fraction errors (with proper multiples of exponents) add. The error in f is ± ∆f. 9.5 Important Points : 9.5.1 When two quantities are added or subtracted the absolute error in the result is the sum of the absolute error in the quantities. 9.5.2 When two quantities are multiplied or divided, the fractional error in the result is the sum of the fractional error in the quantities to be multiplied or to be divided. 9.5.3 If the same quantity x is multiplied together n times (i.e. x n), then the fractional error in x n^ is n times the fractional error in x,

i.e. ± n x

∆x

Examples based on

Errors Ex.27 In an experiment to determine acceleration due to gravity by simple pendulum, a student commit positive error in the measurement of length and 3% negative error in the measurement of time period. The percentage error in the value of g will be- (A) 7% (B) 10% (C) 4% (D) 3%

Sol. We know T = k

g

l

∴ T 2 = k' (^)  

g

l ⇒ g = k' (^2) T

l

g

∆g × 100 =

∆l l

× 100 +

T

2 ∆T

× 100

= 1% + 2 x 3% = 7% Hence correct answer is (A)

10. MEASURMENTS OF LENGTH, MASS & TIME

10.1 Distance of a hill : To find the distance of a hill, a gun is fired towards the hill and the time interval t between the instant of firing the gun and the instant of hearing the echo of the gun is determined. Clearly, during this time interval sound first travels towards the hill from the place of firing and then back from the hill to the place of firing. If v be the velocity of sound, and s the distance of hill from the place of firing , then 2 s = v × t

or s = 2

v ×t

10.2 Distance of moon : A laser beam is a source of very intense, monochromatic and unidirectional beam. By sending a laser beam towards the moon instead of sound waves, the echo method becomes useful in finding the distance of moon from the earth. If t is the total time taken by laser beam in going towards moon and back, then distance of moon

from the earth’s surface is given by : S = (^2)

c ×t

Where c = 3 × 10 8 m/s ; is the velocity of laser beam. 10.3 Thickness of matter sheet : For finding the thickness of some matter sheet, a signal from point A on the front surface of sheet is sent to the back surface. The signal gets reflected from point C on the back surface and is again received back at point B on the front surface. If the time interval between the instants of sending the signal from point A and receiving the signal back at B, is t, then thickness of sheet

S = 2

C × t (where C is the velocity of signal)

10.4 Distance of submerged objects or submarines in sea (Sound navigation and ranging - Sonar) Sonar in a instrument which uses ultrasonic waves (waves having frequency > 20,000 Hz) to detect and locate the submerged objects, submarines etc. in sea. Ultrasonic waves produced from a transmitter are sent towards the

distant objects under water. When the object comes in the direction of ultrasonic waves, then the waves are reflected back from it. Measuring the time interval t between the instants the ultrasonic waves are sent and received back, the distance S of the object can be calculated by the relation.

S = 2

C × t (where C is the velocity of ultrasonic waves) 10.5 Distance of aeroplane (Radio detection and Ranging - Radar). Radar is an instrument which uses. radiowaves for detecting and locating an aeroplane. Radiowaves produced by a transmitter at the radar station, are sent towards the aeroplane in space. These waves are reflected from the aeroplane. The reflected waves are received by a receiver at the radar station. By noting the time interval between the instants of transmission of waves and their detection, distance of aeroplane can be measured. If t is the required time interval and C the velocity of light (=equal to velocity of radio waves) then distance of aeroplane

S = 2

C ×t

10.6 Triangulation method : This method uses the geometry of the triangle and is useful for measuring the heights in following cases 10.6. Height of a tower or height of an accessible object.

tanθ =

AB

BC

h x h = x tanθ 10.6. Height of a mountain or height of an accessible object :

This method is useful in cases when it becomes impossible to measure the distance between the object and the observation point.

tanθ 1 =

AB

BC

h BC BC = h cot θ 1 Similarly tan θ 2 =

AB

BD

BD = h cot θ 2 BD – BC = x = h (cot θ 2 – cot θ 1 ) 10.7 Parallax method : Parallax (Definition) : When we observe the object P by closing our right and left eye alternately, we observe a shift in the position of object w.r.t the background. This is known as parallax.

θ = x

LR

b x (assuming distance LR as a circular arc of radius x)

x = b θ 10.7. Determination of distance of moon from earth. θ = θ 1 + θ 2

Because θ =

P PP

PM

1 2

Hence PM =

P PP 1 2

θ As astronomical bodies are at very large distances from earth

θ 1

θ 1

θ 2

θ (^2)

P 2 P 1

M

O P

P 1 PP (^2)

P 1 P 2 PM

MO

OM =

P P1 2

θ

17. Significant figures indicate the precision of measurement which depend on least count of the measuring instruments. 18. So far as significant figures are concerned, in mathematical operations like addition and subtraction, the result would be correct upto minimum number of decimal places in any of the quantities involved. However, in multiplication and division, number of significant figures in the result will be limited corresponding to the minimum number of significant figures in any of the quantities involved. To represent the result to a correct number of significant figures, we round off as per the rules already stated. 19. Whenever two measured quantities are multiplied or divided, the maximum possible relative or percentage error in the computed result is equal to the sum of relative or percentage errors in the observed quantities. Therefore maximum possible error in

Z = (^) p

m n C

A B

is :

Z

∆Z ×

= C^100

100 p C B

100 n B A

m ∆A× + ∆ × + ×∆ ×