# Units and Measurement

Science: It came from Latin word "scientia" which meant knowledge, a knowing, expertness or experience. The knowledge gained through the systematic study of the structure and behaviour of the physical and natural world through observation and experiment is termed as Science.

Physics: It came from Greek word "Phusis" which meant "knowledge of nature" .The branch of science that deals with the study of natural phenomena is called physics.

Mechanics: The branch of applied mathematics dealing with motion and forces producing motion is known as mechanics.

UNITS AND MEASUREMENT

Introduction: The process of comparing an unknown physical quantity with a known fixed quantity is known as    Measurement.

Unit: The quantity used as a standard of measurement is called the unit.

Fundamental Units and Derived Units

Physical quantities

Two        types

Fundamental quantities              Derived quantities

The units for these quantities are called fundamental units and derived units.

#Fundamental units: The physical quantities which are independent and cannot be derived from other physical quantities are called fundamental quantities and their units are called fundamental or base units. Examples: mass, time, length, temperature, electric current, intensity of light and quantity of matter.

#Derived Units: The units of physical quantities which are dependent upon the fundamental units are called derived units. Example: Units of speed, force, momentum, velocity, etc.

Characteristics of standard unit

i)It should be well defined.

ii) It should be universally agreed.

iii) It should remain constant.

iv)It should be easily available and reproducible.

System of Units

MKS- system: The system in which the unit of length is measured in metre, unit of mass in kilogram and unit of time in second is known as MKS system.

CGS- system: The system in which the unit of length is measured in centimetre, mass in gram and time in second is known as CGS system.

FPS-system: The system in which the unit of length is measured in foot, mass in pound and time in second is known as FPS system.

SI- units: The system in which the seven basic and two supplementary units are taken as length in meter, mass in kilogram, time in second, temperature in Kelvin, electric current in ampere, luminous intensity in candela and quantity of matter in mole is known as SI units.

Units of Fundamental Quantities

 Fundamental Quantity Unit Symbol length metre m Mass kilogram kg Time second s Temperature Kelvin K Current ampere A Luminous intensity candela cd Amount of substance mole mol

Supplementary Units

 Quantity Unit Symbol Plane angle radian rad Solid angle steradian sr

Definitions of SI Base Units

Metre: The metre is defined as the length of the path travelled by light in a vacuum during a time interval of 1/299,792,458 of a second.

Second: The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium- 133 atoms.

Kilogram: The kilogram is the mass equal to the mass of standard platinum- iridium alloy cylinder(90% platinum and 10% iridium) kept at the International Bureau of Weights and Measures in Paris, France.

i) It is a decimal system.

ii) It is a rational system of units.

iii) It is a coherent system of units.

iv) It is highly useful in practical and theoretical work.

Dimension of a Physical Quantity

The dimension of the physical quantity is defined as the power to which the fundamental quantities are raised to express the physical quantity. The dimension of mass, length and time are represented as [M], [L] and [T] respectively. For example:

i.e. velocity =  DisplacementTime  =[L][T]  = [L][T = [L][T-1] = [M0LT-1]

Hence, the dimension of velocity are, zero in mass, 1 in length and -1 in time.

Dimensional Formula: The dimensional formula is defined as the expression of the physical quantity in terms of its basic unit with proper dimensions. For example, dimensional force is

i.e.  F= [MLT-2]

Dimensional equation: An equation containing physical quantities with dimensional formula is known as dimensional equation. For example:

Dimensional equation of v = u + at is

[M0LT-1] = [M0LT-1] + [M0LT-2][M0L0T-1]=[M0LT-1]

Dimensional Formula of some physical quantities

 S.No. physical quantity Relation with other phy.quantity Dimensional formula SI- unit 1. Area length x breadth [L] x[L]=[M0L2T0] m2 2. Volume length x breadth x height [L]x[L]x[L]=[M0L3T0] m3 3. Density massvolume M[L3] =[ML-3T0] kg m-3 4. Speed or velocity distancetime [L][T]  = [M0LT-1] ms-1 5. Acceleration velocitytime [LT-1][T] =[M0LT-2] ms-2 6. Momentum mass x velocity [M] x [LT-1]=[MLT-1] kg m s-1 7. Force mass x acceleration [M] X [LT-2]=[MLT-2] kg m s-2 8. pressure forcearea [MLT-2][L2]= [ML-1T-2] Nm-2 or pascal(pa) 9. Work force x distance [MLT-2] x [L]=[ML2T-2] J (joule) 10. Energy(mechanical, heat, light, etc) work [ML2T-2] J 11. Power worktime [ML2T-2][T]  =[ML2T-3] W(watt) 12. Gravitational constant(G) force x (distance)2(mass)2 [M-1L3T-2] Nm2kg-2 13. Impulse force x time [MLT-2] x [T]=[MLT-1] Ns 14. Surface tension forcelength [MLT-2][L] = [ML0T-2] Nm-1 15. Coefficient of viscosity forcearea x velocity gradient [ML-1T-1] daP 16. Angle arcradius Dimensionless rad 17. Moment of inertia mass x (distance)2 [ML2T0] kg m2 18. Angular momentum moment of inertia x angular velocity [ML2]x [T-1]= [ML2T-1] kg m2 s-1 19. Torque or couple force x perpendicular distance [MLT-2] x [L]=[ML2T-2] Nm 20. Frequency 1second [T-1] Hz

Principle of Homogeneity

According to this principle, the dimensions of fundamental quantities on left hand side of an equation must be equal to the dimensions of the fundamental quantities on the right hand side of that equation.

Let us consider three quantities A, B and C such that C= A + B. Therefore, according to the principle, the dimensions of C are equal to the dimensions of A and B.

Uses of Dimensional Equations

i) To derive the relation between various physical quantities.

ii) To convert value of physical quantity from one system of unit to another system.

iii) To check the correctness of a physical relation.

iv)To find the dimension of constants in a given relation.

i) To check the correctness of a physical relation: The correctness of a physical relation is checked by using the principle of homogeneity of dimension. According to this principle, if the dimension of M, L and T are same on both sides of an equation, then the relation is corect.

For example: Let an equation v2= u2 + 2as, whose correctness is to be checked dimensionally.

Dimensional equations of v, u, a and s are respectively given below:

v= [LT-1],     u= [LT-1],       a= [LT-2],   s=[L]

The above equation can be written in the dimensional form as

[LT-1]2 = [LT-1]2 + 2[LT-2] [L]

or, [L2T-2] = [L2T-2] + 2[L2T-2]

Here 2 is a dimensionless constant. Hence the dimensions of each term on RHS= dimension on LHS. so, the equation is dimensionally correct.

ii) To derive a relation between various physical quantities: This is possible by making use of principle of homogeneity of dimensions. This is applied in the following example.

To find the time period of a simple pendulum, its time period, t may depend upon (i) mass, m of the pendulum (ii) the length, l and (iii) acceleration due to gravity, g.

i.e.           t   ma

t   lb

gc

Combining these equations , we get

t   ma  lb gc

or,  t= k ma lb gc             ...............(i)

where k is a dimensionless constant of proportionality , and a, b and c are indices to be determined.

Writing down the dimensions on both sides of the equation, we get

[T] = [Ma] [Lb] [LT-2]C

or,   [M0L0T] = [MaLb+cT-2]

Comparing dimensions on both sides, we have

a= 0,        b=  12  ,        c= -1 2

Substituting the values of a, b and c in eqn. (i), we have

t= k m0 l1/2 g-1/2                  = k (lg )1/2

or, t= k lg             ...........(ii)

The value of k=2π  found by experiment, and the period of pendulum.

t= 2πlg

iii) To convert value of physical quantity from one system of units to another system: The method of dimensional analysis can be used to obtain the value of physical quantity in one system, when its value in another system is given.

As discussed earlier, the measure of a physical quantity is given by

X = n u

where, u is the unit and n is the numerical value of the physical quantity for the unit chosen. If u1 and u2 are units for measurement of the physical quantity in two systems and n1 and n2 are the numerical values of the physical quantity for the two systems, then

n1 u1 = n2u2

If a, b and c are the dimensions of the physical quantity in mass, lenght and time, then

n1[M1a L1bT1c]= n2[M2aL2bT2c]

where, M1,   L1,  T1 and M2 , L2  and T2 are the units of mass, length and time in the two systems. therefore,

n2 = n1[M1M2 ]a  [L1L2 ]b  [T1T2 ]c

This equation can be used to find the value of a physical quantity in the new system, when its value in the first system is known.

For example: Convert 1 joule into erg. Joule is the unit of work in SI- units and erg is the unit of work in CGS- system.

Given system (SI-units)                                                         New system (CGS- system)

n1 = 1                                                                                          n2= ?

M1= 1kg                                                                                                 M2= 1 g

L1 = 1m                                                                                       L2= 1 cm

T1= 1s                                                                                           T2= 1 s

Dimensional formula of work = [ML2T-2]. Comparing it with [Ma Lb Tc ], we get

a=1,                           b=2,                           c=-2

Now,            n2= n1 [M1M2] a [L1L2] b [T1T2 ]c

= 1[1kg1g] 1 [1m1cm] 2 [1s1s ]-2

=[1000g1g] 1 [100cm1cm] 2[1]-2

=(1000) (100)2

=107

1 joule= 107 erg.

iv)To determine the dimensions of a constant: Dimension of a constant appearing in a physical equation can be determined as follows:

Newton's law of gravitation gives the force between two masses m1 and m2 separated by a distance r as :

F= Gm1m2r2

where, G s the universal gravitational constant. The dimension of G is determined by using dimensional analysis.

Here,         G= Fr2m1m2

Now, we have

F= [MLT-2],              m1 = m2 = [M],    r= [L]

The dimension of G = dimension of Fr2m1m2  =MLT-2x[L]2M[M]  = [M-1] [L3][T-2]

Thus, G=[M-1 L3T-2]

Hence, the dimensions of G are -1 in mass, 3 in length and -2 in time.

Limitations of Dimensional Analysis

i) It gives no information whether a physical quantity is a scalar or vector,

ii) It does not give information about the dimensionless constant.

iii) We cannot derive the formulae containing trigonometric function, exponential functions, and logarthmic function.

iv) If a quantity depends on more than three factors having dimension, the formula cannot be derived.

Precision and Significant Figures

Measurement of any physical quantity is never absolutely correct. So precision is the degree of exactness and gives the limitation of the measuring instrument. The significant figures of the measure of a physical quantity are all those digits about which we are absolutely sure plus one digit that has a little doubt or uncertainty.

The rules for determing the numbers of significant figures:

a. All non- zero digits are significant. so 32.84 have four significant figures.

b. All zeros between two non- zero digits are significant. Thus, 200.03 gram have five significant figures.

c. All zeros to the right of the decimal point are significant. For example, 134cm, 152.0cm and 162.00cm have three, four and five significant figures respectively.

d. The number of significant figures of significant figures does not depend on the system of the units. For example  31.6cm, 0.172m and 0.00132km all have three significant figures.

Error in Measurement

i) Instrumental Errors

ii)Systematic Errors

iii) Personal Errors

iv) External Errors

v) Least count error

#SOLVED PROBLEMS (VVI FOR NEB EXAM)

1. Check the dimensional consistency of the following relation.

Fs= 12 mv2 - 12 mu2      .........(i)

where F= force, s= distance, m= mass, u = initial velocity, and v= final velocity.

solution:

The relation will be dimensionally consistent if the dimensions of each term are the same.

Dimension of F=[MLT-2]

Dimension of s= [L]

Dimension of u or v =[LT-1]

Dimension of m = [M]

since , the constant  12  has no dimensions, it will not enter into dimensional equation. Putting dimensions of various   physical quantities in eqn. (i), we get

[MLT-2][L] = [M] [LT-1]2 - [M] [LT-1]2

or, [ML2T-2] = [ML2T-2]- [ML2T-2]

or, [ML2T-2] = [ML2T-2]

Hence, the given relation is dimensionally correct.

2.  The density ρ  of the earth and radius R is given by

ρ=3g4πRG

where g is acceleration due to gravity of the earth and G is gravitational constant. Check dimensional consistency of this relation.

solution:

Dimension of L.H.S. = [M][L3]  = [ML-3]

Dimension of R.H.S. = [LT-2]L[M-1L3T-2]  = [ML-3]

Since, dimensions on both sides of the given equation are same, the relation is dimensionally correct.

3. The surface tension T is given by force per unit length. What is the dimension of surface tension?

solution:

we know, the surface tension of a liquid

T= Fl

Dimension of force, F= [MLT-2]

Dimension of length, l = [L]

So, dimension of surface tension, T =[MLT-2L ] = [MT-2]

4. covert density of water 1 g/cm3 (CGS- system) into kg/m3 (MKS - system)

solution:

we know, 1 gm/cm3 =1gm1cm3  =1/1,0001(100)3m3kg = 11,000 kg x (100)31m3  = 1,000,0001,000kg/m3

1 gm/cm3 = 1,000 kg/m3

5. The frequency n of vibration of a strecthed string is a function of its tension F, the length l and mass per unit length m. From the knowledge of dimension, prove that

n1lFm

Solution:

we can express the frequency of vibration as

n α  Fa lb mc

or, n= k Fa lb mc       ........(i)

where k is dimensionless constant and a, b and c are numbers which we want to find.

The dimension of frequency is [T-1], the dimension of F is [MLT-2], the dimension of length is [L] and that of mass    per unit length (m) is [ML-1]. Putting the dimensions of various physical quantities in eqn. (i), we get

[T-1] = [MLT-2]a [L]b [ML-1]c  = [MaLaT-2aLbMcL-c]

or,  [M0] [L0] [T-1] = [Ma+c] [La+b-c] [T-2a]

Equating the powers of [M], [L] and [T] on both sides, we get

a+c =0           .......(ii)

a+b-c = 0        ........(iii)

and        -2a = -1          .........(iv)

From Eq. (iv)

a= 12

putting this value in Eqn. (ii),

12  + c= 0

or, c= -12

so from Eq. (iii), we have

a + b -c =0

or, 12  + b + 12  =0

or, b +1 = 0

or, b= -1

Putting the values of a, b and c in Eq. (i), we get

n= k F1/2 l-1 m-1/2

= klFM

n1lFM

6. check the correctness of the formula t= 2πmk  , where t is the time period, m is the mass and k is the force per unit displacement.

solution:

we have,   t=2 πmk

Since, 2π  is  constant, in dimensional formula, the equation can be written as

t = MMT-2

= T2

= [T]

Hence the given equation, t = 2πmk  is dimensionally correct.

7. A student writes R2GM  for escape velocity of the earth. Check the correctness of the formula by using dimensional analysis.

solution:

The escape velocity, v= R2GM

In dimensional formula, this equation becomes

[LT-1] = LL3T-2  = 1L2T-2

[LT-1] = T2L2

[LT-1] = [TL-1]

which is not true. Hence given equation is dimensionally incorrect.

8. Write the dimensional formula of gravitational constant specific heat and specific latent heat.

solution:

we have, F = Gm1m2r2 , where G is gravitational constant

or, G= Fr2m1m2

G = Fr2m1m2  = M1L1T-1L2M1M1  = M-1L3T-2

Also, Q = msθ , where s is specific heat.

or,  S= Qmθ

S=M1L2T-2M1K1  = [M0L2T-2K-1]

Again, Q = mL, where L is latent heat

or, [latent heat] = Qm = M1L2T-2M1  = [L2T-2] = [M0L2T-2]

9. Taking force, length and time as fundamental quantities, find the dimensional formula for density.

solution:

let ρ  = [ FLb Tc ]        .......(i)

where a,  b and c are dimensions

or, [M1L-3T0] = [(M1L1T-2)aLbTc]

or, [M1L-3T0] = [MaLa+bTc-2a]

Equating indices we get,

a= 1

a+b = -3

or, b= -3-a = -3-1 = -4

c- 2a = 0

∴  c= 2a = 2. 1= 2

putting these values of a, b and c in (i)  we get,

[ρ]=[FL_4T2]

10. Check the correctness of the formulae t= 2πlg  using dimensional analysis, t is the time period of simple pendulum, l is the simple pendulum and g is the acceleration due to gravity.

solution:

Given formula is,

t= 2πlg

Dimension formula of L.H.S.[t] = [T]

Dimension formula of R.H.S [2πlg  ] = LL/T-2  = [T2 ] =[T]

Dimensions of each terms on either side are same. hence, the formula is dimensionally correct.

11. Name any two physical quantities which have the same dimensions. Can a quantity have unit but no dimension? Explain.

Ans: 1st part: The two physical quantities which have same dimensions are work and energy.

we have

work = Force x displacement

i.e. W = F. S

∴  [W] =[MLT-2] x [L]

= [ML2T-2]

K.E. = 12mv2

K.E.=[ 12 mv2] =[ML2T-2]

2nd part: Yes, a quantity may have unit but no dimension e.g. plane angle. The plane angle has no dimension but its S.I. unit is radian.

plane angle = arc lengthradius

i.e. θ= 1r

θ  = LT  = [L0]

∴The plane angle is dimensionless but its S.I.unit is radian.

12. If y = a + bt+ ct2 , where y is the distance and t is the time. What is the dimension and unit of c?

Ans:

Here, y= a + bt+ ct2

From the principle of homogeneity of dimensions, the dimensions of each and every term on either sides of a   correct physical relations are same.

i.e. [y] = [ct2]

or, [c] = yt2

c=LT-2

∴Dimensions of c is LT-2and its SI unit is ms-2.

13. A student writes an expression of the force causing a body of mass (m) to move in a circular motion with a velocity (v) as F= mv2 use the dimensional method to check its correctness.

Ans:

Given, F= mv2

Dimension of F = [MLT-2]

Dimension of m =[M]

Dimension of v= [LT-1]

∴Dimension of RHS=mv2  = [ML2T-2]

Since, dimension of LHS is not equal to dimension of RHS. The given equation is incorrect.

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