Physics class11 topic no. 2

1. Units and
Measurements
…….Vidya Gaude

2. A B
Volume of A is less than B
Compare - measurement
Volume of B is greater than A
 Length
 Breadth
 Height
 Size (Volume)
PHYSICAL
QUANTITES

3. Ashok’s weight is 81.
Ram is 20 taller than Akash.
Aman studies for 3.
UNITS
UNITS
UNITS

4. Ashok’s weight is 81kg.
Ram is 20cm taller than Akash.
Aman studies for 3h.
UNITS
UNITS
UNITS
Measurement = Numerical value Unit

5. Measurement
The comparison of any physical quantity with its
same kind is called measurement.
Units
A definite amount of a physical quantity is taken as its
standard unit.

6. Physical Quantities
All the quantities in terms of which laws of physics are
described, and whose measurement is necessary are
called physical quantities.
 Distance/length
 Time
 Mass
 Speed
 Velocity
 Force
 Acceleration
 Power
 Pressure
 Work
 Energy
 Momentum
 Temperature
 Electric current, etc

7. Physical quantities
Length, mass, time, speed, velocity, force, power, etc
Fundamental quantities
Those physical quantities which are
independent to each other.
Derived quantities
Those physical quantities which
are derived from fundamental
quantities
Length, mass, time,
temperature, etc
Speed, velocity, acceleration,
force, momentum, power, etc
Speed =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒
Acceleration =
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑡𝑖𝑚𝑒
Force= 𝑚𝑎𝑠𝑠 𝑥 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛

8. Physical unit
Fundamental units
Those units which are
independent to each other.
Derived units
Those units which are
derived from other units.
Length- cm or m or km
Mass- g or kg
Time- s or h
Speed and velocity- cm/s or m/s
force = kg m/s or N

9. Systems of Units
Is the complete set of units, both fundamental and derived,
for all kinds of physical quantities.
1. cgs system :
the unit of length is centimetre,
the unit of mass is gram and
the unit of time is second.
2. mks system :
the unit of length is metre,
the unit of mass is kilogram and
the unit of time is second.
3. fps system :
the unit of length is foot,
the unit of mass is pound and
the unit of time is second.
4. SI system(Systamme internationale) :
This system contains seven
fundamental units and two
supplementary units.

10. Fundamental Quantities (base quantities) and its Units
Sr.no. Fundamental
Quantities
SI unit Symbol
1 Length ( width,
height, depth,
distance)
metre m
2 Mass kilogram kg
3 Time second s
4 Temperature kelvin K
5 Electric current ampere A
6 Amount of
substance
mole mol
7 Luminous intensity candela cd

11. Supplementary Quantities and its Units
Sr.no. Supplementary
Quantities
SI unit Symbol
1 Plane angle radian rad
2 Solid angle steradian sr
Plane angle Solid angle

12. Abbreviations in power of 10
Multiple Prefix Symbol
10 deca da
102 hecto h
103 kilo k
106 mega M
109 giga G
1012 tera T
1015 peta P
Multiple Prefix Symbol
10-1 deci d
10-2 centi c
10-3 milli m
10-6 micro  (mu)
10-9 nano n
10-12 pico p
10-15 femto f

13. Least value can measured on a scale
(Least Count): 1mm = 0.1cm
Least Count (LC)
Least value that can be measured on any measuring instrument.

14. Meter scale
Least value can measured : 0.1cm
Vernier Calliper
Least value can measured : 0.01cm
Screw Gauge
Least value can measured : 0.001cm
Diameter?
Length?
Diameter?
Direct methods for the measurement of length

15. Indirect methods for the measurement of length
Measurement of Large Distances
 The distance of a planet or a star from the earth
 Cannot be measured directly with a metre scale
 An important method in such cases is the parallax method.
Parallax:
Is a displacement or difference in the apparent position of
an object viewed along two different lines of sight.
O
A B
 The distance between the two points of
observation is called the basis(b).
 θ - is called the parallax angle or parallactic
angle.
 D - the distance of a far away object O.
b
θ
𝜃 =
𝐿𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑐
𝑟𝑎𝑑𝑖𝑢𝑠
𝜃 =
𝑏
𝐷
D

16. Application of parallax method
1. To measure the distance ( D) of a far away planet.
𝜃 =
𝑏
𝐷
𝐷 =
𝑏
𝜃
2. To determine the size or diameter (d) of the planet.
𝜃 =
𝑑
𝐷
d = 𝐷 𝜃
d
Earth
D
𝜃
A B

17. Measurement of length:
 1 micrometre (m) = 10−6m
 1 angstrom ( A ) = 10−10
m
1 fermi (fm) = 10−15 m
1 astronomical unit (AU) = 1.49 x 1011
m
(average distance between sun and earth)
 1 light year = 9.46 x 1015 m
 1 parsec = 3.08 x 1016 m
Some Practical Units

18. DIMENSIONS
Dimensions of a physical quantity are the powers to which the
fundamental quantities must be raised to represent the given
physical quantity.
Volume = length x breadth x height
= length x length x length
= (𝑙𝑒𝑛𝑔𝑡ℎ)3
[ Volume ] = [𝐿3
]
[ Volume ] = [𝑀0
𝐿3
𝑇0
]
Base quantity Symbol Dimension
Length l L
Mass m M
Time t T

19. DIMENSIONS, DIMENSIONAL FORMULA
AND DIMENSIONAL EQUATION
[ Volume ] = [𝑀0
𝐿3
𝑇0
]
Dimensions Of Volume : 0 in mass
3 in length
0 in time
Dimensional
equation
[𝑀0 𝐿3 𝑇0] − Dimensional formula

20. Area = length x breadth
= length x length
= (𝑙𝑒𝑛𝑔𝑡ℎ)2
[ area ] = [𝐿2
]
[ area ] = [𝑀0 𝐿2 𝑇0]
Dimensions of area : 0 in mass
2 in length
0 in time
Density =
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
=
𝑚𝑎𝑠𝑠
𝑙𝑏ℎ
=
𝑚𝑎𝑠𝑠
(𝑙𝑒𝑛𝑔𝑡ℎ)3
[density] =
𝑀1
𝐿3
[ density] = [𝑀1 𝐿−3 𝑇0]
Dimensions of density : 1 in mass
-3 in length
0 in time

21. Velocity =
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒
=
𝑙𝑒𝑛𝑔𝑡ℎ
𝑡𝑖𝑚𝑒
=
𝐿1
𝑇1
[ Velocity] = [𝑀0 𝐿1 𝑇−1]
Dimensions of velocity : 0 in mass
1 in length
-1 in time
Speed =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒
=
𝑙𝑒𝑛𝑔𝑡ℎ
𝑡𝑖𝑚𝑒
=
𝐿1
𝑇1
[ speed] = [𝑀0 𝐿1 𝑇−1]
Dimensions of speed : 0 in mass
1 in length
-1 in time

22. acceleration =
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑡𝑖𝑚𝑒
=
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒 𝑥 𝑡𝑖𝑚𝑒
=
𝑙𝑒𝑛𝑔𝑡ℎ
(𝑡𝑖𝑚𝑒)2
=
𝐿1
𝑇2
[ acceleration] = [𝑀0 𝐿1 𝑇−2]
Force= mass x acceleration
= 𝑚𝑎𝑠𝑠 𝑥
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑡𝑖𝑚𝑒
= mass x
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒 𝑥 𝑡𝑖𝑚𝑒
=
𝑚𝑎𝑠𝑠 𝑥 𝑙𝑒𝑛𝑔𝑡ℎ
(𝑡𝑖𝑚𝑒)2
=
𝑀1 𝐿1
𝑇2
[force] = [𝑀1 𝐿1 𝑇−2]

23. Work = Force x displacement
[Work] = [𝑀1 𝐿1 𝑇−2] [𝑀0 𝐿1 𝑇0]
= [𝑀1
𝐿2
𝑇−2
]
Power =
𝑊𝑜𝑟𝑘
𝑡𝑖𝑚𝑒
=
[𝑀1 𝐿2 𝑇−2]
[𝑀0 𝐿0 𝑇1]
[Power] = [𝑀1 𝐿2 𝑇−3]Energy = Capacity to do work
[Energy] = [𝑀1
𝐿2
𝑇−2
]
Momentum = Mass x velocity
[Work] = [𝑀1
𝐿0
𝑇0
] [𝑀0
𝐿1
𝑇−1
]
= [𝑀1
𝐿1
𝑇−1
]
Pressure =
𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎
=
[𝑀1 𝐿1 𝑇−2]
[𝑀0 𝐿2 𝑇0]
[Pressure] = [𝑀1 𝐿−1 𝑇−2]

24. Plane angle =
𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑐
𝑟𝑎𝑑𝑖𝑢𝑠
=
[𝑀0 𝐿1 𝑇0]
[𝑀0 𝐿1 𝑇0]
[Plane angle] = [𝑀0 𝐿0 𝑇0]
A Dimensionless quantity is a quantity to which no
physical dimension is assigned.

25. Sr.no. Physical quantity Relation with other
quantities
Dimensional
formula
SI units
1 Area l x b [𝑀0 𝐿2 𝑇0] 𝑚2
2 Volume l x b x h [𝑀0 𝐿3 𝑇0] 𝑚3
3 Density Mass/volume [𝑀1 𝐿−3 𝑇0] kg/𝑚3
4 Speed or velocity Distance/time [𝑀0
𝐿1
𝑇−1
] m/s
5 Acceleration Velocity/time [𝑀0
𝐿1
𝑇−2
] m/𝑠2
6 Force Mass x acceleration [𝑀1
𝐿1
𝑇−2
] kg m/𝑠2
or N
7 Work Force x displacement [𝑀1
𝐿2
𝑇−2
] J
8 Energy Capacity to do work [𝑀1 𝐿2 𝑇−2] J
9 Momentum Mass x velocity [𝑀1
𝐿1
𝑇−1
] kg m/s
10 Power Work/time [𝑀1 𝐿2 𝑇−3] W
11 Pressure Force/area [𝑀1 𝐿−1 𝑇−2] 𝑝𝑎𝑠𝑐𝑎𝑙 𝑃𝑎
12 Angle Length of the arc /
radius
[𝑀0
𝐿0
𝑇0
] radians

26. Homogeneity Principle
If the dimensions of left hand side of an equation are equal
to the dimensions of right hand side of the equation, then the
equation is dimensionally correct.
Mathematically [LHS] = [RHS]

27. Applications of dimensional analysis
1. To check the correctness of a physical relation.
2. To convert value of physical quantity from one system of unit to
another system.
3. To derive the relation between various physical quantities.

28. 1. To check the correctness of a physical relation.
Ex.1. v = u + at
LHS [v] = [𝑀0 𝐿1 𝑇−1]
RHS [u] =[𝑀0 𝐿1 𝑇−1]
[at]= [𝑀0 𝐿1 𝑇−2] [𝑀0 𝐿0 𝑇1]
= [𝑀0 𝐿1 𝑇−1]
Dimensions on both the side is same, the given physical relation is correct.

29. Ex.2. s = ut +
1
2
a 𝑡2
LHS [s] = [𝑀0
𝐿1
𝑇0
]
RHS [ut] =[𝑀0 𝐿1 𝑇−1] [𝑀0 𝐿0 𝑇1]
=[𝑀0 𝐿1 𝑇0]
[
1
2
a 𝑡2 ]= [𝑀0 𝐿1 𝑇−2] [𝑀0 𝐿0 𝑇2]
= [𝑀0
𝐿1
𝑇0
]
Dimensions on both the side is same, the given physical relation is correct.

30. Ex.3.
1
2
m 𝑣2 = mgh
LHS [
1
2
m 𝑣2 ]= [𝑀1 𝐿0 𝑇0] [𝑀0 𝐿2 𝑇−2]
= [𝑀1 𝐿2 𝑇−2]
RHS [mgh] = [𝑀1 𝐿0 𝑇0] [𝑀0 𝐿1 𝑇−2] [𝑀0 𝐿1 𝑇0]
= [𝑀1 𝐿2 𝑇−2]
Dimensions on both the side is same, the given physical relation is correct.

31. Ex.4. T = 2𝜋
𝑙
𝑔
LHS [ T ]= [𝑀0 𝐿0 𝑇1]
RHS [2𝜋
𝑙
𝑔
] =
𝑙
𝑔
1/2
=
𝑀0 𝐿1 𝑇0
[𝑀0 𝐿1 𝑇−2
1/2
= [𝑀0 𝐿0 𝑇2]1/2
= [𝑀0 𝐿0 𝑇1]
Dimensions on both the side is same, the given physical relation is correct.

32. 2. To convert value of physical quantity from one system of unit to
another system.
Q= 𝒏 𝟏 𝒖 𝟏
Q= 𝒏 𝟐 𝒖 𝟐
𝒏 𝟏 𝒖 𝟏 = 𝒏 𝟐 𝒖 𝟐
𝒏 𝟐= 𝒏 𝟏
𝒖 𝟏
𝒖 𝟐
𝒏 𝟐= 𝒏 𝟏
𝑴 𝟏
𝑴 𝟐
𝒂 𝑳 𝟏
𝑳 𝟐
𝒃 𝑻 𝟏
𝑻 𝟐
𝒄

33. Ex.1. Convert 1newton into dyne.
newton is a unit of force.
Dimension of force is [𝑀1 𝐿1 𝑇−2]
a= 1 b = 1 c = - 2
SI to cgs
SI cgs
𝑀1 = 1 𝑘𝑔 = 1000𝑔 𝑀2 = 1 𝑔
𝐿1 = 1 𝑚 = 100𝑐𝑚 𝐿2 = 1 𝑐𝑚
𝑇1 = 1𝑠 𝑇2 = 1𝑠
𝑛1 = 1 𝑛2 = ?
𝒏 𝟐= 𝒏 𝟏
𝑴 𝟏
𝑴 𝟐
𝒂 𝑳 𝟏
𝑳 𝟐
𝒃 𝑻 𝟏
𝑻 𝟐
𝒄
𝒏 𝟐= 1
𝟏𝟎𝟎𝟎𝒈
𝟏𝒈
𝟏 𝟏𝟎𝟎𝒄𝒎
𝟏𝒄𝒎
𝟏 𝟏𝒔
𝟏𝒔
−𝟐
𝒏 𝟐= 1 𝟏𝟎𝟎𝟎 𝟏 𝟏𝟎𝟎 𝟏 𝟏 −𝟐
𝒏 𝟐= 𝟏 𝒙 𝟏𝟎 𝟑 𝒙 𝟏𝟎 𝟐 𝒙 𝟏 = 𝟏𝟎 𝟓
1 newton is 𝟏𝟎 𝟓
dyne

34. 𝒏 𝟐= 𝒏 𝟏
𝑴 𝟏
𝑴 𝟐
𝒂 𝑳 𝟏
𝑳 𝟐
𝒃 𝑻 𝟏
𝑻 𝟐
𝒄
𝒏 𝟐= 1
𝟏𝟎−𝟑 𝒌𝒈
𝟏𝒌𝒈
𝟏
𝟏𝟎−𝟐 𝒎
𝟏𝒎
𝟐
𝟏𝒔
𝟏𝒔
−𝟐
𝒏 𝟐= 1 𝟏𝟎−𝟑 𝟏
𝟏𝟎−𝟐 𝟐
𝟏 −𝟐
𝒏 𝟐= 𝟏 𝒙 𝟏𝟎−𝟑 𝒙 𝟏𝟎−𝟒 𝒙 𝟏 = 𝟏𝟎−𝟕
1erg is 𝟏𝟎−𝟕
joule
cgs SI
𝑀1 = 1 𝑔 = 𝟏𝟎−𝟑 𝒌𝒈 𝑀2 = 1 𝑘𝑔
𝐿1 = 1 𝑐𝑚 = 𝟏𝟎−𝟐 𝒎 𝐿2 = 1 𝑚
𝑇1 = 1𝑠 𝑇2 = 1𝑠
𝑛1 = 1 𝑛2 =?
Ex.2. Convert 1erg into joule.
erg is a unit of work/energy.
Dimension of work is [𝑀1 𝐿2 𝑇−2]
a= 1 b = 2 c = - 2
cgs to SI

35. 3. To derive the relation between various physical quantities.
Ex.1. Derive the relation between the force ( F ) acting on a body of mass (m) with
an acceleration (a).
F  𝑚 𝑎
𝑎 𝑏
F = K 𝑚 𝑎
𝑎 𝑏
[𝑀1
𝐿1
𝑇−2
] = 𝐾 [𝑀1
𝐿0
𝑇0
] 𝑎
[𝑀0
𝐿1
𝑇−2
] 𝑏
[𝑀1 𝐿1 𝑇−2] = 𝐾 [𝑀 𝑎 𝐿0 𝑇0] [𝑀0 𝐿 𝑏 𝑇−2𝑏]
[𝑀1 𝐿1 𝑇−2] = 𝐾 [𝑀 𝑎 𝐿 𝑏 𝑇−2𝑏]
Equating the powers of M L T
a=1
b=1
-2b=-2
F = K 𝑚1
𝑎1
F=ma

36. EX.2: The force (F) acting on a particle (moving uniformly in a circle) depends on the mass
(m) of the particle, its velocity (v) and radius (r) of the circle. Derive dimensionally formula
for force (F).
F ∝ ma vb rc
∴ F = k ma vb rc (where k is constant)
Putting dimensions of each quantity in the equation,
[M1L1T-2] = [M1L0T0]a [M0L1T-1]b [M0L1T0]c
= [MaLb+cT+cT-b]
⇒ a =1, b +c = 1, -b = -2
⇒ a= 1, b = 2, c = -1
∴ F = km1v2r -1
= kmv2/r

37. Ex.3. Consider a simple pendulum, having a bob attached to a string, that oscillates under the
action of the force of gravity. Suppose that the period of oscillation of the simple pendulum
depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the
expression for its time period using method of dimensions.
T  𝑙 𝑎 𝑚 𝑏 𝑔 𝑐
T = 𝐾 𝑙 𝑎
𝑚 𝑏
𝑔 𝑐
[𝑀0 𝐿0 𝑇1] = 𝐾 [𝑀0 𝐿1 𝑇0] 𝑎 [𝑀1 𝐿0 𝑇0] 𝑏 [𝑀0 𝐿1 𝑇−2] 𝑐
[𝑀0 𝐿0 𝑇1] = K [𝑀0 𝐿 𝑎 𝑇0] [𝑀 𝑏 𝐿0 𝑇0] [𝑀0 𝐿𝑐 𝑇−2𝑐]
[𝑀0 𝐿0 𝑇1] = K [𝑀 𝑏 𝐿 𝑎+𝑐 𝑇−2𝑐]
Equating the powers of M L T
b = 0 a+c=0 -2c=1
𝑐 =
−1
2
a = −𝑐 =
1
2
T = 𝐾 𝑙1/2 𝑚0 𝑔−1/2
T = K
𝑙
𝑔

38. Limitations of Dimensional Analysis
1.Dimensionless quantities cannot be determined by this method. Constant
of proportionality cannot be determined by this method.
2.This method is not applicable to trigonometric, logarithmic and
exponential functions.
3.In the case of physical quantities which are dependent upon more than
three physical quantities, this method will be difficult.
4.If one side of the equation contains addition or subtraction of physical
quantities, we cannot use this method to derive the expression.

39. Significant Figures
In the measured value of a physical quantity, the number of digits
about the correctness of which we are sure plus the next doubtful
digit, are called the significant figures.
Length of a wire = 30.5 cm
Mass of ball = 31 g
3
2

40. Rules for Finding Significant Figures
1. All non-zero digits are significant.
Example : 42.3 has three significant figures.
243.4 has four significant figures.
24.123 has five significant figures
2. A zero becomes significant figure if it appears between to non-zero digits.
Example : 5.03 has three significant figures.
1005 has 4 significant figures.
10.05 has 4 significant figures.
3. Trailing zeros or the zeros placed to the right of the number are significant.
Example : 4.330 has four significant figures.
433.00 has five significant figures.
343.000 has six significant figures.

41. 4. All zeros to the right of the last non-zero digit are not significant.
e.g., 6250 has only 3 significant figures.
5. All zeros to the right of a non-zero digit in the decimal part are significant,
e.g., 1.4750 has 5 significant figures.
6. Leading zeros or the zeros placed to the left of the number are never significant.
Example : 0.00325 has only 3 significant figures.
0.0325 has only 3 significant figures.
0.325 has only 3 significant figures.
7. In exponential notation, the numerical portion gives the number of significant
figures.
Example : 1.32 x 10-2 has three significant figures.
1.32 x 104 has three significant figures.

42. Significant Figures in Algebric Operations
(i) In Addition or Subtraction
In addition or subtraction of the numerical values the final result should
retain the least decimal place as in the various numerical values.
(ii) In Multiplication or Division
In multiplication or division of the numerical values, the final result
should retain the least significant figures as the various numerical values.

43. ACCURACY, PRECISION OF INSTRUMENTS AND ERRORS IN MEASUREMENT
The result of every measurement by any measuring
instrument contains some uncertainty. This uncertainty is
called error.
The accuracy of a measurement is a measure of how close
the measured value is to the true value of the quantity.
Precision tells us to what resolution or limit the quantity is
measured.

44. The errors in measurement can be classified as
systematic errors and random errors.
Systematic errors: The systematic errors are those errors that tend to be in one
direction, either positive or negative.
Some of the sources of systematic errors are
 Instrumental errors
 Imperfection in experimental technique or procedure
 Personal errors
Random errors: The random errors are those errors, which occur irregularly and
hence are random with respect to sign and size.
These can arise due to random and unpredictable fluctuations in experimental
conditions
e.g. unpredictable fluctuations in temperature, voltage supply,
mechanical vibrations of experimental set-ups, etc

45. The errors (absolute errors ) in the measurements are
 𝑎1 =𝑎1 − 𝑎 =
 𝑎2 =𝑎2 − 𝑎 =
 𝑎3 =𝑎3 − 𝑎 =
 𝑎4 =𝑎4 − 𝑎 =
.
.
 𝑎 𝑛 =𝑎 𝑛 − 𝑎 =
In successive measurements, the readings turn out to be
𝑎1 , 𝑎2 , 𝑎3, 𝑎4 ,…….. 𝑎 𝑛
Mean /true value
𝑎 𝑚𝑒𝑎𝑛 𝑎 =
𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + ⋯ + 𝑎 𝑛
𝑛
Elimination of errors
(Mean value, absolute error and the relative error)

46. Mean absolute errors
 𝑎 =
𝑎1 + 𝑎2 + 𝑎3 + 𝑎4 + ⋯ + 𝑎 𝑛
𝑛
𝑎 = 𝑎   𝑎
the relative error or the percentage error is
𝛿𝑎 =
 𝑎
𝑎
𝑥 100%

47. Combination of Errors
1. Error in sum of two quantities:
Suppose two physical quantities A and B have measured values
A = A  ΔA
B = B  ΔB
where ΔA and ΔB are their absolute errors.
We wish to find the error ΔZ in the sum,
Z = Z ± ΔZ
Z = A + B
Z ± ΔZ = (A  ΔA) + (B  ΔB)
Z ± ΔZ = A  ΔA + B  ΔB
Z ± ΔZ = A+ B  ΔA  ΔB
Z ± ΔZ = Z  ΔA  ΔB
± ΔZ =  ΔA  ΔB
ΔZ = ΔA + ΔB
The maximum possible error in sum is ΔZ = ΔA + ΔB

48. 2. Error in difference of two quantities:
Suppose two physical quantities A and B have measured values
A = A  ΔA
B = B  ΔB
where ΔA and ΔB are their absolute errors.
We wish to find the error ΔZ in the difference
Z = Z ± ΔZ
Z = A – B
Z ± ΔZ = (A  ΔA) - (B  ΔB)
Z ± ΔZ = A  ΔA - B  ΔB
Z ± ΔZ = A- B  ΔA  ΔB
Z ± ΔZ = Z  ΔA  ΔB
± ΔZ =  ΔA  ΔB
The maximum possible error in difference is ΔZ = ΔA + ΔB
When two quantities are added or subtracted, the absolute error in the final
result is the sum of the absolute errors in the individual quantities.

49. 3. Error in product of two quantities:
Suppose two physical quantities A and B have measured values
A = A  ΔA
B = B  ΔB
Z = Z ± ΔZ
Z = A B
Z ± ΔZ = (A  ΔA)(B  ΔB)
Z ± ΔZ = AB  AΔB  BΔA  ΔA ΔB
Z ± ΔZ = Z  AΔB  BΔA  ΔA ΔB
± ΔZ =  AΔB  BΔA
ΔZ = AΔB + BΔA
Divide above equation by Z
ΔZ
Z
=
AΔB
AB
+
BΔA
𝑨𝑩
ΔZ
Z
=
ΔA
𝑨
+
ΔB
B
The maximum fractional error in product of two quantities

50. 4. Error in division of two quantities:
The maximum fractional error in division of two quantities
ΔZ
Z
=
ΔA
𝑨
+
ΔB
B
5. Error in quantity raised to some power
Z = 𝐴 𝑛
ΔZ
Z
= 𝒏
ΔA
𝑨
Error in the quantity
Z =
𝑃 𝑎 𝑄 𝑏
𝑅 𝑐
ΔZ
Z
= 𝒂
ΔP
𝑷
+ 𝒃
ΔQ
Q
+𝒄
ΔR
R