Chapter 2: Units and Measurement

Units and Systems

Fundamentals of Units

• Unit: Standard reference for measuring physical quantities.
• Base units: Fundamental units for base quantities (length, mass, time, etc.).
• Derived units: Formed by combining base units.
• System of units: Complete set of base and derived units.

Common Systems of Units

• CGS System: Centimetre, gram, second (metric).
• FPS System: Foot, pound, second (British system, non-metric).
• MKS System: Metre, kilogram, second (metric).
• SI System: Internationally accepted system based on 7 base units and 2 supplementary units.

Solved Examples

Ex 1

Identify the base units in the SI system for length, mass, and time.

• Length = metre (m)
• Mass = kilogram (kg)
• Time = second (s)

Practice Set

• Q1: Which of the following is NOT a base unit in the SI system?
  • A) Metre
  • B) Kilogram
  • C) Newton
  • D) Second
• Q2: The FPS system uses which unit for mass?
  • A) Gram
  • B) Pound
  • C) Kilogram
  • D) Ounce
• Q3: Which system is internationally accepted for scientific measurements?
  • A) CGS
  • B) FPS
  • C) SI
  • D) MKS

Answer Key

• Q1: C) Newton - It is a derived unit, not a base unit.
• Q2: B) Pound - FPS system uses pound for mass.
• Q3: C) SI - International System of Units is globally accepted.

Memory Aids

• SI base units: Length (m), Mass (kg), Time (s), Current (A), Temperature (K), Amount (mol), Luminous Intensity (cd).
• CGS = Centimetre, Gram, Second; FPS = Foot, Pound, Second; MKS = Metre, Kilogram, Second.

SI Base Units

Seven Base Quantities and Units

• Length: Metre (m) - distance light travels in \(1/299,792,458\) seconds.
• Mass: Kilogram (kg) - defined by Planck constant.
• Time: Second (s) - duration of 9,192,631,770 periods of Cs-133 radiation.
• Electric Current: Ampere (A) - flow of one coulomb per second.
• Temperature: Kelvin (K) - based on Boltzmann constant.
• Amount of Substance: Mole (mol) - contains \(6.022 \times 10^{23}\) entities.
• Luminous Intensity: Candela (cd) - luminous intensity in a given direction.

Supplementary Units

• Plane angle (radian, rad): \(d\theta = \frac{ds}{r}\)
• Solid angle (steradian, sr): \(d\Omega = \frac{dA}{r^2}\)

Solved Examples

Ex 1

Calculate the plane angle subtended by an arc of length 5 m on a circle of radius 10 m.

• Formula: \(d\theta = \frac{ds}{r} = \frac{5}{10} = 0.5\) rad

Practice Set

• Q1: The SI unit of luminous intensity is:
  • A) Candela
  • B) Lumen
  • C) Lux
  • D) Watt
• Q2: One radian is the angle subtended by an arc equal to:
  • A) Radius
  • B) Diameter
  • C) Circumference
  • D) Chord
• Q3: The solid angle subtended by an area equal to the square of the radius on a sphere is:
  • A) 1 steradian
  • B) 1 radian
  • C) 4π steradians
  • D) π steradians

Answer Key

• Q1: A) Candela - SI unit of luminous intensity.
• Q2: A) Radius - radian defined by arc length equal to radius.
• Q3: A) 1 steradian - solid angle for area = \(r^2\).

Memory Aids

• Radian = arc length / radius.
• Steradian = area on sphere / radius squared.
• Remember seven SI base units and their symbols.

Advantages of SI

Key Features

• Coherent system: Derived units formed without extra factors.
• Rational system: Uses rationalized units for electromagnetic quantities.
• Absolute system: Based on universal constants.
• Metric system: Decimal-based, easy conversions.

Solved Examples

Ex 1

Explain why SI is preferred over FPS for scientific work.

• SI is metric and coherent, making calculations simpler and universal.

Practice Set

• Q1: Which of the following is NOT an advantage of SI units?
  • A) Coherent
  • B) Absolute
  • C) Non-metric
  • D) Rational
• Q2: SI units are based on:
  • A) Arbitrary standards
  • B) Universal constants
  • C) Local measurements
  • D) Historical units
• Q3: The SI system is:
  • A) Non-coherent
  • B) Metric
  • C) Based on FPS units
  • D) Only for commercial use

Answer Key

• Q1: C) Non-metric - SI is metric.
• Q2: B) Universal constants - SI definitions use constants.
• Q3: B) Metric - SI is a metric system.

Memory Aids

• SI = Coherent + Rational + Absolute + Metric.
• Universal constants define SI units.

Errors in Measurement

Types of Errors

• Systematic Errors: Consistent bias in one direction.
• Random Errors: Unpredictable fluctuations, vary in sign and magnitude.
• Least Count Errors: Due to instrument resolution limits.

Error Definitions and Formulas

• Absolute Error: \(|\Delta a| = |a_{measured} - a_{true}|\)
• Mean (Average) Value: \(a_{mean} = \frac{1}{n} \sum_{i=1}^n a_i\)
• Mean Absolute Error: \(\Delta a_{mean} = \frac{1}{n} \sum_{i=1}^n |a_i - a_{mean}|\)
• Relative Error: \(\frac{\Delta a_{mean}}{a_{mean}}\)
• Percentage Error: \(\delta_a = \frac{\Delta a_{mean}}{a_{mean}} \times 100\%\)

Combination of Errors

• Sum/Difference: \(\Delta Z = \Delta A + \Delta B\) for \(Z = A \pm B\)
• Product/Quotient: \(\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}\) for \(Z = AB \text{ or } \frac{A}{B}\)
• Power: \(\frac{\Delta Z}{Z} = n \frac{\Delta A}{A}\) for \(Z = A^n\)
• General: \(\frac{\Delta y}{y} = l \frac{\Delta A}{A} + m \frac{\Delta B}{B} + n \frac{\Delta C}{C}\) for \(y = K \frac{A^l B^m}{C^n}\)

Solved Examples

Ex 1

Calculate maximum error in \(Z = A + B\) if \(\Delta A = 0.1\) and \(\Delta B = 0.2\).

• \(\Delta Z = \Delta A + \Delta B = 0.1 + 0.2 = 0.3\)

Practice Set

• Q1: Which error type is due to instrument resolution?
  • A) Systematic
  • B) Random
  • C) Least Count
  • D) Percentage
• Q2: If \(Z = AB\), the relative error in Z is:
  • A) \(\frac{\Delta A}{A} - \frac{\Delta B}{B}\)
  • B) \(\frac{\Delta A}{A} + \frac{\Delta B}{B}\)
  • C) \(\Delta A + \Delta B\)
  • D) \(\Delta A \times \Delta B\)
• Q3: Percentage error is:
  • A) Absolute error \(\times 100\)
  • B) Relative error \(\times 100\)
  • C) Mean error \(\times 100\)
  • D) Sum of errors \(\times 100\)

Answer Key

• Q1: C) Least Count - due to smallest measurable value.
• Q2: B) \(\frac{\Delta A}{A} + \frac{\Delta B}{B}\) - sum of relative errors.
• Q3: B) Relative error \(\times 100\) - definition of percentage error.

Memory Aids

• Sum/Difference errors add absolute errors.
• Product/Quotient errors add relative errors.
• Power errors multiply relative error by power.
• Percentage error = relative error \(\times 100\%\).

Significant Figures

Definition and Importance

• Digits accurately known in a measurement are called significant figures.
• More significant figures mean higher measurement accuracy.

Rules for Counting Significant Figures

• All non-zero digits are significant.
• Zeros between non-zero digits are significant.
• Leading zeros are not significant.
• Trailing zeros without a decimal point are not significant.
• Powers of 10 are not counted as significant figures.
• Trailing zeros in decimal numbers are significant.

Arithmetic Operations with Significant Figures

• Addition/Subtraction: Result has decimal places equal to the number with least decimal places.
• Multiplication/Division: Result has significant figures equal to the number with least significant figures.

Rounding Off Rules

• If the digit to drop > 5, increase preceding digit by 1.
• If the digit to drop < 5, leave preceding digit unchanged.
• If digit to drop = 5 and preceding digit is even, drop 5.
• If digit to drop = 5 and preceding digit is odd, increase preceding digit by 1.

Solved Examples

Ex 1

Add 734.52 g, 341.4 g, and 0.402 g with correct significant figures.

• Sum = 1076.322 g
• Least decimal places = 1 (from 341.4 g)
• Final result = 1076.3 g

Ex 2

Calculate density = 4.237 g / 2.51 cm³ with correct significant figures.

• Least significant figures = 3 (2.51)
• Density = 1.69 g/cm³

Practice Set

• Q1: How many significant figures are in 0.00450?
  A) 2  B) 3  C) 4  D) 5
• Q2: Result of 12.11 + 0.023 + 1.1 rounded correctly is:
  A) 13.233  B) 13.2  C) 13.23  D) 13.0
• Q3: Multiply 3.00 × 10⁸ m/s by 3.156 × 10⁷ s. The light year is:
  A) 9.47 × 10¹⁵ m  B) 9.5 × 10¹⁵ m  C) 9.47 × 10¹⁴ m  D) 9.5 × 10¹⁴ m

Answer Key

• Q1: B) 3 — Leading zeros not significant; digits 4, 5, 0 (trailing zero in decimal) are significant.
• Q2: B) 13.2 — Least decimal place is 1 (1.1), so round to 1 decimal place.
• Q3: A) 9.47 × 10¹⁵ m — Least significant figures is 3 (3.00), so result has 3 significant figures.

Memory Aids

• Non-zero digits always count.
• Zeros between non-zero digits count.
• Leading zeros never count.
• Trailing zeros count only if decimal present.
• Add/Subtract: Match decimal places.
• Multiply/Divide: Match significant figures.
• Rounding: >5 round up, <5 stay, 5 depends on even/odd.

Least Count

Definition

• Smallest value measurable accurately by an instrument.

Least Count of Vernier Callipers

• Main Scale Division (M.S.D) = M units.
• Vernier Scale Division (V.S.D) = V units.
• Length of ‘a’ main scale divisions = length of ‘b’ vernier scale divisions.
• Formula: \( L.C = \frac{(b - a)}{b} \times M \)
• Vernier Constant = Least Count = (M - V)

Least Count of Screw Gauge

• Least Count = \( \frac{\text{Pitch}}{\text{Number of divisions on circular scale}} \)
• Pitch = Distance moved by screw per full rotation.
• Pitch = Distance moved on linear scale / Number of full rotations.

Solved Examples

Ex 1

Calculate least count if 10 main scale divisions = 9 vernier scale divisions and M.S.D = 1 mm.

• a = 10, b = 9, M = 1 mm
• \( L.C = \frac{(9 - 10)}{9} \times 1 = \frac{-1}{9} = -0.111 \) mm (absolute value)
• Least Count = 0.111 mm

Ex 2

Screw gauge with pitch 0.5 mm and 50 divisions on circular scale. Find least count.

• Least Count = 0.5 mm / 50 = 0.01 mm

Practice Set

• Q1: If 20 main scale divisions equal 19 vernier scale divisions and M.S.D = 1 mm, least count is:
  A) 0.05 mm  B) 0.1 mm  C) 0.2 mm  D) 0.25 mm
• Q2: A screw gauge has pitch 0.5 mm and 100 divisions on circular scale. Least count is:
  A) 0.005 mm  B) 0.01 mm  C) 0.02 mm  D) 0.05 mm
• Q3: If pitch is 1 mm and least count is 0.01 mm, number of divisions on circular scale is:
  A) 50  B) 75  C) 100  D) 125

Answer Key

• Q1: A) 0.05 mm — \( L.C = \frac{(19-20)}{19} \times 1 = 0.0526 \) mm approx.
• Q2: A) 0.005 mm — Least count = 0.5/100 = 0.005 mm.
• Q3: C) 100 — Number of divisions = Pitch / Least Count = 1/0.01 = 100.

Memory Aids

• Least Count = Smallest measurable unit.
• Vernier: \( L.C = \frac{b - a}{b} \times M \).
• Screw Gauge: Least Count = Pitch / Number of divisions.
• Pitch = Distance per full rotation.

Dimensions of Physical Quantities

Definition and Concept

• Physical quantities are expressed in terms of base quantities: Mass (M), Length (L), Time (T).
• Dimension of a quantity = powers (exponents) of base quantities in its formula.
• Example: Force = Mass × Acceleration = M × L × T^-2 → Dimensions: [MLT^-2]
• Energy = Force × Distance = M × L × T^-2 × L = M × L² × T^-2 → Dimensions: [ML²T^-2]

Solved Examples

• Ex 1: Find dimensions of velocity.
  Velocity = Distance / Time = L / T → [LT^-1]
• Ex 2: Find dimensions of pressure.
  Pressure = Force / Area = [MLT^-2] / L² = M L^-1 T^-2

Practice Set

• Q1: What are the dimensions of acceleration?
  A) [LT^-1]   B) [LT^-2]   C) [MLT^-2]   D) [T^-1]
• Q2: The dimensions of power are:
  A) [ML²T^-3]   B) [ML²T^-2]   C) [MLT^-2]   D) [M²L²T^-3]
• Q3: If force has dimensions [MLT^-2] and displacement has dimensions [L], what are the dimensions of work done?
  A) [MLT^-2]   B) [ML²T^-2]   C) [ML²T^-1]   D) [M²L²T^-2]

Answer Key

• Q1: B) [LT^-2]. Acceleration = change in velocity/time = L/T².
• Q2: A) [ML²T^-3]. Power = Work done/time = Energy/time = [ML²T^-2]/T.
• Q3: B) [ML²T^-2]. Work = Force × Displacement = [MLT^-2] × [L].

Memory Aids

• Force: [MLT^-2] → M (mass), L (length), T squared inverse.
• Energy/Work: [ML²T^-2] → Force × Length.
• Power: Energy/time → [ML²T^-3].

Dimensional Formulae and Dimensional Equations

Definition and Basics

• Dimensional Formula: Expresses a physical quantity in terms of base quantities (Mass [M], Length [L], Time [T]).
• Example: Volume = [M⁰L³T⁰], Velocity = [M⁰L T^-1]
• Dimensional Equation: Equation equating a physical quantity to its dimensional formula.
• Example: Force [F] = [M L T^-2], Density [ρ] = [M L^-3 T⁰]
• Dimensionless Quantities: Quantities without dimensions, e.g., pure numbers, ratios, exponential, logarithmic, and trigonometric functions.

Solved Examples

Ex 1: Find the dimensional formula of speed.

• Speed = Distance / Time
• Distance = [L], Time = [T]
• Dimensional formula = [L T^-1]

Ex 2: Write the dimensional equation of force.

• Force = Mass × Acceleration
• Mass = [M], Acceleration = [L T^-2]
• Dimensional formula = [M L T^-2]

Practice Set

Q1: What is the dimensional formula of pressure?

• A. [M L^-1 T^-2]
• B. [M L^-2 T^-1]
• C. [M L T^-2]
• D. [M L^-3 T⁰]

Q2: Which of the following is a dimensionless quantity?

• A. Velocity
• B. Density
• C. Refractive Index
• D. Force

Q3: The dimensional formula of energy is?

• A. [M L² T^-2]
• B. [M L T^-1]
• C. [M L² T^-1]
• D. [M L³ T^-2]

Answer Key

Q1: A. Pressure = Force/Area = [M L T^-2] / [L²] = [M L^-1 T^-2]

Q2: C. Refractive index is a ratio of speeds, hence dimensionless.

Q3: A. Energy = Force × Distance = [M L T^-2] × [L] = [M L² T^-2]

Memory Aids

• Force = M × L × T^-2 (Think: mass × acceleration)
• Velocity = L × T^-1 (Distance over time)
• Dimensionless quantities include ratios and functions like sin, log, e^x

Dimensional Analysis and Applications

Key Concepts

• Dimensional Consistency: All terms in a physical equation must have the same dimensions (Principle of Homogeneity).
• Unit Conversion: Use formula \( n_2 = n_1 \times \frac{Q_1}{Q_2} \) to convert quantities between unit systems.
• Deriving Relations: Dimensional analysis helps find relations between physical quantities by equating dimensions.
• Dimensions of Constants: Constants in equations can have dimensions deduced from other quantities.

Solved Examples

Ex 1: Check dimensional consistency of Work done = Force × Displacement.
Dimensions: [ML^2T^{-2}] = [MLT^{-2}] × [L] = [ML^2T^{-2}] ✓

Ex 2: Convert 1 Newton (N) to dynes.
Using \( x = 1 \times \frac{Kg}{g} \times \frac{m}{cm} \times \left(\frac{s}{s}\right)^{-2} = 1000 \times 100 = 10^5 \) dynes.

Ex 3: Derive time period \( T \) of simple pendulum depending on length \( l \) and gravity \( g \).
Assume \( T = K l^b g^c \). Equate dimensions:
\( [T] = [L]^b [LT^{-2}]^c = L^{b+c} T^{-2c} \)
Equate powers: \( 1 = -2c \Rightarrow c = -\frac{1}{2} \), \( 0 = b + c \Rightarrow b = \frac{1}{2} \)
So, \( T = K \sqrt{\frac{l}{g}} \).

Practice Set

Q1: Which principle ensures dimensions on both sides of an equation are equal?
A) Conservation of Energy
B) Principle of Homogeneity
C) Newton's Second Law
D) Law of Conservation of Mass

Q2: Convert 2 N to dynes. (1 N = 10^5 dynes)
A) 2 × 10^3 dynes
B) 2 × 10^5 dynes
C) 2 × 10^6 dynes
D) 2 × 10^4 dynes

Q3: Time period \( T \) of a pendulum depends on length \( l \) and gravity \( g \). Which is the correct dimensional formula?
A) \( T = K \sqrt{\frac{l}{g}} \)
B) \( T = K \frac{l}{g} \)
C) \( T = K l g \)
D) \( T = K \sqrt{l g} \)

Answer Key

Q1: B
Principle of Homogeneity states dimensions must be same on both sides.

Q2: C
2 N = 2 × 10^5 dynes = 2 × 10^5 dynes (Option C matches 2 × 10^6 dynes is incorrect, correct is 2 × 10^5 dynes, so option B is correct.) Correction: Option B is correct.

Q3: A
Derived relation shows \( T \propto \sqrt{\frac{l}{g}} \).

Memory Aids

• Check dimensions term-wise for equation correctness.
• Unit conversion: multiply by ratio of units raised to their powers.
• Pendulum time period formula: \( T \sim \sqrt{l/g} \).
• Constants' dimensions found by balancing equation dimensions.