Physics relies on accurate measurement of physical quantities. Every theory, experiment, and technological application begins with defining what we measure and how precisely we measure it. This chapter establishes the foundational concepts of units, significant figures, dimensional formulas, and dimensional analysis, enabling students to interpret numerical data correctly and apply it to real‑world problems.

Introduction

Measurement is the comparison of an unknown physical quantity with a known standard. Physics recognises two categories of quantities:

  • Fundamental quantities – cannot be broken into simpler quantities (length, mass, time, etc.)
  • Derived quantities – defined in terms of fundamental quantities (velocity, force, momentum, etc.)

A measurement always consists of:

  1. A numerical magnitude, and
  2. A unit, representing the standard used.

Accurate measurement reduces ambiguity, ensures consistent communication, and forms the basis of scientific laws.

Examples of Physical Measurements

  • Speed of a moving car → measured in m/s
  • Mass of an object → measured in kg
  • Temperature of a solution → measured in kelvin (K)

Errors in Measurement

Every measurement involves some inaccuracy. Errors may arise due to:

  • Instrument limitations
  • Observer reaction time
  • Environmental factors
  • Improper calibration

Physics attempts to minimise these errors using precise instruments and repeated observations.

The International System of Units (SI Units)

To maintain global uniformity, the SI system was established. It contains seven base quantities that form the building blocks of all physical measurements.

SI Base Units

Physical QuantitySI UnitSymbol
Lengthmetrem
Masskilogramkg
Timeseconds
Electric currentampereA
TemperaturekelvinK
Amount of substancemolemol
Luminous intensitycandelacd

Supplementary and Derived Units

Derived units are formed from base units. Examples:

  • Frequency → hertz (Hz) = s⁻¹
  • Force → newton (N) = kg·m/s²
  • Energy → joule (J) = kg·m²/s²
  • Power → watt (W) = J/s

SI Prefixes

To represent very large or very small quantities:

  • kilo (k) = 10³
  • mega (M) = 10⁶
  • giga (G) = 10⁹
  • milli (m) = 10⁻³
  • micro (µ) = 10⁻⁶
  • nano (n) = 10⁻⁹

These prefixes simplify numerical values while maintaining precision.

Standards of Measurement

  • The standard metre is defined by the distance travelled by light in vacuum in a specific fraction of a second.
  • The kilogram standard is defined by the Planck constant.
  • The second is defined using atomic transition frequencies.

These definitions ensure extreme accuracy across the world.

Significant Figures

Significant figures communicate the precision of a measured quantity. They include all certain digits plus one doubtful digit.

Rules for Determining Significant Figures

  1. All non‑zero digits are significant.
    Example: 456 → 3 significant figures.
  2. Zeros between significant digits are significant.
    Example: 205 → 3 significant figures.
  3. Leading zeros are not significant.
    Example: 0.0034 → 2 significant figures.
  4. Trailing zeros are significant if decimal exists.
    Example: 20.00 → 4 significant figures.

Significant Figures in Calculations

  • Addition/Subtraction → Least number of decimal places.
  • Multiplication/Division → Least number of significant digits.

Solved Examples

Example 1: Identify significant figures

Number: 0.02050
→ 4 significant figures

Example 2: Multiplication

(3.21 × 0.0021)
Least significant figures = 2
= 0.006741 → 0.0067

Dimensions of Physical Quantities

Dimensions express how a quantity depends on base quantities.

Common Dimensional Expressions

  • Velocity → [LT⁻¹]
  • Acceleration → [LT⁻²]
  • Force → [MLT⁻²]
  • Momentum → [MLT⁻¹]
  • Pressure → [ML⁻¹T⁻²]
  • Energy → [ML²T⁻²]

Uses of Dimensions

  • Checking the correctness of formulas
  • Determining units
  • Deriving relations among quantities
  • Converting units

Example

Find dimensions of power:
Power = Work/Time = (ML²T⁻²)/T = [ML²T⁻³]

Dimensional Formulae and Dimensional Equations

A dimensional formula represents a physical quantity in terms of M, L, and T.

Examples

  • Density = Mass/Volume = M / L³ → [ML⁻³]
  • Pressure = Force/Area = (MLT⁻²)/L² → [ML⁻¹T⁻²]
  • Young’s modulus = Stress/Strain → [ML⁻¹T⁻²]

Verification of Physical Equations

A relation is dimensionally correct if dimensions match on both sides.

Example

Verify: v = u + at
Dimensions:

  • v → [LT⁻¹]
  • u → [LT⁻¹]
  • at → [LT⁻¹]

Equation is dimensionally correct.

Dimensional Analysis and Its Applications

Dimensional analysis is a powerful tool for:

  • Checking the validity of equations
  • Deriving new relations
  • Changing units

1. Checking Validity

Equation: KE = ½ mv²
Left side: [ML²T⁻²]
Right side: [M] × [L²T⁻²]
→ Matches → Correct.

2. Deriving a Formula

Time period of a pendulum depends on L and g.
Assume: T = k Lᵃ gᵇ
Dimensions:
Left: [T]
Right: [Lᵃ (LT⁻²)ᵇ] = L^{a+b} T^{-2b}
Equate powers → a + b = 0, −2b = 1 → b = −½, a = ½
Thus: T = k √(L/g)

3. Converting Units

1 N = 1 kg·m/s²
In CGS → dyne
= 10⁵ dyne

JEE/NEET Level Numerical Problems

Problem 1: Dimensional Analysis (JEE Main)

The escape velocity v of a planet depends on gravitational constant G, mass M, and radius R. Derive its formula.

Assume: v = k Gᵃ Mᵇ Rᶜ Dimensions:

  • v: [LT⁻¹]
  • G: [M⁻¹L³T⁻²]
  • M: [M]
  • R: [L]

Equating powers of M, L, T:

  • Mass: −a + b = 0
  • Length: 3a + c = 1
  • Time: −2a = −1 → a = ½ → b = ½ → c = −½ 
  • Thus: v = k √(GM/R)

Problem 2: Significant Figures (NEET)

Compute (4.52 × 0.030) with correct significant figures.

  • 4.52 → 3 significant figures
  • 0.030 → 2 significant figures Multiplication result must have 2 significant figures. 4.52 × 0.030 = 0.1356 ≈ 0.14

Problem 3: Unit Conversion (JEE Main)

Convert 1 Pa (pascal) into CGS. 1 Pa = 1 N/m²
= (10⁵ dyne) / (10² cm)²
= 10⁻¹ dyne/cm²

Problem 4: Dimensions (NEET)

Find the dimensions of the universal gas constant R. PV = nRT → R = PV/nT = [ML⁻¹T⁻²][L³] / [1 × T] = [ML²T⁻²K⁻¹]

FAQs

Q1. Why do we need significant figures?

They reflect the precision of measurements and prevent false accuracy in reporting numerical results.

Q2. What is the purpose of dimensional analysis?

Dimensional analysis helps verify the correctness of physical equations, derive relations between quantities, convert units, and check the consistency of formulas.

Q3. Can a dimensionally correct equation be physically incorrect?

Yes. For example, v = u + at² is dimensionally correct but physically incorrect for uniformly accelerated motion. Dimensional accuracy does not guarantee physical validity.

Q4. Why are SI units preferred globally?

SI units provide uniformity, clarity, and universal acceptance. Scientific communication becomes easier when all measurements follow the same standard.

Q5. How do errors affect measurements in physics?

Errors can arise from instrument limitations, observer reaction, or external factors. They affect the accuracy and precision of results. Using better instruments and repeating measurements reduces error margins.

Q6. Why must derived units be expressed in terms of base units?

Expressing derived units in terms of base units ensures clarity, dimensional consistency, and ease of verification or conversion.

Q7. What is the difference between accuracy and precision?

Accuracy refers to closeness to the true value, while precision refers to the consistency of repeated measurements. A measurement can be precise but not accurate.

Q8. Why is dimensional analysis not applicable to trigonometric or exponential functions?

Trigonometric, exponential, and logarithmic functions require dimensionless arguments. Hence, dimensional analysis cannot be used to validate such relations.

Conclusion

Units and measurements form the backbone of physics. Understanding significant figures ensures clarity in reporting data, while dimensions and dimensional analysis allow us to verify equations, derive relationships, and simplify unit conversions. Mastering these fundamental ideas equips students with the analytical tools required throughout their scientific journey.

At Deeksha Vedantu, we ensure that students not only learn these principles but also apply them confidently in numerical problems, experiments, and competitive examinations-building a strong conceptual foundation that supports their success in Class 11, Class 12, and beyond.

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