Units & Measurement
NCERT Class 11 Physics · Chapter 1

In physics, every measurement requires a comparison with a known standard. The standard quantity chosen for comparison is called a unit. Thus, measurement of any physical quantity involves comparing it with a definite, internationally accepted reference value.

For example:

• The length of a table may be measured as 2 metres.
• The mass of an object may be 5 kilograms.
• The duration of an event may be 10 seconds.

In each case, the number indicates how many times the unit is contained in the measured quantity.

Why Units are Important

• They ensure uniformity of measurement across the world.
• They allow scientists to communicate results precisely.
• They make it possible to compare experimental results.
• They form the basis of scientific calculations and equations.

Some physical quantities are so basic that they cannot be expressed in terms of other quantities. These are called fundamental (or base) quantities.

The units assigned to these quantities are known as fundamental units. All other units in physics are derived from them.

The globally accepted system used in science is the International System of Units (SI). It defines seven base quantities.

Seven SI Base Quantities

1. Length — metre (m)
   Measures distance between two points.
   Example: height of a building, distance between cities.
2. Mass — kilogram (kg)
   Measures the quantity of matter in an object. Example: mass of a book, mass of Earth.
3. Time — second (s)
   Measures duration of events. Example: period of oscillation of a pendulum.
4. Electric Current — ampere (A)
   Rate of flow of electric charge.
5. Thermodynamic Temperature — kelvin (K)
   Measures thermal energy of particles.
6. Amount of Substance — mole (mol)
   Used to count microscopic entities such as atoms and molecules.
7. Luminous Intensity — candela (cd)
   Measures brightness of a light source.

Many physical quantities in physics cannot be described using a single fundamental unit. Instead, their units are obtained by combining two or more SI base units. Such units are called derived units.

Therefore,

The physical quantities measured using these units are called derived quantities.

How Derived Units Are Formed

Derived units are obtained using the mathematical relations between physical quantities. Since most physical laws relate several quantities together, their units also become combinations of base units.

• If a quantity involves length divided by time, its unit becomes metre per second.
• If a quantity involves mass multiplied by acceleration, its unit combines kilogram, metre and second.
• If a quantity involves force acting through distance, its unit combines newton and metre.

This systematic construction ensures that all measurements remain consistent within the International System of Units (SI).

Examples of Derived Units

Throughout the history of science, different countries used different measurement systems. This created confusion because the same physical quantity could be expressed using different units.

To solve this problem, scientists adopted a single globally accepted system called the International System of Units (SI). It ensures that measurements made in different laboratories and countries remain consistent, comparable, and universally understood.

Earlier Systems of Units

Before the adoption of SI units, scientists used several systems of units. The most important ones are described below.

CGS System (Centimetre–Gram–Second)

In the CGS system:

• Length is measured in centimetre (cm)
• Mass is measured in gram (g)
• Time is measured in second (s)

This system was widely used in early physics research, particularly in electromagnetism and mechanics. However, many derived units became inconveniently small, which limited its practical use.

FPS System (Foot–Pound–Second)

In the FPS system:

• Length → foot (ft)
• Mass → pound (lb)
• Time → second (s)

This system was commonly used in engineering applications in countries like the United States and the United Kingdom. Because it is not based on powers of ten, it is inconvenient for scientific calculations.

MKS System (Metre–Kilogram–Second)

The MKS system uses:

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

This system was easier to use because it follows a decimal structure. It later became the foundation for the modern SI system.

Base Units of the SI System

The SI system defines seven fundamental quantities. All other physical quantities are derived from these base units.

SI Prefixes

Physical quantities often involve extremely large or extremely small values. To express these conveniently, SI uses prefixes representing powers of ten.

• Tera (T) = \(10^{12}\)
• Giga (G) = \(10^{9}\)
• Mega (M) = \(10^{6}\)
• Kilo (k) = \(10^{3}\)
• Centi (c) = \(10^{-2}\)
• Milli (m) = \(10^{-3}\)
• Micro (μ) = \(10^{-6}\)
• Nano (n) = \(10^{-9}\)

Rules for Writing SI Units

To maintain consistency in scientific writing, certain conventions must be followed when writing SI units.

• Unit symbols are written in lowercase letters (m, kg, s), except when named after scientists such as Newton (N) or Joule (J).
• Unit symbols are never pluralised (10 kg, not 10 kgs).
• A space is always left between the number and the unit (example: 5 m, 12 kg, 20 s).
• Compound units are written using powers or division, for example \(m\,s^{-1}\) or \(m/s\).

In physics, every measurement contains some degree of uncertainty because no measuring instrument can provide perfectly exact values. The precision of a measurement depends on the smallest division of the measuring instrument used.

To represent the precision of measured quantities, scientists use significant figures. They indicate which digits in a measurement are reliable and which digit is the estimated one.

Example of Significant Figures

Suppose the length of a pencil is measured using a scale and the reading obtained is 12.4 cm.

• Digits 1 and 2 are certain.
• The digit 4 is estimated.

Therefore, the measurement \(12.4\,\text{cm}\) contains three significant figures.

Rules for Counting Significant Figures

1. All non-zero digits are significant.
   • 234 → 3 significant figures
   • 7.81 → 3 significant figures
2. Zeros between non-zero digits are significant.
   • 101 → 3 significant figures
   • 2.05 → 3 significant figures
3. Leading zeros are not significant.
   • 0.0025 → 2 significant figures
   • 0.040 → 2 significant figures
4. Trailing zeros after a decimal point are significant.
   • 2.300 → 4 significant figures
   • 0.0600 → 3 significant figures
5. Trailing zeros in whole numbers without a decimal point may be ambiguous.
   • 1500 → may have 2, 3, or 4 significant figures
   To avoid confusion, scientific notation is used.

Scientific Notation and Significant Figures

Scientific notation clearly shows the number of significant figures because all meaningful digits appear before the power of ten.

• \(1.50 \times 10^3\) → 3 significant figures
• \(1.500 \times 10^3\) → 4 significant figures
• \(1.5 \times 10^3\) → 2 significant figures

Significant Figures in Calculations

When performing calculations with measured quantities, the final answer must reflect the precision of the least precise measurement.

• Addition or Subtraction
  The result must have the same number of decimal places as the measurement with the least decimal places.
  
  Example: \(12.11 + 0.3 = 12.4\)
• Multiplication or Division
  The result must contain the same number of significant figures as the quantity with the least significant figures.
  
  Example: \(4.56 \times 1.4 = 6.4\)

When results of measurements or calculations contain more digits than justified by the precision of the data, the extra digits must be removed. This process is called rounding off.

Rounding off ensures that the final numerical value reflects the correct precision of the measurement and does not give a false impression of accuracy.

Why Rounding Off is Important

• Maintains consistency with significant figures.
• Prevents writing numbers with unrealistic precision.
• Ensures correct reporting of experimental results.
• Essential when performing calculations in physics problems.

Rules for Rounding Off Numbers

Suppose a number is to be rounded to a specific number of significant figures. The following rules are applied.

• Rule 1
  If the digit to be dropped is less than 5, the preceding digit remains unchanged.
  
  Example: \[ 2.743 \approx 2.74 \] (Rounded to three significant figures)
• Rule 2
  If the digit to be dropped is greater than 5, the preceding digit is increased by 1.
  
  Example: \[ 5.786 \approx 5.79 \]
• Rule 3
  If the digit to be dropped is exactly 5, the rounding rule depends on the preceding digit.
  • If the preceding digit is even, it remains unchanged.
  • If the preceding digit is odd, it is increased by 1.

  This method is called the round-to-even rule and prevents systematic rounding errors.

  Examples: \[ 2.345 \approx 2.34 \] (4 is even) \[ 2.355 \approx 2.36 \] (5 is odd)

Example from Physics Calculations

Suppose the calculated value of a quantity is \(3.6782\) but the measurement allows only three significant figures.

\[ 3.6782 \approx 3.68 \]

Each side of a cube is measured to be \(7.203\,\text{m}\). Find the total surface area and the volume of the cube, expressed with the correct number of significant figures.

Given

Side of cube \[ a = 7.203\ \text{m} \] The measurement \(7.203\) contains **four significant figures**. Therefore, all final answers must also contain **four significant figures**.

1. Total Surface Area of Cube

The total surface area of a cube is \[ SA = 6a^2 \] Substituting the value of \(a\) \[ \begin{aligned} SA &= 6(7.203)^2 \\ &= 6 \times 7.203 \times 7.203 \\ &= 311.299254 \end{aligned} \] Rounding to **four significant figures** \[ SA = 311.3\ \text{m}^2 \]

2. Volume of Cube

Volume of cube is \[ V = a^3 \] Substituting the value of \(a\) \[ \begin{aligned} V &= (7.203)^3 \\ &= 7.203 \times 7.203 \times 7.203 \\ &= 373.714754427 \end{aligned} \] Rounding to **four significant figures** \[ V = 373.7\ \text{m}^3 \]

A substance has a mass of \(5.74\ \mathrm{g}\) and occupies a volume of \(1.2\ \mathrm{cm^3}\). Calculate its density and express the answer with the correct number of significant figures.

Step 1: Identify Significant Figures

• Mass = \(5.74\ \mathrm{g}\) → 3 significant figures
• Volume = \(1.2\ \mathrm{cm^3}\) → 2 significant figures

In multiplication or division, the final answer must have the same number of significant figures as the quantity with the least significant figures.

Therefore, the final density must be expressed using 2 significant figures.

Step 2: Apply the Density Formula

Density is defined as

\[ d = \frac{\text{Mass}}{\text{Volume}} \] Substituting the given values \[ \begin{aligned} d &= \frac{5.74}{1.2} \\ &= 4.7833333333 \end{aligned} \] Rounding to 2 significant figures \[ d = 4.8\ \mathrm{g\,cm^{-3}} \]

Every physical quantity can be expressed in terms of fundamental quantities. The powers to which the fundamental quantities must be raised to represent a physical quantity are called its dimensions.

In mechanics, three fundamental quantities are used:

• Length → \([L]\)
• Mass → \([M]\)
• Time → \([T]\)

Using these fundamental dimensions, many physical quantities can be expressed.

The principle of homogeneity of dimensions states that in any valid physical equation, the dimensions of every term must be the same. In other words, the dimensions of the left-hand side (LHS) of an equation must be identical to the dimensions of the right-hand side (RHS).

This principle ensures that a physical equation remains valid irrespective of the system of units used.

Example

Consider the equation of motion:

\[ v = u + at \]

Dimensions of velocity \(v\) and initial velocity \(u\):

\[ [v] = [u] = [LT^{-1}] \]

Dimensions of \(at\):

\[ [a][t] = [LT^{-2}][T] = [LT^{-1}] \]

Since the dimensions of all terms are the same, the equation satisfies the principle of homogeneity.

Dimensional analysis is an important mathematical tool in physics. It helps scientists understand relationships between physical quantities and verify whether equations are correct.

Main Applications

• Checking the correctness of equations

  A physical equation is correct only if the dimensions of both sides of the equation are the same.

  Example: \[ F = ma \] LHS: \([MLT^{-2}]\) RHS: \([M][LT^{-2}] = [MLT^{-2}]\)
• Deriving relations between physical quantities

  When the exact formula of a physical relationship is unknown, dimensional analysis can help determine how one quantity depends on others.

  Example: Period of a pendulum depends on length and gravitational acceleration.
• Converting units from one system to another

  Dimensional relations allow conversion of physical quantities between systems such as CGS, MKS, and SI.

  Example: Converting force from dyne (CGS) to newton (SI).

Consider the equation \[ \frac{1}{2}mv^2 = mgh \] where \(m\) is the mass of the body, \(v\) is its velocity, \(g\) is the acceleration due to gravity, and \(h\) is the height. Verify whether this equation is dimensionally correct.

Step 1: Write the Given Equation

\[ \frac{1}{2}mv^2 = mgh \]

The numerical constant \( \frac{1}{2} \) is dimensionless, so it does not affect dimensional analysis.

Step 2: Dimensions of the Left-Hand Side (LHS)

LHS:

\[ \frac{1}{2}mv^2 \]

Dimension of mass \(m\):

\[ [M] \]

Dimension of velocity \(v\):

\[ [LT^{-1}] \]

Dimension of squared velocity \(v^2\):

\[ [L^2T^{-2}] \]

Therefore, dimensions of LHS become:

\[ [M][L^2T^{-2}] = [ML^2T^{-2}] \]

Step 3: Dimensions of the Right-Hand Side (RHS)

RHS:

\[ mgh \]

Dimension of mass \(m\):

\[ [M] \]

Dimension of gravitational acceleration \(g\):

\[ [LT^{-2}] \]

Dimension of height \(h\):

\[ [L] \]

Combining the dimensions:

\[ [M][LT^{-2}][L] = [ML^2T^{-2}] \]

Step 4: Comparison of Dimensions

Dimensions of LHS:

\[ [ML^2T^{-2}] \]

Dimensions of RHS:

\[ [ML^2T^{-2}] \]

Since both sides have identical dimensions, the equation satisfies the principle of homogeneity.

Consider a simple pendulum consisting of a bob attached to a string that oscillates under the action of gravity. Assume that the time period \(T\) of the pendulum depends on its length \(l\), the mass of the bob \(m\), and the acceleration due to gravity \(g\).

Using the method of dimensional analysis, derive the expression for the time period of the pendulum.

Step 1: Assume the relation

If \(T\) depends on \(l\), \(m\), and \(g\), we can write

\[ T \propto l^a m^b g^c \] or \[ T = k\, l^a m^b g^c \] where \(k\) is a dimensionless constant.

Step 2: Write dimensions of each quantity

• Time \(T\) → \([T]\)
• Length \(l\) → \([L]\)
• Mass \(m\) → \([M]\)
• Acceleration due to gravity \(g\) → \([LT^{-2}]\)

Step 3: Substitute dimensions

\[ T = k\, l^a m^b g^c \] \[ [T] = [L]^a [M]^b [LT^{-2}]^c \] \[ [T] = [L^a][M^b][L^cT^{-2c}] \] \[ [T] = [L^{a+c} M^b T^{-2c}] \]

Step 4: Compare powers of fundamental dimensions

Comparing dimensions on both sides:

For mass \(M\)

\[ b = 0 \]

For length \(L\)

\[ a + c = 0 \]

For time \(T\)

\[ -2c = 1 \] Solving: \[ c = -\frac{1}{2} \] \[ a + \left(-\frac{1}{2}\right) = 0 \] \[ a = \frac{1}{2} \]

Step 5: Substitute values of \(a\), \(b\), and \(c\)

\[ T = k\, l^{1/2} m^0 g^{-1/2} \] \[ T = k \sqrt{\frac{l}{g}} \]

Step 6: Final expression

Experimentally it is found that \[ k = 2\pi \] Therefore, \[ T = 2\pi \sqrt{\frac{l}{g}} \]

The following key points summarise the important concepts of Units and Measurements in physics. These points are useful for quick revision before examinations such as board exams, JEE, and NEET.

• Physics is a quantitative science based on the measurement of physical quantities. Certain quantities are chosen as fundamental or base quantities, such as length, mass, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity.
• Each base quantity is defined with respect to a standard reference called a unit. Examples include metre (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol), and candela (cd).
• Physical quantities derived from base quantities are called derived quantities. Their units are combinations of base units, such as velocity (\(m\,s^{-1}\)), force (newton), and energy (joule).
• A complete collection of base units and derived units forms a system of units. The most widely used system today is the International System of Units (SI).
• The SI system is based on seven base units and is accepted internationally for scientific, industrial, and everyday measurements.
• Some derived quantities have special names for convenience, such as newton (force), joule (energy), and watt (power).
• Each SI unit has a standard symbol. Examples include \(m\) for metre, \(kg\) for kilogram, \(s\) for second, \(A\) for ampere, and \(N\) for newton.
• Very large or very small physical quantities are often expressed using scientific notation and SI prefixes, which simplify calculations and indicate measurement precision.
• Proper conventions must be followed when writing symbols for physical quantities, SI units, and prefixes in scientific work.
• During calculations involving physical quantities, units should be treated algebraically so that the final result is expressed in the correct unit.
• Measured and calculated values must be reported using the correct number of significant figures. Rules of rounding off should also be followed.
• The dimensions of physical quantities describe their dependence on fundamental quantities such as mass, length, and time.
• Dimensional analysis is useful for checking the consistency of equations, deriving relationships between physical quantities, and converting units between different systems.
• A dimensionally consistent equation may not always represent the correct physical relation, but a dimensionally inconsistent equation is definitely incorrect.

The following table provides a quick summary of the most important concepts from NCERT Physics Class 11 Chapter 1 — Units and Measurements. It is designed for rapid revision before board exams, JEE, NEET, and other competitive examinations.

Concept Flow Map

Quick Revision Table

Dimensional Formula Tree