Class 11 Physics Chapter 1: Units and Measurements — Notes & Solutions

01 / Chapter Theory

Units & Measurement

Official NCERT syllabus: base quantities, modern 2018 SI definitions, significant figures, arithmetic operations, rounding off rules, and dimensional analysis with stepwise worked examples.

6 SECTIONS

Chapter Navigation

1.1Introduction 1.2The International System of Units 1.3Significant Figures & Rounding 1.4Dimensions of Physical Quantities 1.5Dimensional Formulae & Equations 1.6Dimensional Analysis & Applications

1.1

NCERT Section

Introduction

Fundamental Concept

1.1.1 Measurement and Units

Measurement of any physical quantity involves comparison with a basic, arbitrarily chosen, internationally accepted reference standard called a unit.

The result of a measurement is expressed by a numerical measure accompanied by a unit:

Physical Quantity (Q) = n × u
(where n is the numerical value and u is the standard unit)

Figure 1.0: Physical quantity invariance: Q = n₁u₁ = n₂u₂ (1 m = 100 cm). Smaller unit ⇔ larger numerical value.

Rule: As the size of the unit u changes, the numerical value n changes inversely such that n₁u₁ = n₂u₂ = constant.

Classification

1.1.2 Base vs. Derived Quantities

Base Quantities: Independent physical quantities that cannot be defined in terms of other quantities (e.g., length, mass, time). Their units are base units.
Derived Quantities: Quantities expressed as combinations of base quantities (e.g., speed, force, density). Their units are derived units.
System of Units: A complete set of units containing both base and derived units.

1.2

NCERT Section

The International System of Units

Historical Overview

1.2.1 Evolution of Metric Systems

System of Units	Length	Mass	Time
CGS System	centimetre (cm)	gram (g)	second (s)
FPS (British) System	foot (ft)	pound (lb)	second (s)
MKS System	metre (m)	kilogram (kg)	second (s)

The internationally accepted system is the Système International d'Unités (SI). Revised by the General Conference on Weights and Measures (CGPM) in November 2018, SI is decimal-based, making conversions simple and universal.

Table 1.1 • Core Reference

1.2.2 SI Base Quantities and Modern Definitions (2018 Revision)

Base Quantity	Name	Symbol	Modern SI Definition (CGPM Revision)
Length	metre	m	Defined by fixing the speed of light in vacuum c to be 299,792,458 m s⁻¹, where second is defined in terms of caesium frequency Δν_Cs.
Mass	kilogram	kg	Defined by fixing the Planck constant h to be 6.62607015 × 10⁻³⁴ J s (kg m² s⁻¹).
Time	second	s	Defined by fixing the caesium-133 ground-state hyperfine transition frequency Δν_Cs to be 9,192,631,770 Hz (s⁻¹).
Electric Current	ampere	A	Defined by fixing the elementary charge e to be 1.602176634 × 10⁻¹⁹ C (A s).
Thermodynamic Temperature	kelvin	K	Defined by fixing the Boltzmann constant k to be 1.380649 × 10⁻²³ J K⁻¹ (kg m² s⁻² K⁻¹).
Amount of Substance	mole	mol	Contains exactly 6.02214076 × 10²³ elementary entities (fixed value of Avogadro constant N_A in mol⁻¹). Entities must be specified.
Luminous Intensity	candela	cd	Defined by fixing the luminous efficacy of 540 × 10¹² Hz monochromatic radiation K_cd to be 683 lm W⁻¹.

Supplementary Units

1.2.3 Plane Angle and Solid Angle

Plane angle (dθ): Ratio of arc length ds to radius r. Unit is radian (rad).

dθ = ds / r (dimensionless: [L]/[L] = 1)
Solid angle (dΩ): Ratio of intercepted area dA to radius squared r². Unit is steradian (sr).

dΩ = dA / r² (dimensionless: [L²]/[L²] = 1)

Figure 1.1: Description of (a) plane angle dθ (rad) and (b) solid angle dΩ (sr).

Table 1.2 • Reference

1.2.4 Non-SI Units Retained for General Use

Name	Symbol	Value in SI Units
minute	min	60 s
hour	h	60 min = 3600 s
day	d	24 h = 86,400 s
year	y / yr	365.25 d = 3.156 × 10⁷ s
degree (angle)	°	(π/180) rad ≈ 1.745 × 10⁻² rad
litre	L / l	1 dm³ = 10⁻³ m³
tonne	t	10³ kg = 1000 kg
bar (pressure)	bar	10⁵ Pa = 10⁵ N m⁻²
standard atmosphere	atm	1.013 × 10⁵ Pa = 101.3 kPa
curie (activity)	Ci	3.7 × 10¹⁰ Bq (s⁻¹)
barn (area)	b	10⁻²⁸ m² = 100 fm²
hectare (area)	ha	1 hm² = 10⁴ m²

Reference

1.2.5 SI Multiples and Sub-multiples

Factor	Prefix	Symbol	Factor	Prefix	Symbol
10¹⁸	exa	E	10⁻¹	deci	d
10¹⁵	peta	P	10⁻²	centi	c
10¹²	tera	T	10⁻³	milli	m
10⁹	giga	G	10⁻⁶	micro	μ
10⁶	mega	M	10⁻⁹	nano	n
10³	kilo	k	10⁻¹²	pico	p
10²	hecto	h	10⁻¹⁵	femto	f
10¹	deca	da	10⁻¹⁸	atto	a

1.3

NCERT Section

Significant Figures

Precision & Rules

1.3.1 Significant Figures

The reported result of a measurement includes all digits known reliably plus the first uncertain digit. These are called significant figures.

Rules for Determining Significant Figures:

All non-zero digits are significant. (e.g. 2875 → 4 s.f.).
All zeros between two non-zero digits are significant. (e.g. 2.005 → 4 s.f.).
If number < 1, zeros to the right of the decimal point and left of the first non-zero digit are not significant. (e.g. 0.002308 → 4 s.f.).
Trailing zeros in a number without a decimal point are NOT significant. (e.g. 123000 → 3 s.f.).
Trailing zeros in a number with a decimal point ARE significant. (e.g. 3.500 → 4 s.f., 0.06900 → 4 s.f.).
Scientific Notation (a × 10^b) removes ambiguity. All digits in base number a (1 ≤ a < 10) are significant. Exponent b is the order of magnitude.
Exact Numbers / Multipliers (e.g. 2 in r = d/2, or π) have infinite significant figures.

Figure 1.2: Anatomy of significant figures in 0.0020450 kg (Total: 5 Significant Figures).

Rule: Changing units does not alter the number of significant figures in a measurement.

Arithmetic Rules

1.3.2 Arithmetic Operations & Rounding Rules

1. Multiplication and Division:

The result retains as many significant figures as the measurement with the least significant figures.

Density = 4.237 g (4 s.f.) / 2.51 cm³ (3 s.f.) = 1.688047... → 1.69 g cm⁻³

2. Addition and Subtraction:

The result retains as many decimal places as the number with the least decimal places.

436.32 g + 227.2 g (1 decimal place) + 0.301 g = 663.821 g → 663.8 g

3. Rounding Off Rules:

If dropped digit is > 5 → raise preceding digit by 1 (e.g. 2.746 → 2.75).
If dropped digit is < 5 → leave preceding digit unchanged (e.g. 1.743 → 1.74).
If dropped digit is exactly 5:
- Preceding digit is even → drop 5 (e.g. 2.745 → 2.74).
- Preceding digit is odd → raise preceding digit by 1 (e.g. 2.735 → 2.74).

Multi-step Rule: In multi-step calculations, retain one extra digit in intermediate steps and round off only at the end.

Example 1.1 • Surface Area & Volume of a Cube

NCERT Solved

Each side of a cube is measured to be $7.203\text{ m}$. What are the total surface area and the volume of the cube to appropriate significant figures?

Final Results: Surface Area $= 311.3\text{ m}^2$ | Volume $= 373.7\text{ m}^3$ ($4\text{ s.f.}$)

Step 1: Identify given precision: Measured edge length $a = 7.203\text{ m}$ (contains 4 significant figures). The calculated surface area and volume must be rounded to 4 significant figures.

Step 2: Calculate Surface Area: $$A = 6 a^2 = 6 \times (7.203\text{ m})^2 = 6 \times 51.883209\text{ m}^2 = 311.299254\text{ m}^2$$ Rounding to $4\text{ s.f.}$ gives $311.3\text{ m}^2$.

Step 3: Calculate Volume: $$V = a^3 = (7.203\text{ m})^3 = 373.714754\text{ m}^3$$ Rounding to $4\text{ s.f.}$ gives $373.7\text{ m}^3$.

Example 1.2 • Density with Significant Figures

NCERT Solved

$5.74\text{ g}$ of a substance occupies $1.2\text{ cm}^3$. Express its density by keeping the significant figures in view.

Final Result: Density $\rho = 4.8\text{ g cm}^{-3}$ ($2\text{ s.f.}$)

Step 1: Input precision: Mass $m = 5.74\text{ g}$ (3 significant figures). Volume $V = 1.2\text{ cm}^3$ (2 significant figures → limits result precision).

Step 2: Division and rounding: $$\rho = \frac{m}{V} = \frac{5.74\text{ g}}{1.2\text{ cm}^3} = 4.783333\dots\text{ g cm}^{-3} \approx 4.8\text{ g cm}^{-3}$$

1.4

NCERT Section

Dimensions of Physical Quantities

Dimensional Nature

1.4.1 Base Dimensions

The nature of a physical quantity is described by its dimensions, denoted with square brackets [ ]:

Base Quantity	Dimension Symbol	Base Quantity	Dimension Symbol
Length	[L]	Thermodynamic Temperature	[K]
Mass	[M]	Luminous Intensity	[cd]
Time	[T]	Amount of Substance	[mol]
Electric Current	[A]	-	-

Definition: The dimensions of a physical quantity are the powers (exponents) to which base quantities are raised to represent that quantity.

Volume = $[\text{L}] \times [\text{L}] \times [\text{L}] = [\text{M}^0 \text{L}^3 \text{T}^0]$
Force = $[\text{M}] \times [\text{L}\text{T}^{-2}] = [\text{M}\text{L}\text{T}^{-2}]$

Figure 1.3: Stepwise derivation of mechanics dimensional formulae from base dimensions [M], [L], [T].

1.5

NCERT Section

Dimensional Formulae & Equations

Reference Table

1.5.1 Physical Quantities and Dimensional Formulae

Physical Quantity	Relation	SI Unit	Dimensional Formula
Area	Length × Breadth	m²	[M⁰ L² T⁰]
Volume	Length × Breadth × Height	m³	[M⁰ L³ T⁰]
Density	Mass / Volume	kg m⁻³	[M L⁻³ T⁰]
Velocity	Displacement / Time	m s⁻¹	[M⁰ L T⁻¹]
Acceleration	Velocity / Time	m s⁻²	[M⁰ L T⁻²]
Linear Momentum	Mass × Velocity	kg m s⁻¹	[M L T⁻¹]
Force	Mass × Acceleration	N (kg m s⁻²)	[M L T⁻²]
Work / Energy	Force × Displacement	J (N m)	[M L² T⁻²]
Power	Work / Time	W (J s⁻¹)	[M L² T⁻³]
Pressure / Stress	Force / Area	Pa (N m⁻²)	[M L⁻¹ T⁻²]
Surface Tension	Force / Length	N m⁻¹	[M L⁰ T⁻²]
Gravitational Const (G)	Force × Distance² / Mass²	N m² kg⁻²	[M⁻¹ L³ T⁻²]
Planck's Constant (h)	Energy / Frequency	J s	[M L² T⁻¹]
Electric Potential	Work / Charge	V (J C⁻¹)	[M L² T⁻³ A⁻¹]
Electrical Resistance	Potential / Current	Ω	[M L² T⁻³ A⁻²]

1.6

NCERT Section

Dimensional Analysis and Applications

Application 1

1.6.1 Checking Dimensional Consistency

Principle of Homogeneity: Physical quantities can be added or subtracted only if they have the same dimensions. Every term on both sides of a valid equation must have identical dimensions.

Pure numbers and ratios (refractive index, strain) are dimensionless.
Arguments of trigonometric ($\sin\theta$), logarithmic ($\ln x$), and exponential ($e^x$) functions must be dimensionless.

Rule: A dimensionally inconsistent equation is definitely wrong. A dimensionally consistent equation is not guaranteed to be physically exact (it cannot detect dimensionless constants like 1/2 or 2π).

Example 1.3 • Testing Equation Correctness

NCERT Solved

Check whether the equation $\frac{1}{2} m v^2 = m g h$ is dimensionally correct, where $m$ is mass, $v$ is velocity, $g$ is acceleration due to gravity, and $h$ is height.

Conclusion: Dimensions of $\text{LHS} = \text{RHS} \implies$ Equation is dimensionally correct.

Step 1: Determine dimensions of LHS: $$\left[\frac{1}{2} m v^2\right] = [\text{M}] \times [\text{L}\text{T}^{-1}]^2 = [\text{M}\text{L}^2\text{T}^{-2}]$$

Step 2: Determine dimensions of RHS: $$[m g h] = [\text{M}] \times [\text{L}\text{T}^{-2}] \times [\text{L}] = [\text{M}\text{L}^2\text{T}^{-2}]$$

Step 3: Compare both sides: $$\text{LHS} = \text{RHS} = [\text{M}\text{L}^2\text{T}^{-2}]$$ The equation is dimensionally correct.

Example 1.4 • Ruling Out Incorrect Formulae

NCERT Solved

The SI unit of energy is $\text{J} = \text{kg m}^2\text{ s}^{-2}$; that of speed $v$ is $\text{m s}^{-1}$ and acceleration $a$ is $\text{m s}^{-2}$. Which formulae for kinetic energy ($K$) can be ruled out on dimensional grounds:
(a) $K = m^2 v^3$ | (b) $K = \frac{1}{2} m v^2$ | (c) $K = m a$ | (d) $K = \frac{3}{16} m v^2$ | (e) $K = \frac{1}{2} m v^2 + m a$

Conclusion: Formulae (a), (c), and (e) are ruled out.

Step 1: State target dimension of Kinetic Energy: $$[K] = [\text{M}\text{L}^2\text{T}^{-2}]$$

Step 2: Evaluate dimensions of each formula: $$(a) \quad [m^2 v^3] = [\text{M}^2\text{L}^3\text{T}^{-3}] \neq [K] \implies \textbf{Ruled out}$$ $$(b) \quad \left[\frac{1}{2} m v^2\right] = [\text{M}\text{L}^2\text{T}^{-2}] = [K] \implies \textbf{Dimensionally valid}$$ $$(c) \quad [m a] = [\text{M}\text{L}\text{T}^{-2}] \neq [K] \implies \textbf{Ruled out}$$ $$(d) \quad \left[\frac{3}{16} m v^2\right] = [\text{M}\text{L}^2\text{T}^{-2}] = [K] \implies \textbf{Dimensionally valid}$$ $$(e) \quad \left[\frac{1}{2} m v^2 + m a\right] \text{ adds } [\text{M}\text{L}^2\text{T}^{-2}] \text{ to } [\text{M}\text{L}\text{T}^{-2}] \implies \textbf{Violates homogeneity (Ruled out)}$$

Example 1.5 • Deriving Time Period of a Simple Pendulum

NCERT Solved

Derive an expression for the time period $T$ of a simple pendulum assuming it depends on length $l$, mass of the bob $m$, and acceleration due to gravity $g$.

Final Result: $T = k \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{l}{g}}$

Figure 1.4: Parameters for simple pendulum period: $l\ [\text{L}]$, $g\ [\text{L}\text{T}^{-2}]$, and mass $m\ [\text{M}]$ (exponent $= 0$).

Step 1: Set up dimensional relation with unknown exponents: $$T = k \cdot l^x \cdot g^y \cdot m^z$$

Step 2: Substitute dimensions: $$[\text{M}^0\text{L}^0\text{T}^1] = [\text{L}]^x \cdot [\text{L}\text{T}^{-2}]^y \cdot [\text{M}]^z = [\text{M}]^z [\text{L}]^{x+y} [\text{T}]^{-2y}$$

Step 3: Equate exponents on both sides: $$\text{For M: } z = 0$$ $$\text{For T: } -2y = 1 \implies y = -\frac{1}{2}$$ $$\text{For L: } x + y = 0 \implies x = -y = \frac{1}{2}$$

Step 4: Formulate final equation: $$T = k\, l^{1/2} g^{-1/2} m^0 \implies T = k \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{l}{g}}$$

02 / Self-Assessment Lab

Concept Check Questions

Select any option to reveal stepwise explanations for every individual choice, detailing why the correct answer is valid and why the alternatives are incorrect.

20 QUESTIONS

Progress: 0 / 20 Answered

Score: 0

Q01 UNANSWERED

Which of the following physical quantities is an SI base quantity?

Option A is incorrect. Force is a derived quantity (F = ma) with unit newton (N = kg m s⁻²).

Option B is correct. Luminous intensity is one of the seven fundamental SI base quantities, measured in candela (cd).

Option C is incorrect. Work is a derived quantity (W = F · d) with unit joule (J = kg m² s⁻²).

Option D is incorrect. Electric charge is derived (Q = I × t, unit: C = A s); electric current (ampere) is the base quantity.

Core Rule

The seven SI base quantities are Length (m), Mass (kg), Time (s), Electric Current (A), Temperature (K), Amount of Substance (mol), and Luminous Intensity (cd).

Q02 UNANSWERED

In the revised 2018 SI system, the kilogram is defined by fixing the numerical value of which fundamental constant?

Option A is incorrect. The speed of light c (299,792,458 m s⁻¹) defines the metre.

Option B is correct. The kilogram is defined by fixing the Planck constant h to 6.62607015 × 10⁻³⁴ kg m² s⁻¹.

Option C is incorrect. The Boltzmann constant k (1.380649 × 10⁻²³ J K⁻¹) defines the kelvin.

Option D is incorrect. The Avogadro constant N_A (6.02214076 × 10²³ mol⁻¹) defines the mole.

Core Rule

Since November 2018, all SI base units are defined by fixing exact numerical values of natural physical constants.

Q03 UNANSWERED

Both radian (rad) and steradian (sr) are:

Option A is incorrect. Angles are ratios of like geometric lengths and have no dimension of length.

Option B is incorrect. Solid angle is area / radius² ([L²]/[L²] = 1), so dimensions cancel out.

Option C is correct. Radian (ds/r) and steradian (dA/r²) are dimensionless [M⁰ L⁰ T⁰] quantities with designated SI units.

Option D is incorrect. They represent planar arc angles and spherical solid angles.

Core Rule

A quantity can possess an official SI unit while being completely dimensionless (e.g. radian and steradian).

Q04 UNANSWERED

How many significant figures are in 0.0023080 m?

Option A is incorrect. Omits the captured zero and the trailing decimal zero.

Option B is incorrect. The trailing zero after decimal indicates measured precision and must be counted.

Option C is correct. Leading zeros (0.00) are not significant. The digits '2', '3', '0' (between 3 and 8), '8', and the trailing '0' (after decimal) are all significant (5 s.f.).

Option D is incorrect. Leading zeros before the first non-zero digit are never significant.

Core Rule

Leading zeros establish magnitude only. Trailing zeros after a decimal point indicate measurement precision and are significant.

Q05 UNANSWERED

When 2.745 and 2.735 are rounded off to 3 significant figures, the results are respectively:

Option A is correct. For 2.745, preceding digit '4' is even → drop 5 → 2.74. For 2.735, preceding digit '3' is odd → raise by 1 → 2.74.

Option B is incorrect. 2.745 has an even preceding digit (4), so it stays 2.74.

Option C is incorrect. 2.735 has an odd preceding digit (3), so it must be raised to 2.74.

Option D is incorrect. Inverts the even/odd rounding rule.

Core Rule

When dropping an exact 5: if preceding digit is even, leave unchanged; if odd, raise by 1.

Q06 UNANSWERED

The sum of numbers 436.32 g, 227.2 g and 0.301 g to appropriate significant figures is:

Option A is incorrect. Contains 3 decimal places, violating the least decimal places rule.

Option B is incorrect. 227.2 g has only 1 decimal place, limiting the final result.

Option C is correct. In addition/subtraction, the result is limited to 1 decimal place (matching 227.2 g). 436.32 + 227.2 + 0.301 = 663.821 → 663.8 g.

Option D is incorrect. Unnecessarily rounds to 0 decimal places.

Core Rule

Addition and subtraction precision is dictated by decimal places, not significant figure count.

Q07 UNANSWERED

The mass of a body is 5.74 g and its volume is 1.2 cm³. Its density with proper significant figures is:

Option A is incorrect. 4.78 has 3 s.f., but volume (1.2) has only 2 s.f.

Option B is correct. Division rule: density = 5.74 / 1.2 = 4.7833... g cm⁻³. Since volume has 2 s.f., round to 2 s.f. → 4.8 g cm⁻³.

Option C is incorrect. Unrounded division output implies unwarranted precision.

Option D is incorrect. 5 has only 1 s.f., losing legitimate measured precision.

Core Rule

Multiplication and division results retain the least number of significant figures present in any input factor.

Q08 UNANSWERED

What is the dimensional formula of Universal Gravitational Constant (G)?

Option A is correct. From F = G(m₁m₂)/r²: G = F r² / m² = [M L T⁻²][L²] / [M²] = [M⁻¹ L³ T⁻²].

Option B is incorrect. [M L² T⁻²] is Work / Energy.

Option C is incorrect. Power of length is 3 and time is −2.

Option D is incorrect. Mass exponent is −1, not +1.

Core Rule

Substitute base dimensions into defining formulas to deduce dimensional expressions for physical constants.

Q09 UNANSWERED

Which pair of physical quantities has identical dimensions?

Option A is incorrect. Force is [M L T⁻²] while Torque is [M L² T⁻²].

Option B is correct. Both Work (F · d) and Torque (τ = r × F) have dimensional formula [M L² T⁻²].

Option C is incorrect. Power is Energy/time: [M L² T⁻³] vs [M L² T⁻²].

Option D is incorrect. Stress is [M L⁻¹ T⁻²], whereas Strain is dimensionless [M⁰ L⁰ T⁰].

Core Rule

Quantities with identical dimensions may represent physically distinct scalar or vector concepts.

Q10 UNANSWERED

If an equation fails the dimensional consistency test, what can be concluded with certainty?

Option A is incorrect. An equation with mismatched dimensions cannot be physically or numerically valid.

Option B is correct. By the principle of homogeneity, if dimensions differ across terms, the equation is definitely wrong.

Option C is incorrect. Dimensional errors represent invalid physical equations.

Option D is incorrect. Constants cannot resolve dimensional mismatches.

Core Rule

Dimensional consistency is necessary, but not sufficient, to prove physical validity.

Q11 UNANSWERED

The arguments of trigonometric, exponential, and logarithmic functions in physics must be:

Option A is incorrect. Trigonometric functions cannot take dimensional inputs.

Option B is incorrect. In $\sin(\omega t)$, $\omega t$ must be dimensionless ([T⁻¹][T] = 1).

Option C is correct. Mathematical function expansions require pure, dimensionless arguments.

Option D is incorrect. They must be dimensionless ratios.

Core Rule

In expressions like $y = A \sin(kx - \omega t)$, $kx$ and $\omega t$ must evaluate to pure numbers.

Q12 UNANSWERED

In deriving T = k l^x g^y m^z for a simple pendulum, the value of z is found to be:

Option A is correct. LHS [T¹] contains [M⁰]; equating mass powers yields $z = 0$, proving time period is mass-independent.

Option B is incorrect. $x = 1/2$ is the exponent of length $l$.

Option C is incorrect. $y = -1/2$ is the exponent of gravitational acceleration $g$.

Option D is incorrect. Pendulum period does not depend on mass.

Core Rule

Dimensional analysis reveals independence from variables when their power solves to zero ($m^0$).

Q13 UNANSWERED

1 light year (ly) is approximately equal to:

Option A is incorrect. $1.496 \times 10^{11}$ m is 1 Astronomical Unit (1 AU).

Option B is correct. 1 ly = $(3 \times 10^8\text{ m s}^{-1}) \times (365.25 \times 86,400\text{ s}) = 9.46 \times 10^{15}\text{ m}$.

Option C is incorrect. $3.08 \times 10^{16}$ m is 1 parsec (pc).

Option D is incorrect. $3.84 \times 10^8$ m is the mean Earth-Moon distance.

Core Rule

1 AU ($1.496 \times 10^{11}\text{ m}$) < 1 ly ($9.46 \times 10^{15}\text{ m}$) < 1 parsec ($3.08 \times 10^{16}\text{ m}$).

Q14 UNANSWERED

What are the dimensions of Planck's constant (h)?

Option A is correct. From $E = h\nu$, $h = E/\nu = [M L^2 T^{-2}] / [T^{-1}] = \mathbf{[M L^2 T^{-1}]}$ (identical to Angular Momentum).

Option B is incorrect. [M L² T⁻²] represents Energy.

Option C is incorrect. [M L T⁻¹] represents Linear Momentum.

Option D is incorrect. [M L² T⁻³] represents Power.

Core Rule

Planck's constant ($h$) and Angular Momentum ($L = r \times p$) share identical dimensions [M L² T⁻¹].

Q15 UNANSWERED

What is the order of magnitude of the diameter of the Earth (1.28 × 10⁷ m)?

Option A is correct. Since 1.28 ≤ 5, it rounds to 1 × 10⁷ m, giving order of magnitude 10⁷ m.

Option B is incorrect. Only values > 5 round up to $10^1$.

Option C is incorrect. $10^6$ m is one order too small.

Option D is incorrect. Order of magnitude is stated as a pure power of 10.

Core Rule

For $a \times 10^b$: if $a \le 5$, order is $10^b$; if $a > 5$, order is $10^{b+1}$.

Q16 UNANSWERED

The dimensional formula for electric potential is:

Option A is correct. Potential $V = \text{Work} / \text{Charge} = [M L^2 T^{-2}] / [A T] = \mathbf{[M L^2 T^{-3} A^{-1}]}$.

Option B is incorrect. Missing time in charge denominator ($Q = I \times t$).

Option C is incorrect. [M L T⁻³ A⁻¹] is Electric Field.

Option D is incorrect. [M L² T⁻³ A⁻²] is Resistance.

Core Rule

Potential = Work / Charge; Resistance = Potential / Current = [M L² T⁻³ A⁻²].

Q17 UNANSWERED

Which of the following equations is dimensionally INCORRECT?

Option A is dimensionally correct. Each term has dimension [L T⁻¹].

Option B is dimensionally correct. Each term has dimension [L].

Option C is INCORRECT (hence the right choice). $v^2$ has [L² T⁻²], but $2a/s = [L T^{-2}]/[L] = [T^{-2}]$, violating homogeneity. Correct relation is $v^2 = u^2 + 2as$.

Option D is dimensionally correct. Impulse matches change in momentum [M L T⁻¹].

Core Rule

All terms combined by addition or subtraction must have identical dimensions.

Q18 UNANSWERED

Why must intermediate results in multi-step calculations retain one extra digit?

Option A is incorrect. Computations cannot alter physical instrument least counts.

Option B is correct. Retaining an extra digit prevents cumulative truncation errors in intermediate steps.

Option C is incorrect. Arithmetic does not affect dimensions.

Option D is incorrect. Measured experimental data have finite significant figures.

Core Rule

Round off only at the final step to preserve arithmetic precision.

Q19 UNANSWERED

1 barn is a unit of nuclear cross-section area equal to:

Option A is correct. Table 1.2 defines 1 barn (b) = 10⁻²⁸ m² = 100 fm².

Option B is incorrect. $10^{-15}$ m is 1 femtometre (length), not area.

Option C is incorrect. $10^4$ m² is 1 hectare.

Option D is incorrect. $10^{-10}$ m is 1 Ångström.

Core Rule

1 barn ($10^{-28}\text{ m}^2$) is a non-SI unit retained for nuclear physics cross-section measurements.

Q20 UNANSWERED

Dimensional analysis cannot deduce a formula if the quantity depends on:

Option A is incorrect. Product-type dependencies with 3 variables are easily solvable.

Option B is correct. The power-product method ($Q = k A^x B^y C^z$) cannot deduce relations containing additive composite terms.

Option C is incorrect. Lengths and velocities can be combined using powers.

Option D is incorrect. Force ($F = ma$) is a product of mass and acceleration.

Core Rule

Dimensional analysis cannot determine dimensionless constants, exponential/trigonometric expressions, or additive composite formulas.

03 / Textbook Solutions

NCERT Exercises 1.1 – 1.17

Complete stepwise solutions for every textbook exercise question in NCERT Class 11 Physics Chapter 1 Reprint 2026-27.

17 QUESTIONS

Ex 1.1 Unit Conversions

1.1 Fill in the blanks:

(a)

The volume of a cube of side 1 cm is equal to _____ m³

(b)

The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to _____ (mm)²

(c)

A vehicle moving with a speed of 18 km h⁻¹ covers _____ m in 1 s

(d)

The relative density of lead is 11.3. Its density is _____ g cm⁻³ or _____ kg m⁻³.

Prerequisite: ↗ 1.2 The International System of Units

Answers: (a) 10⁻⁶ m³ | (b) 1.5 × 10⁴ (mm)² | (c) 5 m | (d) 11.3 g cm⁻³ or 1.13 × 10⁴ kg m⁻³

(a) Volume of cube:
Side a = 1 cm = 10⁻² m.
Volume V = a³ = (10⁻² m)³ = 10⁻⁶ m³.

(b) Surface area of cylinder:
r = 2.0 cm = 20 mm, h = 10.0 cm = 100 mm.
Total surface area A = 2πr(r + h) = 2 × 3.1416 × 20 × (20 + 100) = 15,079.6 mm² → 1.5 × 10⁴ (mm)² (to 2 s.f.).

(c) Speed conversion:
Speed v = 18 km h⁻¹ = 18 × (1000 m / 3600 s) = 5 m s⁻¹.
Distance in 1 s = v × t = 5 m s⁻¹ × 1 s = 5 m.

(d) Density from Relative Density:
In CGS: Density = 11.3 × 1 g cm⁻³ = 11.3 g cm⁻³.
In SI: Density = 11.3 × 10³ kg m⁻³ = 1.13 × 10⁴ kg m⁻³.

Ex 1.2 Unit Conversions

1.2 Fill in the blanks by suitable conversion of units:

(a)

1 kg m² s⁻² = _____ g cm² s⁻²

(b)

1 m = _____ ly

(c)

3.0 m s⁻² = _____ km h⁻²

(d)

G = 6.67 × 10⁻¹¹ N m² (kg)⁻² = _____ (cm)³ s⁻² g⁻¹.

Prerequisite: ↗ 1.2.4 Non-SI Units Retained

Answers: (a) 10⁷ | (b) 1.06 × 10⁻¹⁶ | (c) 3.89 × 10⁴ | (d) 6.67 × 10⁻⁸

(a) Energy conversion (Joule to Erg):
1 kg m² s⁻² = (10³ g) × (10⁴ cm²) × s⁻² = 10⁷ g cm² s⁻².

(b) Metre to Light-year:
1 ly = (3 × 10⁸ m s⁻¹) × (365.25 × 86400 s) = 9.46 × 10¹⁵ m.
1 m = 1 / (9.46 × 10¹⁵) ly = 1.06 × 10⁻¹⁶ ly.

(c) Acceleration conversion:
3.0 m s⁻² = 3.0 × (10⁻³ km) × (3600 s / 1 h)² = 3.89 × 10⁴ km h⁻².

(d) Gravitational Constant conversion:
G = 6.67 × 10⁻¹¹ m³ s⁻² kg⁻¹ = 6.67 × 10⁻¹¹ × (10² cm)³ × s⁻² × (10³ g)⁻¹
= 6.67 × 10⁻¹¹ × 10⁶ × 10⁻³ = 6.67 × 10⁻⁸ (cm)³ s⁻² g⁻¹.

Ex 1.3 System of Units

1.3 A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1 J = 1 kg m² s⁻². Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α⁻¹ β⁻² γ² in terms of the new units.

Prerequisite: ↗ 1.6 Dimensional Analysis

Proven: n₂ = 4.2 α⁻¹ β⁻² γ² [new units]

Step 1: State dimensions of energy.
[Heat Energy] = [M¹ L² T⁻²] → a = 1, b = 2, c = −2.

Step 2: Apply conversion formula.
n₂ = n₁ × [M₁ / M₂]^a [L₁ / L₂]^b [T₁ / T₂]^c

Step 3: Substitute values.
n₂ = 4.2 × [1 kg / α kg]¹ × [1 m / β m]² × [1 s / γ s]⁻² = 4.2 α⁻¹ β⁻² γ².

Ex 1.4 Conceptual Precision

1.4 Explain this statement clearly:
“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”. In view of this, reframe the following statements wherever necessary:
(a) atoms are very small objects
(b) a jet plane moves with great speed
(c) the mass of Jupiter is very large
(d) the air inside this room contains a large number of molecules
(e) a proton is much more massive than an electron
(f) the speed of sound is much smaller than the speed of light.

Prerequisite: ↗ 1.1 Measurement Reference Standard

Explanation: A dimensional quantity has physical meaning only when compared against a defined standard or relative scale.

Reframed Statements:
• (a) Atoms are very small compared to the size of a pinhead (atomic size ≈ 10⁻¹⁰ m).
• (b) A jet plane moves with great speed compared to an express train.
• (c) The mass of Jupiter is very large compared to the mass of the Earth (≈ 318 times).
• (d) The air inside this room contains a very large number of molecules compared to a mole (≈ 10²⁷ molecules).
• (e) A proton is about 1836 times more massive than an electron.
• (f) The speed of sound (≈ 340 m s⁻¹) is much smaller than the speed of light (≈ 3 × 10⁸ m s⁻¹).

Ex 1.5 Relative Units

1.5 A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Prerequisite: ↗ 1.1 Physical Quantities

Answer: Distance = 500 new units of length

Step 1: Identify given quantities in new system.
Speed of light c = 1 new unit s⁻¹.
Time t = 8 min 20 s = (8 × 60 + 20) s = 500 s.

Step 2: Calculate distance.
Distance d = c × t = 1 new unit s⁻¹ × 500 s = 500 new units of length.

Ex 1.6 Least Count & Precision

1.6 Which of the following is the most precise device for measuring length:
(a) a vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light?

Prerequisite: ↗ 1.3 Least Count and Precision

Answer: (c) Optical instrument is the most precise.

Step 1: Calculate least count of each instrument.
• (a) Vernier callipers: Least count = 1 mm / 20 = 0.05 mm = 5 × 10⁻⁵ m.
• (b) Screw gauge: Least count = 1 mm / 100 = 0.01 mm = 10⁻⁵ m.
• (c) Optical instrument: Precision ≈ wavelength of visible light λ ≈ 6000 Å = 6 × 10⁻⁷ m = 0.0006 mm.

Step 2: Compare least counts.
The optical instrument has the minimum least count (10⁻⁷ m) and is therefore the most precise.

Ex 1.7 Measurement & Magnification

1.7 A student measures the thickness of a human hair by looking at it through a microscope of magnification 100. He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm. What is the estimate on the thickness of hair?

Prerequisite: ↗ 1.3 Measurement Error

Answer: Estimated thickness = 0.035 mm = 35 μm

Step 1: State relationship between magnification and size.
Magnification M = Observed Image Width / Actual Object Thickness.

Step 2: Calculate actual thickness.
Actual thickness = 3.5 mm / 100 = 0.035 mm = 35 μm.

Ex 1.8 Experimental Precision

1.8 Answer the following:
(a) You are given a thread and a metre scale. How will you estimate the diameter of the thread?
(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale. Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
(c) The mean diameter of a thin brass rod is to be measured by vernier callipers. Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?

Prerequisite: ↗ 1.3 Uncertainty & Random Errors

(a) Diameter of a thread:
Wind the thread closely in N touching turns along a rod. Measure the total length L with the metre scale. The diameter is d = L / N.

(b) Arbitrary increase in screw gauge accuracy:
No. Increasing divisions reduces theoretical least count, but human eye resolution, backlash error, zero error, and mechanical imperfections set a physical limit to accuracy.

(c) 100 vs 5 measurements:
Random errors fluctuate positively and negatively. In the arithmetic mean of 100 observations, random errors tend to cancel out. The standard error is reduced by a factor of √N.

Ex 1.9 Linear vs Areal Magnification

1.9 The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement?

Prerequisite: ↗ 1.1 Geometric Relations

Answer: Linear magnification ≈ 94.1

Step 1: Convert screen area to cm².
A_screen = 1.55 m² = 1.55 × 10⁴ cm² = 15,500 cm².

Step 2: Compute areal magnification.
m_area = 15,500 cm² / 1.75 cm² = 8857.14.

Step 3: Compute linear magnification.
m_linear = √(m_area) = √(8857.14) ≈ 94.1.

Ex 1.10 Significant Figures

1.10 State the number of significant figures in the following:
(a) 0.007 m² | (b) 2.64 × 10²⁴ kg | (c) 0.2370 g cm⁻³ | (d) 6.320 J | (e) 6.032 N m⁻² | (f) 0.0006032 m²

Prerequisite: ↗ 1.3.1 Rules for Significant Figures

Value	Significant Figures	Reason
(a) 0.007 m²	1	Leading zeros are not significant.
(b) 2.64 × 10²⁴ kg	3	Digits in coefficient 2.64 are significant; power of 10 is ignored.
(c) 0.2370 g cm⁻³	4	Trailing zero after decimal point is significant.
(d) 6.320 J	4	Trailing zero after decimal point is significant.
(e) 6.032 N m⁻²	4	Zero between non-zero digits is significant.
(f) 0.0006032 m²	4	Leading zeros are not significant; internal zero is significant.

Ex 1.11 Significant Figures in Geometry

1.11 The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Prerequisite: ↗ 1.3.2 Arithmetic with Significant Figures

Answers: Total Surface Area = 8.72 m² (3 s.f.) | Volume = 0.0855 m³ (3 s.f.)

Step 1: State dimensions and identify least s.f.
l = 4.234 m (4 s.f.), b = 1.005 m (4 s.f.), t = 2.01 cm = 0.0201 m (3 s.f.).

Step 2: Compute Total Surface Area.
A = 2(lb + bt + tl) = 2(4.25517 + 0.0202005 + 0.0851034) = 8.720948 m² → 8.72 m² (3 s.f.).

Step 3: Compute Volume.
V = l × b × t = 4.234 × 1.005 × 0.0201 m³ = 0.0855289 m³ → 0.0855 m³ (3 s.f.).

Ex 1.12 Decimal Places Rule

1.12 The mass of a box measured by a grocer’s balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?

Prerequisite: ↗ 1.3.2 Decimal Place Rule

Answers: (a) Total mass = 2.34 kg | (b) Mass difference = 0.02 g

(a) Total mass:
2.30 kg + 0.02015 kg + 0.02017 kg = 2.34032 kg.
Since 2.30 kg has only 2 decimal places, round to 2 decimal places → 2.34 kg.

(b) Difference in piece masses:
20.17 g − 20.15 g = 0.02 g (2 decimal places).

Ex 1.13 Homogeneity & Relativity

1.13 A relation in physics relates moving mass m to rest mass m₀, speed v, and speed of light c. A student writes:

m = m₀ / (1 − v²)^1/2

Guess where to put the missing c.

Prerequisite: ↗ 1.6.1 Principle of Homogeneity

Answer: Place c² under v² → m = m₀ / [1 − (v² / c²)]^1/2

Step 1: Check dimensions of LHS and numerator.
[m] = [M] and [m₀] = [M]. The denominator must be dimensionless.

Step 2: Apply homogeneity to denominator.
In (1 − v²), the number 1 is dimensionless. Thus v² must be divided by a quantity having dimension of velocity squared, which is c².

Step 3: State correct formula.
m = m₀ / √(1 − v²/c²).

Ex 1.14 Atomic Scale

1.14 The unit of length on the atomic scale is an angstrom (Å: 1 Å = 10⁻¹⁰ m). The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m³ of a mole of hydrogen atoms?

Prerequisite: ↗ 1.2 Avogadro Number & Moles

Answer: Total atomic volume = 3.15 × 10⁻⁷ m³

Step 1: Calculate volume of one atom (r = 0.5 × 10⁻¹⁰ m).
V₁ = (4/3)πr³ = (4/3) × 3.1416 × (0.5 × 10⁻¹⁰ m)³ = 5.236 × 10⁻³¹ m³.

Step 2: Multiply by Avogadro's number.
V_mole = N_A × V₁ = (6.022 × 10²³) × (5.236 × 10⁻³¹ m³) = 3.15 × 10⁻⁷ m³.

Ex 1.15 Molar vs. Atomic Volume

1.15 One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1 Å). Why is this ratio so large?

Prerequisite: ↗ 1.2 Table 1.2 Practical Units (Litre)

Answer: Ratio ≈ 7.1 × 10⁴

Step 1: State molar volume in SI units.
V_molar = 22.4 L = 2.24 × 10⁻² m³.

Step 2: State molecular volume for 1 mole.
With radius r = 0.5 Å, V_molecules ≈ 3.15 × 10⁻⁷ m³.

Step 3: Compute ratio & explain.
Ratio = (2.24 × 10⁻²) / (3.15 × 10⁻⁷) ≈ 7.1 × 10⁴.
The ratio is large because gas molecules are separated by distances much larger than their molecular dimensions; most of the macroscopic gas volume is empty space.

Ex 1.16 Parallax & Relative Motion

1.16 Explain this observation clearly:
If you look out of the window of a fast moving train, nearby trees seem to move rapidly in a direction opposite to the train’s motion, but distant objects (hill tops, the Moon) seem to be stationary.

Prerequisite: ↗ 1.2.3 Plane Angle dθ = ds / r

₁) Large ω = v/r₁ → Fast relative sweep Distant Objects (huge r₂) ω ≈ 0 → Appears stationary

Figure 1.5: Angular velocity of line of sight ω = dθ/dt ≈ v/r. Near objects sweep rapidly; distant objects appear stationary.

Explanation: Perception of motion depends on the rate of change of the angle of the line of sight (dθ/dt ≈ v/r).

Nearby objects (small r): Subtend a large angular change dθ in a short time, appearing to sweep rapidly backwards.

Distant objects (very large r): Subtend an extremely small angular shift dθ ≈ 0, making their direction appear virtually unchanged (stationary).

Ex 1.17 Density of the Sun

1.17 The Sun is a hot plasma with inner core temperature > 10⁷ K and outer surface at 6000 K. In what range do you expect the mass density of the Sun to be (solids/liquids or gases)? Check from data: mass = 2.0 × 10³⁰ kg, radius = 7.0 × 10⁸ m.

Prerequisite: ↗ 1.1 Derived Density Formula

Answer: Average density ≈ 1.39 × 10³ kg m⁻³ (in the range of solids and liquids)

Step 1: Calculate volume of the Sun.
V = (4/3)πR³ = (4/3) × 3.1416 × (7.0 × 10⁸ m)³ = 1.437 × 10²⁷ m³.

Step 2: Calculate density.
ρ = M / V = (2.0 × 10³⁰ kg) / (1.437 × 10²⁷ m³) ≈ 1.39 × 10³ kg m⁻³.

Step 3: Physical interpretation.
The density of the Sun (≈ 1.4 × 10³ kg m⁻³) is in the range of liquids and solids (water is 1.0 × 10³ kg m⁻³), not gases, due to strong inward gravitational compression.

04 / Rapid Reference

Chapter Summary & Formulas

A comprehensive cheat sheet containing definitions, unit conversion formulas, significant figures, error combination rules, dimensional tables, conditions, and exam tips.

100% SYLLABUS

1.1 & 1.2 • Units & Systems

Physical Quantities & The SI System

A physical quantity is any property that can be measured. It is represented as $Q = n \cdot u$ (where $n$ is the numerical value and $u$ is the unit).

Unit Invariance Principle: $n_1 u_1 = n_2 u_2 = \text{constant}$. Thus, $n \propto \frac{1}{u}$. Larger unit $\implies$ smaller numerical value.

The 7 SI Base Quantities

Base Quantity	SI Unit	Symbol	Dimension	NCERT Definition Standard (Modern 2018)
Length	metre	$\text{m}$	$[\text{L}]$	Defined by taking the fixed numerical value of the speed of light in vacuum $c = 299,792,458\text{ m/s}$.
Mass	kilogram	$\text{kg}$	$[\text{M}]$	Defined by taking the fixed numerical value of Planck's constant $h = 6.62607015 \times 10^{-34}\text{ J s}$.
Time	second	$\text{s}$	$[\text{T}]$	Defined by the caesium frequency $\Delta \nu_{\text{Cs}} = 9,192,631,770\text{ Hz}$.
Electric Current	ampere	$\text{A}$	$[\text{I}]$ or $[\text{A}]$	Defined by taking the fixed numerical value of elementary charge $e = 1.602176634 \times 10^{-19}\text{ C}$.
Thermodynamic Temp	kelvin	$\text{K}$	$[\text{K}]$ or $[\theta]$	Defined by taking the fixed numerical value of Boltzmann constant $k = 1.380649 \times 10^{-23}\text{ J/K}$.
Amount of Substance	mole	$\text{mol}$	$[\text{N}]$	Contains exactly $6.02214076 \times 10^{23}$ elementary entities (Avogadro's number $N_A$).
Luminous Intensity	candela	$\text{cd}$	$[\text{J}]$	Luminous intensity of a source of monochromatic radiation of frequency $540 \times 10^{12}\text{ Hz}$ and radiant intensity of $1/683\text{ W/sr}$.

Supplementary SI Units

Plane Angle ($d\theta$): $d\theta = \frac{ds}{r}\text{ (radian, rad)}$. Full circle $= 2\pi\text{ rad} = 360^\circ$. Condition: Always convert degrees to radians in physics equations: $1^\circ = \frac{\pi}{180}\text{ rad} \approx 1.745 \times 10^{-2}\text{ rad}$.
Solid Angle ($d\Omega$): $d\Omega = \frac{dA}{r^2}\text{ (steradian, sr)}$. Full sphere $= 4\pi\text{ sr}$.

Trick: Plane angle and Solid angle are dimensionless ($[\text{M}^0\text{L}^0\text{T}^0]$) but possess physical units. They are the only such quantities in physics!

1.3 • Significant Figures & Rounding

Rules for Significant Figures (S.F.)

Significant figures represent the digits in a measured value that are known with certainty plus one digit that is uncertain.

Rule Type	Counting Strategy	Examples
Non-Zero Digits	All non-zero digits are significant.	$142.8\text{ cm}$ has 4 S.F.
Sandwiched Zeros	Zeros between non-zero digits are always significant.	$100.08\text{ s}$ has 5 S.F.
Leading Zeros	Zeros to the left of the first non-zero digit are never significant.	$0.0025\text{ kg}$ has 2 S.F. ($2, 5$)
Trailing Zeros (No Decimal)	Trailing zeros in a number without a decimal point are NOT significant.	$4800\text{ m}$ has 2 S.F. ($4, 8$)
Trailing Zeros (With Decimal)	Trailing zeros in a number with a decimal point ARE significant.	$4.800\text{ V}$ has 4 S.F.; $0.0560\text{ g}$ has 3 S.F.
Exact Numbers & Constants	Exact counts or mathematical constants have infinite significant figures.	$2\pi$, $c$, or "50 apples" $\rightarrow$ $\infty$ S.F.

Rounding Rules for "Exactly 5" (The Even-Odd Rule)

When the dropped digit is exactly $5$ (or $5$ followed only by zeros):

If the preceding digit is odd $\rightarrow$ increase it by 1 (e.g., $4.735 \rightarrow 4.74$).
If the preceding digit is even $\rightarrow$ leave it unchanged (e.g., $4.725 \rightarrow 4.72$).

Mnemonic: "Odd Opens the door (adds 1), Even stands Equal (stays same)."

Arithmetic Operations Rules

Operation	Rule	Example
Addition & Subtraction	Match the minimum decimal places.	$436.32 + 227.2 = 663.52 \rightarrow \mathbf{663.5}$ (1 decimal place)
Multiplication & Division	Match the minimum significant figures.	$4.237 \div 2.51 = 1.68804 \rightarrow \mathbf{1.69}$ (3 S.F.)

Exam Tip: In multi-step calculations, retain one extra significant figure in intermediate calculations to avoid accumulated rounding errors. Round off only the final answer.

1.3 (Syllabus) • Error Analysis

Errors in Measurement & Combination of Errors

Error is the uncertainty in a measurement. If $a_1, a_2, \dots, a_n$ are $n$ measurements, then:

Mean Value ($a_{\text{mean}}$): $a_{\text{mean}} = \frac{1}{n}\sum_{i=1}^{n} a_i$
Absolute Error ($\Delta a_i$): $\Delta a_i = |a_{\text{mean}} - a_i|$ (Always positive)
Mean Absolute Error ($\Delta a_{\text{mean}}$): $\Delta a_{\text{mean}} = \frac{1}{n}\sum_{i=1}^{n} |\Delta a_i|$
Relative (Fractional) Error: $\delta a = \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}}$
Percentage Error: $\% \text{ Error} = \frac{\Delta a_{\text{mean}}}{a_{\text{mean}}} \times 100\%$

Propagation (Combination) of Errors

Mathematical Form	Error Formula	Rule Description & Conditions
Sum: $Z = A + B$	$\Delta Z = \Delta A + \Delta B$	Absolute errors add up. Condition: Even in subtraction, the absolute errors MUST be added, never subtracted.
Difference: $Z = A - B$	$\Delta Z = \Delta A + \Delta B$
Product: $Z = A \cdot B$	$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$	Relative (fractional) errors add up. Condition: Multiply by 100 to convert to percentage errors directly: $\%_Z = \%_A + \%_B$.
Quotient: $Z = \frac{A}{B}$	$\frac{\Delta Z}{Z} = \frac{\Delta A}{A} + \frac{\Delta B}{B}$
Powers / General: $Z = \frac{A^p B^q}{C^r}$	$\frac{\Delta Z}{Z} = p\frac{\Delta A}{A} + q\frac{\Delta B}{B} + r\frac{\Delta C}{C}$	Multiply fractional errors by exponents. Trick: The power $r$ in denominator still contributes positively ($+ r\frac{\Delta C}{C}$) to maximize error.

Common Trap: If $y = \sin(x)$ or $y = \ln(x)$, standard algebraic propagation rules do not apply. Use calculus differentiation: $dy = \cos(x) dx \implies \Delta y = |\cos(x)| \Delta x$.

1.4, 1.5, 1.6 • Dimensional Analysis

Dimensions, Homogeneity & Applications

Dimensions are the powers to which the base quantities are raised to represent a physical quantity. Mechanics quantities are expressed in terms of $[\text{M}]$, $[\text{L}]$, $[\text{T}]$.

Principle of Homogeneity of Dimensions

A physical equation is correct only if the dimensions of all the terms on both sides of the equation are the same. We can only add or subtract quantities with identical dimensions.

Function Arguments Constraint: The arguments of trigonometric ($\sin \theta, \cos \theta$), exponential ($e^x$), and logarithmic ($\ln x$) functions must be completely dimensionless ($[\text{M}^0\text{L}^0\text{T}^0]$).

Three Primary Applications & Formulas

Checking Consistency: Verify if LHS dimensions $=$ RHS dimensions. (Necessary but not sufficient: equations can be dimensionally correct but numerically wrong due to dimensionless constants).
Unit Conversion: Convert numerical value $n_1$ in unit system 1 to $n_2$ in unit system 2: $$n_2 = n_1 \left[\frac{M_1}{M_2}\right]^a \left[\frac{L_1}{L_2}\right]^b \left[\frac{T_1}{T_2}\right]^c$$
Deducing Relations: Set up power-law relations $Q \propto x^a y^b z^c$ and solve for exponents by equating dimensions.

Limitations of Dimensional Analysis

Cannot determine dimensionless proportionality constants (e.g., the factor of $2\pi$ in $T = 2\pi\sqrt{l/g}$).
Cannot derive relationships involving addition/subtraction terms (e.g., $s = ut + \frac{1}{2}at^2$).
Cannot derive relationships involving trigonometric, logarithmic, or exponential functions.
If a physical quantity depends on more than three independent physical quantities in mechanics, we cannot solve for exponents (since we only have 3 equations from $[\text{M}]$, $[\text{L}]$, and $[\text{T}]$).

Twin Dimensions Reference (Highly Asked in Exams)

Physical Quantities Group	Dimensional Formula	Common SI Unit
Work, Energy, Heat, Torque, Moment of Force	$[\text{M L}^2\text{ T}^{-2}]$	$\text{J}$ or $\text{N m}$
Pressure, Stress, Young's Modulus, Energy Density	$[\text{M L}^{-1}\text{ T}^{-2}]$	$\text{Pa}$ or $\text{N m}^{-2}$ or $\text{J m}^{-3}$
Impulse, Linear Momentum	$[\text{M L T}^{-1}]$	$\text{kg m s}^{-1}$ or $\text{N s}$
Planck's Constant ($h$), Angular Momentum ($L$)	$[\text{M L}^2\text{ T}^{-1}]$	$\text{J s}$ or $\text{kg m}^2\text{ s}^{-1}$
Frequency, Angular Velocity, Decay Constant, Velocity Gradient	$[\text{M}^0\text{L}^0\text{T}^{-1}]$	$\text{s}^{-1}$ or $\text{Hz}$
Surface Tension, Surface Energy, Spring Constant	$[\text{M L}^0\text{ T}^{-2}]$	$\text{N m}^{-1}$ or $\text{J m}^{-2}$
Strain, Refractive Index, Relative Permittivity ($\epsilon_r$), Angle	$[\text{M}^0\text{L}^0\text{T}^0]$ (Dimensionless)	No units (except Angle which has rad)

Summary Sheet

Mnemonics & Rapid Exam Tips

Mnemonic for SI Base Units: "Loud Music Troubles Every Teacher's Attentive Listening" $\rightarrow$ Length, Mass, Time, Electricity (Current), Temperature, Amount of substance, Luminous intensity.
Significant Figures Checklist: Decimal point present? Trailing zeros COUNT. Decimal point absent? Trailing zeros DO NOT count. Leading zeros NEVER count.
Error Combinations: Never subtract errors! Whether you add or subtract the main quantities, or multiply/divide them, their absolute/relative errors always accumulate to make the maximum possible error.
Dimensionless Constants: Constants like $2\pi$, $e$, $\log(x)$, and numbers in formulas are dimensionally invisible. Never try to derive them using dimensional analysis.
Homogeneity Check: In $y = a \sin(bt - cx)$, the quantity $bt$ must be dimensionless ($[b] = [\text{T}^{-1}]$) and $cx$ must be dimensionless ($[c] = [\text{L}^{-1}]$), and the entire term $a \sin(bt-cx)$ has dimensions of $[a]$.

05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to calculate score and review answers.

24 QUESTIONS

1. How many base units are defined in the SI system?

2. The SI unit of luminous intensity is:

3. 1 cm³ is equal to:

4. Number of significant figures in 0.007 m² is:

5. 3.500 has how many significant figures?

6. The dimensional formula of volume is:

7. The dimensions of Force are:

8. If an equation has a dimensional mismatch, the equation is definitely:

1. 18 km h⁻¹ converted to SI unit equals:

2. 4.237 ÷ 2.51 reported to correct significant figures is:

3. 436.32 + 227.2 + 0.301 should be reported as:

4. 3.0 m s⁻² in km h⁻² is:

5. If c = 1 new unit/s and light travels for 500 s, the distance is:

6. Hair image width 3.5 mm under 100× magnification gives actual thickness:

7. For addition and subtraction, final precision is determined by:

8. Which of the following is correct regarding dimensional consistency?

1. For simple pendulum period T = k l^x g^y m^z, the exponents are:

2. Why cannot dimensional analysis distinguish between ½mv² and (3/16)mv²?

3. For l = 16.2 ± 0.1 cm (0.6%) and b = 10.1 ± 0.1 cm (1.0%), the combined percentage uncertainty in area is:

4. Why retain an extra digit in intermediate multi-step calculations?

5. Which of the following is completely dimensionless?

6. For x = x₀ + v₀t + ½at², the dimension of the term v₀t is:

7. A 1.55 m² projected image from 1.75 cm² slide gives linear magnification closest to:

8. Dimensional analysis cannot determine:

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