📘 Lecture No. 1: Introduction to Units and Dimensions (JEE 2026)

🔷 1. Introduction

Physics is the study of nature and its laws. To understand these laws quantitatively, we need measurement. For accurate measurement, we must define units for physical quantities.

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🔷 2. Physical Quantities

A Physical Quantity is a quantity that can be measured and expressed in terms of a unit.

Types:

·         Fundamental Quantities: Independent quantities (e.g., mass, length, time).

·         Derived Quantities: Formed using fundamental quantities (e.g., velocity, force).

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🔷 3. Units

A unit is a standard reference used to measure physical quantities.

✅ Types of Units:

1.      Fundamental Units (7 in number – as per SI system)

2.      Derived Units (e.g., m/s, N, J, etc.)

3.      Supplementary Units (like radian and steradian)

✅ Systems of Units:

·         CGS: centimeter, gram, second

·         MKS: meter, kilogram, second

·         FPS: foot, pound, second

·         SI (Standard International): globally accepted

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🔷 4. SI Base Units (7 Units Table)

Physical Quantity	Unit	Symbol
Length	meter	m
Mass	kilogram	kg
Time	second	s
Temperature	kelvin	K
Electric current	ampere	A
Luminous intensity	candela	cd
Amount of substance	mole	mol

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🔷 5. Dimensions

The dimensions of a physical quantity show how it depends on fundamental quantities.

✅ Dimensional Formula

It is expressed in terms of powers of fundamental quantities.

Example:

·         Velocity = distance/time = [M⁰ L¹ T⁻¹]

·         Force = mass × acceleration = [M¹ L¹ T⁻²]

✅ Dimensional Equation

It is an equation showing the relation between derived and fundamental quantities using dimensional formula.

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🔷 6. Uses of Dimensional Analysis

1.      Checking the correctness of physical equations

2.      Converting units from one system to another

3.      Deriving formulae

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🔷 7. Limitations of Dimensional Analysis

·         Cannot determine dimensionless constants (like 1/2, π)

·         Only valid for equations that are dimensionally homogeneous

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🎯 Important Formulas

Quantity	Dimensional Formula
Velocity	[M⁰ L¹ T⁻¹]
Acceleration	[M⁰ L¹ T⁻²]
Force (F=ma)	[M¹ L¹ T⁻²]
Work (W=F×d)	[M¹ L² T⁻²]
Power (P=W/t)	[M¹ L² T⁻³]
Pressure	[M¹ L⁻¹ T⁻²]

📝 MCQs: Units and Dimensions

1. Which of the following pairs have the same dimensions?

A. Force and Pressure
B. Work and Torque
C. Energy and Power
D. Momentum and Impulse

Answer: ✅ D. Momentum and Impulse

Explanation: Both have dimensions [M¹L¹T⁻¹]

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2. Dimensional formula of universal gravitational constant (G) is:

A. [M⁰L³T⁻²]
B. [M⁻¹L³T⁻²]
C. [M⁻²L³T⁻¹]
D. [M⁻¹L²T⁻²]

Answer: ✅ B. [M⁻¹L³T⁻²]

From Newton's Law: F = Gm₁m₂/r² ⇒ G = Fr²/m₁m₂

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3. What is the dimensional formula for pressure?

A. [M¹L⁻¹T⁻²]
B. [M¹L¹T⁻²]
C. [M⁰L⁻²T²]
D. [M¹L²T⁻³]

Answer: ✅ A. [M¹L⁻¹T⁻²]

Pressure = Force / Area = [M¹L¹T⁻²]/[L²] = [M¹L⁻¹T⁻²]

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4. Which of the following is not a derived unit?

A. Joule
B. Meter
C. Pascal
D. Newton

Answer: ✅ B. Meter

Meter is a fundamental unit (SI unit of length)

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5. The dimensional formula of Planck's constant (h) is:

A. [M¹L¹T⁻¹]
B. [M¹L²T⁻¹]
C. [M⁰L²T⁻³]
D. [M¹L²T⁻²]

Answer: ✅ B. [M¹L²T⁻¹]

From E = hν ⇒ h = E/ν = [ML²T⁻²]/[T⁻¹] = [ML²T⁻¹]

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🧮 Numerical Problems

Q1. Convert a speed of 72 km/h into m/s using dimensional method.

Solution:
1 km = 1000 m, 1 hour = 3600 s
Speed = 72 × (1000/3600) = ✅ 20 m/s

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Q2. A force of 10 N acts on a body of mass 2 kg. Calculate its acceleration.

Formula:
F = ma ⇒ a = F/m = 10/2 = ✅ 5 m/s²

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Q3. The energy E of a particle moving in a circular orbit depends on radius (r), mass (m), and angular velocity (ω). Use dimensional analysis to find relation.

Let:
E ∝ mᵃ rᵇ ωᶜ
⇒ [ML²T⁻²] = [M]ᵃ [L]ᵇ [T⁻¹]ᶜ

Equating dimensions:

• M: a = 1
• L: b = 2
• T: –c = –2 ⇒ c = 2

So, ✅ E ∝ m r² ω²

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Q4. The unit of force is 1 N. Express it in CGS units.

Solution:
1 N = 1 kg·m/s² = 10⁵ dyne
✅ Answer: 1 N = 10⁵ dyne

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Q5. Check if the equation v = u + at is dimensionally correct.

LHS: v → [L T⁻¹]
RHS: u + at → [L T⁻¹] + [L T⁻² × T] = [L T⁻¹]
✅ Correct (both sides have same dimensions)

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