What is dimensional analysis in physics?

Dimensional analysis is a method to check the correctness of equations, derive relationships between physical quantities, and convert units. It uses fundamental dimensions [M], [L], [T], [I], [K], [mol], and [cd] to express all physical quantities.

What are the 7 fundamental quantities in SI system?

The 7 fundamental quantities in SI system are: Length (meter), Mass (kilogram), Time (second), Electric Current (ampere), Temperature (kelvin), Amount of Substance (mole), and Luminous Intensity (candela).

How many significant figures are in 0.00340?

The number 0.00340 has 3 significant figures. Leading zeros are not significant, but trailing zeros after decimal point are significant. So the significant figures are 3, 4, and 0.

What is the dimensional formula of force?

The dimensional formula of force is [MLT⁻²]. This is derived from Newton's second law F = ma, where mass has dimension [M] and acceleration has dimension [LT⁻²].

What are the limitations of dimensional analysis?

Limitations of dimensional analysis include: cannot find dimensionless constants (like π, e, 1/2), cannot distinguish between scalar and vector quantities, cannot work with equations having multiple terms of same dimensions, and cannot handle trigonometric, exponential, or logarithmic functions.

Units and Dimensions for JEE 2025-26 | Complete Notes with Formulas & Solved Examples

Q: What is the principle of homogeneity in dimensional analysis?

The principle of homogeneity states that in a correct physical equation, the dimensions of all terms on both sides must be identical. Only quantities with the same dimensions can be added or subtracted.

Units and Dimensions Jee notes, Formulas, PYQs — Units and Dimensions JEE Notes, Formulas, PYQs

Physical Quantities

A physical quantity is any property of a material or system that can be quantified by measurement. Understanding the classification of physical quantities is fundamental to all of Physics and forms the basis of dimensional analysis.

1.1 Classification of Physical Quantities

Fundamental Quantities

Quantities that are independent and cannot be expressed in terms of other quantities.

Length (L)
Mass (M)
Time (T)
Temperature (K)
Electric Current (I)
Amount of Substance (mol)
Luminous Intensity (cd)

Derived Quantities

Quantities that are dependent and can be expressed in terms of fundamental quantities.

Area = Length²
Volume = Length³
Velocity = Length/Time
Force = Mass × Acceleration
Energy = Force × Distance
Power = Energy/Time

💡 JEE Memory Trick

Remember the 7 fundamental quantities using: "My Large Tiger Can Eat Little Ants"
Mass, Length, Time, Current, (Thermodynamic) Temperature, Luminous Intensity, Amount of Substance

1.2 Scalar vs Vector Quantities

Aspect	Scalar Quantity	Vector Quantity
Definition	Has only magnitude	Has both magnitude and direction
Examples	Mass, Time, Temperature, Energy, Speed	Displacement, Velocity, Force, Momentum
Addition Rule	Simple arithmetic (a + b)	Vector addition (triangle/parallelogram law)
Representation	Single number with unit	Arrow notation or bold letter

SI Units System

The International System of Units (SI) is the modern form of the metric system and is the most widely used system of measurement. It consists of 7 base units from which all other units are derived.

2.1 The Seven SI Base Units

Quantity	SI Unit	Symbol	Dimension
Length	meter	m	[L]
Mass	kilogram	kg	[M]
Time	second	s	[T]
Electric Current	ampere	A	[I]
Temperature	kelvin	K	[K]
Amount of Substance	mole	mol	[mol]
Luminous Intensity	candela	cd	[cd]

2.2 Important Derived Units (JEE Specific)

Physical Quantity	SI Unit	Symbol	In Base Units
Force	newton	N	kg·m·s⁻²
Energy/Work	joule	J	kg·m²·s⁻²
Power	watt	W	kg·m²·s⁻³
Pressure	pascal	Pa	kg·m⁻¹·s⁻²
Frequency	hertz	Hz	s⁻¹
Electric Charge	coulomb	C	A·s
Potential Difference	volt	V	kg·m²·s⁻³·A⁻¹
Resistance	ohm	Ω	kg·m²·s⁻³·A⁻²

2.3 SI Prefixes (Powers of 10)

Large Multiples

deca (da)10¹

hecto (h)10²

kilo (k)10³

mega (M)10⁶

giga (G)10⁹

tera (T)10¹²

Small Multiples

deci (d)10⁻¹

centi (c)10⁻²

milli (m)10⁻³

micro (μ)10⁻⁶

nano (n)10⁻⁹

pico (p)10⁻¹²

⚠️ Common JEE Traps

kg is the base unit, not gram (only base unit with prefix)
Temperature: Kelvin (K), not °K or degrees Kelvin
Symbol case matters: m = meter, M = mega prefix
Space between number and unit: 5 m, not 5m

Dimensions & Dimensional Formulas

The dimension of a physical quantity is the power to which the fundamental units must be raised to represent that quantity. Dimensional formulas are essential for checking equations, deriving relationships, and converting units.

3.1 Seven Fundamental Dimensions

[M]

Mass

[L]

Length

[T]

Time

[I]

Current

[K]

Temperature

[mol]

Amount

[cd]

Luminous Int.

[M⁰L⁰T⁰]

Dimensionless

3.2 Dimensional Formulas - Complete List

Physical Quantity	Formula/Definition	Dimensional Formula
Area	Length × Breadth	[M⁰L²T⁰]
Volume	Length × Breadth × Height	[M⁰L³T⁰]
Density	Mass/Volume	[ML⁻³T⁰]
Velocity	Displacement/Time	[M⁰LT⁻¹]
Acceleration	Velocity/Time	[M⁰LT⁻²]
Force	Mass × Acceleration	[MLT⁻²]
Momentum	Mass × Velocity	[MLT⁻¹]
Impulse	Force × Time	[MLT⁻¹]
Work/Energy	Force × Distance	[ML²T⁻²]
Power	Work/Time	[ML²T⁻³]
Pressure	Force/Area	[ML⁻¹T⁻²]
Surface Tension	Force/Length	[M¹L⁰T⁻²]
Angular Velocity	Angle/Time	[M⁰L⁰T⁻¹]
Torque	Force × Distance	[ML²T⁻²]
Moment of Inertia	Mass × (Distance)²	[ML²T⁰]
Gravitational Constant (G)	F = Gm₁m₂/r²	[M⁻¹L³T⁻²]
Planck's Constant (h)	E = hν	[ML²T⁻¹]
Gas Constant (R)	PV = nRT	[ML²T⁻²K⁻¹mol⁻¹]

3.3 Dimensionless Quantities (JEE Favorite)

Pure Numbers (No Dimensions)

Angle

(radian)

Solid Angle

(steradian)

Strain

(ΔL/L)

Refractive Index

(c/v)

Relative Density

(ρ/ρ_water)

π, e, sin θ, cos θ

(Mathematical)

📝 Solved Example 1

Question: Find the dimensional formula of (a) Kinetic Energy (b) Coefficient of Viscosity

Solution:

(a) Kinetic Energy:

KE = \frac{1}{2}mv^2

Dimensions: [M] × [LT⁻¹]² = [M] × [L²T⁻²]

\[[KE] = [ML²T⁻²]\]

(b) Coefficient of Viscosity (η):

From Stokes' Law: F = 6πηrv

\eta = \frac{F}{rv}

Dimensions: [MLT⁻²] / ([L] × [LT⁻¹]) = [MLT⁻²] / [L²T⁻¹]

[\eta] = [ML⁻¹T⁻¹]

🎯 Quick Derivation Tricks

Energy Family: All have [ML²T⁻²] - KE, PE, Heat, Work
Per unit X: Multiply dimension by [X⁻¹] - Energy per unit volume = [ML⁻¹T⁻²]
Time rate: Add [T⁻¹] - Power = Energy/time = [ML²T⁻³]
Per unit area: Divide by [L²] - Pressure = Force/Area = [ML⁻¹T⁻²]

Dimensional Analysis

Dimensional analysis is one of the most powerful tools in Physics. It helps verify equations, derive formulas, and convert units. This is a high-scoring topic in JEE with direct 4-6 marks guaranteed.

4.1 Principle of Homogeneity

The Golden Rule

In any correct physical equation, the dimensions of all terms on both sides must be identical.

Example: s = ut + ½at²

LHS: [s] = [L]

RHS Term 1: [ut] = [LT⁻¹][T] = [L] ✓

RHS Term 2: [½at²] = [LT⁻²][T²] = [L] ✓

All terms have [L] → Equation is dimensionally correct!

⚠️ Important Note

Dimensionally correct ≠ Physically correct
An equation can be dimensionally correct but have wrong numerical constants.
Example: s = ut + at² is dimensionally correct but physically wrong (missing ½)

4.2 Applications of Dimensional Analysis

✅

1. Check Correctness

Verify if a given equation is dimensionally valid

🔄

2. Derive Formulas

Find relationships between physical quantities

🔢

3. Convert Units

Transform quantities between different unit systems

4.3 Method to Derive Formulas

📝 Solved Example 2 (JEE Main 2022 Type)

Question: The time period T of a simple pendulum depends on length L and acceleration due to gravity g. Derive the formula for T using dimensional analysis.

Solution:

Step 1: Assume T depends on L and g

T \propto L^a g^b

\[T = k L^a g^b\]

(where k is a dimensionless constant)

Step 2: Write dimensions

[T] = [L]^a [LT^{-2}]^b

[M^0L^0T^1] = [L^{a+b}T^{-2b}]

Step 3: Compare powers on both sides

For M: 0 = 0 (satisfied automatically)

For L: 0 = a + b → a = -b

For T: 1 = -2b → b = -1/2

Therefore: a = 1/2

Step 4: Substitute values

T = k L^{1/2} g^{-1/2}

T = k\sqrt{\frac{L}{g}}

Note: Dimensional analysis gives us k ≈ 1, but actual value k = 2π can only be found by calculus/experiment.

📝 Solved Example 3 (JEE Advanced Pattern)

Question: The velocity v of a particle depends on time t as v = A sin(Bt + C). Find dimensions of A, B, and C.

Solution:

For A:

Since v = A sin(…), and sin function is dimensionless

\[[v] = [A]\]

[A] = [LT^{-1}]

For B:

Argument of sin must be dimensionless (angle)

\[[Bt] = [M^0L^0T^0]\]

\[[B][T] = [M^0L^0T^0]\]

[B] = [T^{-1}]

For C:

C is added to Bt, so must have same dimension as Bt

[C] = [M^0L^0T^0]\text{ (Dimensionless)}

4.4 Limitations of Dimensional Analysis

❌ What Dimensional Analysis CANNOT Do

Cannot find dimensionless constants (like 2π, ½, e, etc.)
Cannot distinguish between scalar and vector quantities
Cannot handle equations with multiple terms of same dimensions (e.g., s = ut + ½at²)
Cannot work with trigonometric, exponential, or logarithmic functions
Cannot determine if equation is exact - only checks dimensional correctness

Significant Figures

Significant figures indicate the precision of a measurement. Understanding significant figures is crucial for experimental physics and error analysis - a guaranteed 2-3 marks topic in JEE.

5.1 Rules for Counting Significant Figures

Rule	Example	Significant Figures
All non-zero digits are significant	1234	4 S.F.
Zeros between non-zero digits are significant	1002	4 S.F.
Leading zeros are NOT significant	0.0025	2 S.F.
Trailing zeros after decimal ARE significant	2.500	4 S.F.
Trailing zeros in whole number (ambiguous)	2500	2 or 3 or 4 S.F.
Scientific notation removes ambiguity	2.50 × 10³	3 S.F. (clear)
Exact numbers have infinite S.F.	12 eggs, π = 3.14159...	∞ S.F.

💡 Quick Memory Trick

"Atlantic Pacific Rule" for zeros:

Atlantic (A comes before P): If decimal is Absent, start counting from Pacific (right) side
Pacific (P comes after A): If decimal is Present, start counting from Atlantic (left) side

Example 1: 0.00340 (decimal Present → start from left)

0.00340 → 3 S.F.

Example 2: 34000 (decimal Absent → start from right)

34000 → 2 S.F. (ambiguous, use scientific notation)

5.2 Arithmetic Operations with Significant Figures

Addition & Subtraction

Result should have decimal places equal to the least precise measurement

12.11 (2 decimal places)

+ 18.0 (1 decimal place) ←least

+ 1.012 (3 decimal places)

= 31.1 (1 decimal place)

Multiplication & Division

Result should have S.F. equal to the least number of S.F. in any factor

2.5 (2 S.F.) ←least

× 3.42 (3 S.F.)

× 4.125 (4 S.F.)

= 35 (2 S.F.)

📝 Solved Example 4

Question: Calculate with proper significant figures: (2.36 × 10⁴) + (4.5 × 10³) - (3.25 × 10²)

Solution:

Step 1: Convert to same power of 10

2.36 × 10⁴ = 23.6 × 10³

4.5 × 10³ = 4.5 × 10³

3.25 × 10² = 0.325 × 10³

Step 2: Perform addition/subtraction

= (23.6 + 4.5 - 0.325) × 10³

= 27.775 × 10³

Step 3: Round to least decimal places

4.5 has only 1 decimal place (least precise)

\text{Answer: } 27.8 \times 10^3 = 2.78 \times 10^4

5.3 Rounding Off Rules

Rounding Rules (JEE Standard)

Rule 1: If digit after rounding position is < 5

Drop it, keep preceding digit same

3.142 → 3.14 (rounding to 2 decimals)

Rule 2: If digit after rounding position is > 5

Drop it, increase preceding digit by 1

3.147 → 3.15 (rounding to 2 decimals)

Rule 3: If digit is exactly 5 (or 5000...)

Make preceding digit even (round to nearest even number)

3.145 → 3.14 (4 is even)

3.155 → 3.16 (make 5→6 even)

Error Analysis & Measurements

No measurement is perfectly accurate. Understanding errors, their types, and how they propagate is essential for experimental physics and forms 10-15% of JEE Physics numerical questions.

6.1 Types of Errors

1. Systematic Errors

Consistent errors in the same direction (always + or always -)

Instrumental errors
Zero error
Calibration errors
Personal errors

✓ Can be eliminated by correction

2. Random Errors

Unpredictable fluctuations (+ or - randomly)

Environmental changes
Observational errors
Electronic noise
Human limitations

✓ Reduced by repeated measurements

3. Gross Errors

Human mistakes or negligence

Wrong reading
Calculation mistakes
Wrong recording
Procedural errors

✗ Must be avoided by care

6.2 Absolute, Relative & Percentage Errors

Absolute Error

\Delta a = |a_{\text{measured}} - a_{\text{true}}|

Has same unit as quantity

Relative Error

\frac{\Delta a}{a_{\text{true}}}

Dimensionless (no unit)

Percentage Error

\frac{\Delta a}{a_{\text{true}}} \times 100\%

Expressed in %

6.3 Error Propagation Formulas (Most Important)

Operation	Formula	Error Propagation
Addition	Z = A + B	ΔZ = ΔA + ΔB
Subtraction	Z = A - B	ΔZ = ΔA + ΔB
Multiplication	Z = A × B	ΔZ/Z = ΔA/A + ΔB/B
Division	Z = A / B	ΔZ/Z = ΔA/A + ΔB/B
Power	Z = Aⁿ	ΔZ/Z = n(ΔA/A)
General Function	Z = AᵖBᵍCʳ	ΔZ/Z = p(ΔA/A) + q(ΔB/B) + r(ΔC/C)

🎯 Golden Rules for Error Propagation

Addition/Subtraction: Add ABSOLUTE errors (ΔZ = ΔA + ΔB)
Multiplication/Division: Add RELATIVE errors (ΔZ/Z = ΔA/A + ΔB/B)
Power/Root: Multiply relative error by power (ΔZ/Z = n × ΔA/A)
Constants: Have zero error (Δk = 0)

📝 Solved Example 5 (JEE Main 2023 Type)

Question: The resistance R = V/I where V = (100 ± 5) volts and I = (10 ± 0.2) amperes. Calculate percentage error in R.

Solution:

Given: V = 100 V, ΔV = 5 V

I = 10 A, ΔI = 0.2 A

R = V/I (division operation)

For division, add relative errors:

\frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I}

\frac{\Delta R}{R} = \frac{5}{100} + \frac{0.2}{10}

\frac{\Delta R}{R} = 0.05 + 0.02 = 0.07

Percentage error:

\text{Percentage error} = 0.07 \times 100 = 7\%

📝 Solved Example 6 (JEE Advanced Pattern)

Question: The period of a pendulum is T = 2π√(L/g). If L is measured with 2% error and g with 3% error, find percentage error in T.

Solution:

Given: ΔL/L = 2% = 0.02, Δg/g = 3% = 0.03

Step 1: Rewrite formula without constants

T = L^{1/2} g^{-1/2}

Step 2: Apply power rule for error propagation

\frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta L}{L} + \left|-\frac{1}{2}\right|\frac{\Delta g}{g}

(Take absolute value of power)

Step 3: Substitute values

\frac{\Delta T}{T} = \frac{1}{2}(0.02) + \frac{1}{2}(0.03)

\frac{\Delta T}{T} = 0.01 + 0.015 = 0.025

\text{Percentage error in T} = 2.5\%

Unit Conversions

Converting between different unit systems is a practical skill tested frequently in JEE. Master the conversion factors and dimensional methods to solve these problems in under 30 seconds.

7.1 Common Conversion Factors

Length Conversions

1 inch= 2.54 cm

1 foot= 30.48 cm

1 mile= 1.609 km

1 angstrom (Å)= 10⁻¹⁰ m

1 light year= 9.46 × 10¹⁵ m

1 AU= 1.496 × 10¹¹ m

Mass Conversions

1 pound (lb)= 0.4536 kg

1 ounce= 28.35 g

1 tonne= 1000 kg

1 u (atomic mass)= 1.66 × 10⁻²⁷ kg

1 quintal= 100 kg

Time Conversions

1 hour= 3600 s

1 day= 86400 s

1 year= 3.156 × 10⁷ s

1 shake= 10⁻⁸ s

Energy Conversions

1 eV= 1.6 × 10⁻¹⁹ J

1 calorie= 4.186 J

1 kWh= 3.6 × 10⁶ J

1 erg= 10⁻⁷ J

7.2 Dimensional Method for Conversion

📝 Solved Example 7

Question: Convert 1 newton into dynes using dimensional analysis.

Solution:

Step 1: Write dimensional formula

[Force] = [MLT⁻²]

Step 2: Use conversion formula

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

where a=1, b=1, c=-2 (from [MLT⁻²])

Step 3: SI to CGS conversion

M₁ = 1 kg, M₂ = 1 g = 10⁻³ kg

L₁ = 1 m, L₂ = 1 cm = 10⁻² m

T₁ = 1 s, T₂ = 1 s

n_2 = 1 \times \left[\frac{1}{10^{-3}}\right]^1 \left[\frac{1}{10^{-2}}\right]^1 \left[\frac{1}{1}\right]^{-2}

n_2 = 10^3 \times 10^2 \times 1 = 10^5

1 \text{ newton} = 10^5 \text{ dynes}

💡 Quick Conversion Tricks

SI to CGS: Most derived units differ by powers of 10
Energy units: 1 J = 10⁷ erg (remember this!)
Pressure units: 1 Pa = 10 dyne/cm²
Power units: 1 W = 10⁷ erg/s
km/h to m/s: Multiply by 5/18 (or divide by 3.6)

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

✓ Dimensional Analysis: 40%
✓ Error Analysis: 30%
✓ Significant Figures: 20%
✓ Unit Conversions: 10%

JEE Advanced (Last 5 Years)

✓ Deriving formulas: 35%
✓ Error propagation: 35%
✓ Checking correctness: 20%
✓ Mixed concepts: 10%

Weightage Analysis

JEE Main: 4-8 marks (1-2 questions)

JEE Advanced: 6-12 marks (2-3 questions)

Difficulty Level: Easy to Medium

Time Required: 2-3 hours practice

Download Complete PYQ PDF (2015-2024)

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Write the dimensional formula of: (a) Pressure (b) Energy (c) Power (d) Force constant
Check if the equation s = ut + ½at² is dimensionally correct.
Convert 72 km/h to m/s using dimensional analysis.
Find number of significant figures in: (a) 0.00340 (b) 2.500 (c) 34000
Calculate 2.5 × 3.42 with correct significant figures.
If L = (5.0 ± 0.2) m and B = (2.0 ± 0.1) m, find error in area.
Which of the following is dimensionless: angle, strain, refractive index?
Write SI units of: (a) Momentum (b) Impulse (c) Torque

Level 2: Intermediate (JEE Main/Advanced)

Derive the formula for time period of simple pendulum T = 2π√(L/g) using dimensional analysis.
Find dimensional formula of coefficient of viscosity η from Stokes' law F = 6πηrv.
The velocity v of sound in a medium depends on elasticity E and density ρ. Find v using dimensions.
If F = (100 ± 2) N and d = (5.0 ± 0.1) m, find percentage error in work W = Fd.
Check dimensional correctness: v² - u² = 2as + kt where k is a constant. Find dimension of k.
Convert 1 joule to erg using dimensional method.
The area A = (5.2 ± 0.1) cm². Find percentage error in A.
Round off to 3 significant figures: (a) 2.7482 (b) 0.007089 (c) 3.1415

Level 3: Advanced (JEE Advanced/Olympiad)

Find dimensions of a and b in equation: F = ax + bt² where F is force, x is distance, t is time.
Planck's constant h, speed of light c, and gravitational constant G can form a quantity with dimension of length. Find it.
In the formula P = (a/V²)(b - V), P is pressure and V is volume. Find dimensions of a and b.
The density ρ of a sphere is ρ = (12 ± 0.3) g/cm³ and radius r = (2.0 ± 0.1) cm. Find maximum percentage error in mass.
Using dimensional analysis, show that time period of spring-mass system depends on √(m/k).
If energy E, velocity v, and force F are fundamental, find dimensional formula of mass.
The Van der Waals equation is (P + a/V²)(V - b) = RT. Find dimensions of a and b.
In measurement of g by simple pendulum, if L has 2% error and T has 3% error, find error in g.

Get Solutions PDF with Step-by-Step Working

Year	JEE Main	JEE Advanced	Topic Focus
2024	1 Question (4 marks)	2 Questions (6 marks)	Error propagation, Dimensional analysis
2023	2 Questions (8 marks)	1 Question (3 marks)	Significant figures, Unit conversion
2022	1 Question (4 marks)	2 Questions (7 marks)	Deriving formulas, Homogeneity

📑 Quick Navigation - Units and Dimensions

Units & Dimensions JEE Main & Advanced 2025-26

Physical Quantities

1.1 Classification of Physical Quantities

Fundamental Quantities

Derived Quantities

💡 JEE Memory Trick

1.2 Scalar vs Vector Quantities

SI Units System

2.1 The Seven SI Base Units

2.2 Important Derived Units (JEE Specific)

2.3 SI Prefixes (Powers of 10)

Large Multiples

Small Multiples

⚠️ Common JEE Traps

Dimensions & Dimensional Formulas

3.1 Seven Fundamental Dimensions

3.2 Dimensional Formulas - Complete List

3.3 Dimensionless Quantities (JEE Favorite)

Pure Numbers (No Dimensions)

📝 Solved Example 1

🎯 Quick Derivation Tricks

Dimensional Analysis

4.1 Principle of Homogeneity

The Golden Rule

⚠️ Important Note

4.2 Applications of Dimensional Analysis

1. Check Correctness

2. Derive Formulas

3. Convert Units

4.3 Method to Derive Formulas

📝 Solved Example 2 (JEE Main 2022 Type)

📝 Solved Example 3 (JEE Advanced Pattern)

4.4 Limitations of Dimensional Analysis

❌ What Dimensional Analysis CANNOT Do

Significant Figures

5.1 Rules for Counting Significant Figures

💡 Quick Memory Trick

5.2 Arithmetic Operations with Significant Figures

Addition & Subtraction

Multiplication & Division

📝 Solved Example 4

5.3 Rounding Off Rules

Rounding Rules (JEE Standard)

Error Analysis & Measurements

6.1 Types of Errors

1. Systematic Errors

2. Random Errors

3. Gross Errors

6.2 Absolute, Relative & Percentage Errors

Absolute Error

Relative Error

Percentage Error

6.3 Error Propagation Formulas (Most Important)

🎯 Golden Rules for Error Propagation

📝 Solved Example 5 (JEE Main 2023 Type)

📝 Solved Example 6 (JEE Advanced Pattern)

Unit Conversions

7.1 Common Conversion Factors

Length Conversions

Mass Conversions

Time Conversions

Energy Conversions

7.2 Dimensional Method for Conversion

📝 Solved Example 7

💡 Quick Conversion Tricks

📝 Previous Year Questions Analysis

JEE Main (Last 5 Years)

JEE Advanced (Last 5 Years)

Top 10 Most Repeated Question Types

Weightage Analysis

🎯 Practice Problem Set

Level 1: Basic (JEE Main Standard)

Level 2: Intermediate (JEE Main/Advanced)

Level 3: Advanced (JEE Advanced/Olympiad)

Related Physics Notes

Kinematics

Vectors

Mathematical Tools

Units and Dimensions - Complete Guide for JEE 2025-26