Complete Notes on Dimensional Analysis & Formulae
Why Dimensional Analysis is Important
Dimensional analysis is a critical tool in physics for checking equation correctness, deriving formulae, and converting units. It’s a must-know for competitive exams like NEET and JEE, where it’s frequently tested in numerical problems and derivations.
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1. Definition
Dimensional analysis verifies the consistency of physical equations by ensuring all terms have the same dimensions (e.g., [M] for mass, [L] for length, [T] for time).
2. Key Formulas with Units and Dimensions
Below is a comprehensive list of dimensional formulae for key physical quantities, along with their defining formulas and SI units:
| Physical Quantity | Formula | SI Unit | Dimensional Formula |
|---|---|---|---|
| Velocity | Distance / Time | m/s | [L][T]-1 |
| Acceleration | Velocity / Time | m/s2 | [L][T]-2 |
| Force | Mass × Acceleration | Newton (N) | [M][L][T]-2 |
| Momentum | Mass × Velocity | kg·m/s | [M][L][T]-1 |
| Work/Energy | Force × Distance | Joule (J) | [M][L]2[T]-2 |
| Power | Work / Time | Watt (W) | [M][L]2[T]-3 |
| Pressure | Force / Area | Pascal (Pa) | [M][L]-1[T]-2 |
| Density | Mass / Volume | kg/m3 | [M][L]-3 |
| Impulse | Force × Time | N·s | [M][L][T]-1 |
| Angular Velocity | Angle / Time | rad/s | [T]-1 |
| Frequency | 1 / Time | Hertz (Hz) | [T]-1 |
| Gravitational Constant | F = G(m1m2/r2) | N·m2/kg2 | [M]-1[L]3[T]-2 |
| Surface Tension | Force / Length | N/m | [M][T]-2 |
| Viscosity | Force / (Area × Velocity Gradient) | Pa·s | [M][L]-1[T]-1 |
| Planck’s Constant | Energy / Frequency | J·s | [M][L]2[T]-1 |
| Electric Charge | Current × Time | Coulomb (C) | [I][T] |
| Electric Potential | Work / Charge | Volt (V) | [M][L]2[T]-3[I]-1 |
3. Real-Life Example
- Aerospace Engineering: In rocket design, engineers use dimensional analysis to verify that the thrust (force, [M][L][T]-2) matches the fuel consumption rate equations, ensuring the rocket achieves the desired acceleration ([L][T]-2).
4. Common Exam Questions and Answers
Answer: Let’s analyze each term:
- v² (final velocity squared): [L][T]-1 × [L][T]-1 = [L]2[T]-2
- u² (initial velocity squared): [L][T]-1 × [L][T]-1 = [L]2[T]-2
- 2as: 2 is dimensionless; a (acceleration) = [L][T]-2, s (distance) = [L]; so 2as = [L][T]-2 × [L] = [L]2[T]-2
- All terms have the same dimensions ([L]2[T]-2), so the equation is dimensionally correct.
Answer: Energy (work) = Force × Distance
- Force = [M][L][T]-2
- Distance = [L]
- Energy = [M][L][T]-2 × [L] = [M][L]2[T]-2
Answer: Velocity has dimensions [L][T]-1, while force has [M][L][T]-2. Since their dimensions are different, they cannot be added directly, as per the principle of homogeneity.
5. Diagram/Table
The table above serves as a reference diagram summarizing the dimensional formulae for key physical quantities, making it easy to visualize and memorize.
6. Quick Tips/Tricks
- Principle of Homogeneity: Ensure all terms in an equation have the same dimensions for correctness.
- Derivation Trick: Break down complex quantities into basic ones (e.g., Power = Work/Time = [M][L]2[T]-3).
- Memorization Tip: Group similar dimensions (e.g., Work, Kinetic Energy, and Potential Energy all have [M][L]2[T]-2).
- NEET/JEE Hack: Use dimensional analysis to cross-check answers in numerical problems or derive unknown constants like Planck’s constant ([M][L]2[T]-1).
- Limitation: Cannot determine dimensionless constants (e.g., π or Reynolds number).
Note: Dimensional analysis is a powerful tool for verifying equations, deriving new relations, and unit conversions, but it doesn’t provide numerical coefficients or dimensionless quantities.