Saturday, June 1, 2024

CBSE Class XI Physics: Dimensional Analysis Notes

Chapter: Units and Measurements

Topic: Dimensions and Dimensional Analysis

1. Dimensions of Physical Quantities

Dimensions of a physical quantity are the powers to which the fundamental units (base quantities) must be raised to represent that quantity.

1.1 Fundamental Quantities and Their Dimensions

  • Length (L): Dimension symbol [L][L]
  • Mass (M): Dimension symbol [M][M]
  • Time (T): Dimension symbol [T][T]
  • Electric Current (I): Dimension symbol [I][I]
  • Thermodynamic Temperature (θ): Dimension symbol [θ][θ]
  • Amount of Substance (N): Dimension symbol [N][N]
  • Luminous Intensity (J): Dimension symbol [J][J]

1.2 Dimensional Formula

The dimensional formula of a physical quantity is an expression that shows how and which of the fundamental quantities are included in that quantity. It is written in terms of the symbols for base quantities.

Example:

  • Force (F): [F]=[MLT2][F] = [M L T^{-2}]
  • Energy (E): [E]=[ML2T2][E] = [M L^2 T^{-2}]

2. Dimensional Analysis

Dimensional analysis is a technique used to check the correctness of equations, convert units, and derive relationships between physical quantities.

2.1 Principle of Homogeneity

The principle of homogeneity states that an equation is dimensionally correct if the dimensions of all the terms on both sides of the equation are the same.

Example: For the equation s=ut+12at2s = ut + \frac{1}{2}at^2:

  • Dimensions of ss (displacement): [L][L]
  • Dimensions of utut (initial velocity ×\times time): [LT1][T]=[L][L T^{-1}][T] = [L]
  • Dimensions of 12at2\frac{1}{2}at^2 : [LT2][T2]=[L][L T^{-2}][T^2] = [L]

Since all terms have the same dimension, the equation is dimensionally correct.

2.2 Applications of Dimensional Analysis

  1. Checking the Correctness of Equations: Ensure that both sides of the equation have the same dimensions.

  2. Deriving Relations: If a physical quantity depends on multiple factors, we can use dimensional analysis to derive a relation among them.

  3. Converting Units: Conversion of units from one system to another using dimensional formula.

3. Examples of Dimensional Formulas

3.1 Velocity (v)

Velocity is the rate of change of displacement: v=LTv = \frac{L}{T} Dimensional Formula: [v]=[LT1][v] = [L T^{-1}]

3.2 Acceleration (a)

Acceleration is the rate of change of velocity: a=LT1Ta = \frac{L T^{-1}}{T} Dimensional Formula: [a]=[LT2][a] = [L T^{-2}]

3.3 Force (F)

Force is mass times acceleration: F=M×a=M×LT2F = M \times a = M \times \frac{L}{T^2} Dimensional Formula: [F]=[MLT2][F] = [M L T^{-2}]

3.4 Work/Energy (W/E)

Work is force times displacement: W=F×L=MLT2×LW = F \times L = M L T^{-2} \times L Dimensional Formula: [W/E]=[ML2T2][W/E] = [M L^2 T^{-2}]

3.5 Power (P)

Power is work done per unit time: P=WT=ML2T2TP = \frac{W}{T} = \frac{M L^2 T^{-2}}{T} Dimensional Formula: [P]=[ML2T3][P] = [M L^2 T^{-3}]

3.6 Pressure (P)

Pressure is force per unit area: P=FL2=MLT2L2P = \frac{F}{L^2} = \frac{M L T^{-2}}{L^2} Dimensional Formula: [P]=[ML1T2][P] = [M L^{-1} T^{-2}]

4. Limitations of Dimensional Analysis

  1. No Information about Dimensionless Constants: Dimensional analysis cannot determine dimensionless constants (e.g., kk in F=kxF = kx).

  2. No Verification of Exactness: A dimensionally correct equation may not be precisely accurate in real-life scenarios.

  3. Applicable Only to Physical Quantities Expressed as Products: It applies only where quantities can be expressed as products of the fundamental units.

5. Practice Problems

  1. Check the dimensional correctness of the equation v=u+at


    • v
      : [LT1][L T^{-1}]

    • u
      : [LT1][L T^{-1}]

    • a \cdot t
      : [LT2][T]=[LT1][L T^{-2}] \cdot [T] = [L T^{-1}]
  2. Derive the dimensional formula for gravitational constant
    G
    in the equation F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}


    • F
      : [MLT2][M L T^{-2}]
    • m1m_1: [M][M]
    • m2m_2: [M][M]
    • r2r^2: [L2][L^2]

    Thus, GG has the dimensional formula: [G]=[F][r2][m1][m2]=[MLT2][L2][M][M]=[M1L3T2][G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{[M L T^{-2}][L^2]}{[M][M]} = [M^{-1} L^3 T^{-2}]

Summary

  • Dimensions are the powers to which the fundamental units must be raised to represent a physical quantity.
  • Dimensional analysis helps in verifying equations, deriving relations, and converting units.
  • The principle of homogeneity is crucial for checking the correctness of equations.
  • Dimensional analysis has limitations, including its inability to provide exact numerical constants and its applicability limited to physical quantities expressible as products of fundamental units.

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