Applications (Uses) of Dimensional Analysis-Principle of Homogeneity-Unit Coversion

Dimensional analysis is a powerful tool in physics used to analyze physical quantities, derive units, and verify equations. It is based on the principle of homogeneity, which states that only quantities with the same dimensions can be equated.

Uses of Dimensional Analysis

(1) To Find the Unit of a Physical Quantity in a Given System

• Write the definition or formula for the physical quantity and determine its dimensions.
• Replace mass (M), length (L), and time (T) with fundamental units of the required system to obtain the unit.
• Example:
  • Work = Force × Displacement
  • Dimensional Formula: \begin{array}{l} [W] = [MLT^{-2}] \times [L] = [ML^2T^{-2}] \end{array}
  • Units:
    • C.G.S. System: erg (g cm²/s²)
    • M.K.S. System: joule (kg m²/s²)

(2) To Find Dimensions of Physical Constants or Coefficients

• Any physical constant can be analyzed using its equation and substituting the dimensional formulae of other quantities.
• Examples:
  • Gravitational Constant (G):
    • Newton’s Law: \begin{array}{l} F = G \frac{m_1 m_2}{r^2} \end{array}
    • Dimensional Formula: \begin{array}{l} [G] = [M^{-1}L^3T^{-2}] \end{array}
  • Planck’s Constant (h):
    • Planck’s Equation: \begin{array}{l} E = h \nu \end{array}
    • Dimensional Formula: \begin{array}{l} [h] = [ML^2T^{-1}] \end{array}
  • Coefficient of Viscosity (η):
    • Poiseuille’s Equation: \begin{array}{l} \eta = \frac{\pi r^4 P}{8l V/t} \end{array}
    • Dimensional Formula: \begin{array}{l} [\eta] = [ML^{-1}T^{-1}] \end{array}

(3) To Convert a Physical Quantity from One System of Unit to Another

• The measure of a physical quantity remains constant across different unit systems.
• General formula: \begin{array}{l} n_1 u_1 = n_2 u_2 \end{array}
• Example:
  • Convert Newton (N) to Dyne:
    • \begin{array}{l} 1 N = 1 \text{ kg} \cdot \text{m/s}^2 \end{array}
    • Using dimensional conversion: \begin{array}{l} 1 N = 10^5 \text{ Dyne} \end{array}
  • Convert Gravitational Constant (G) from C.G.S. to M.K.S.:
    • In C.G.S. system: \begin{array}{l} G = 6.67 \times 10^{-8} \text{ cm}^3/\text{g} \cdot \text{s}^2 \end{array}
    • Conversion results in SI value: \begin{array}{l} 6.67 \times 10^{-11} \text{ m}^3/\text{kg} \cdot \text{s}^2 \end{array}

(4) To Check Dimensional Correctness of an Equation(Principle of Homogeneity)

• According to the Principle of Homogeneity, dimensions of all terms on both sides of an equation must be the same.
• Example:
  • Checking the equation: \begin{array}{l} F = \frac{mv^2}{r} \end{array}
    • LHS: \begin{array}{l} [F] = [MLT^{-2}] \end{array}
    • RHS: \begin{array}{l} [M] \times [L^2T^{-2}] \times [L^{-1}] = [MLT^{-2}] \end{array}
    • Since both sides match, the equation is dimensionally correct.

For example –  

We have a physics equation,  

S = ut + 1/2at²

LHS is distance,  

So dimension of LHS = L¹M⁰T⁰

RHS,

In dimensional frame,

= [u][t] + [a][t]²

= [LT^-1][T] + [LT^-2][T]²

= [L] + [L]

Both the terms of RHS are equal to LHS, therefore this equation is dimensionally correct and the homogeneity principle of dimensional analysis holds.  

Example : Is v = at dimensionally correct?

Solution:

Here, LHS: velocity, v 

v = [LT^-1]

And, RHS = at i.e. acceleration × time

Acceleration = [LT^-2]

Time = [T]

Therefore, at = [LT^-1]

Hence,

LHS = RHS

and the given equation is dimensionally correct.

Problem : You remember πr² and 2πr as formulas of area and circumference of circle, but you are unable to recall which formula is for area and which for circumference. Find using dimensional analysis.

Solution:

   Formula 1 – πr² = [π].[r]²

   As π is a constant, so it is dimensionless

   So, =  1. [L]²

   Radius has the dimension of length, therefore formula 1 has dimensional formula of area.

   Area = πr²

   Formula 2 – 2πr = 2.[π].[r]

   As π is a constant, so it is dimensionless

   So, =  2.1. [L] = dimension of length

   Radius has the dimension of length, therefore formula 2 has dimensional formula of circumference.

   Circumference = 2πr

(5) As a Research Tool to Derive New Relations

If a physical quantity depends on others in a product form, dimensional analysis helps derive its relation.

Lets derive the formula for centripetal force, F, acting on an object moving in a uniform circle.

We know that, centripetal force acting on a particle depends on its mass(m), velocity with which its moving(v) and the radius of the circle(r).

 F = m^av^br^c

 Writing the dimensions of all quantities

 [MLT-2] = M^a[LT^-1]^b L^c

 On simplifying,  

 [MLT^-2] = M^a[L^b+cT^-b]

 Using the principle of homogeneity,

 a = 1,  

 b + c = 1

 b = 2

 On solving, we get

 a = 1, b = 2, c = -1

Therefore the centripetal force, F is

F= k (mv²)/r

• Example:
  • Time period of a simple pendulum: \begin{array}{l} T = K m^x l^y g^z \end{array}
    • Using dimensions: x = 0, y = 1/2, z = -1/2
    • Thus: \begin{array}{l} T = K \sqrt{\frac{l}{g}} \end{array}
    • Experimentally, \begin{array}{l} T = 2\pi \sqrt{\frac{l}{g}} \end{array}
  • Stoke’s Law:
    • Viscous force on a sphere: \begin{array}{l} F = K \eta^x r^y v^z \end{array}
    • Solving for x, y, z, we get: \begin{array}{l} F = 6\pi \eta r v \end{array}

Example

Convert the unit of G, which is gravitational constant,

G = 6.67 x 10^-11Nm²/kg² in CGS system.

Solution: 

Since, we have

G = 6.67 x 10-¹¹ Nm²/kg²

Converting kg into grams, 1 kg = 1000 gms

= 6.67 x 10-¹¹ x 10⁸ x 10³ cm³/g¹ s²

= 6.67 x 10⁸  cm³/g¹ s2

Frequently Asked Questions (FAQs)

1. What is the principle of homogeneity in dimensional analysis?
2. How does dimensional analysis help in unit conversion?
3. Can dimensional analysis determine numerical constants?

Worksheet

1. Derive the dimensional formula of surface tension.
2. Verify the equation using dimensional analysis : s = ut + ½ at²

4. Do You Know?

✅ The SI unit of viscosity is Pascal-second (Pa·s). ✅ Planck’s constant relates energy and frequency. ✅ Dimensional analysis helped discover Bohr’s atomic model.

Important Points for Quick Revision

✅ Dimensions are unique for each physical quantity.
✅ The Principle of Homogeneity states that both sides of an equation must have the same dimensions.
✅ Conversion factors are derived using dimensional equations.
✅ Dimensional analysis cannot determine dimensionless numerical constants like π or 1/2.
✅ It is useful in unit conversions, verifying equations, and deriving new formulas.

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