Question

Explain the principle of homogeniety of dimensions. What are its uses? Give example.

Answer 1

The principle of homogeneity of dimensions states that the dimensions of all the terms in a physical expression should be the same. 

For example, in the physical expression v² = u² + 2as, the dimensions of v² , u² and 2 as are the same and equal to [L² T^-2 ]. 

(i) To convert a physical quantity from one system of units to another: 

This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant, i.e, = constant,n₁ [u₁] = constant 

(or) n₁ [u₁] = n₂ [u₂].

Consider a physical quantity which has dimension ‘a’ in mass, ‘b’ in length and ‘c’ in time. If the fundamental units in one system are M₁, L₁ and T₁ and the other system are M₂, L₂ and T₂ respectively, then we can write,

We have thus converted the numerical value of physical quantity from one system of units into the other system.

Example: Convert 76 cm of mercury pressure into Nm^-2 using the method of dimensions. 

Solution: In cgs system 76 cm of mercury pressure = 76 × 13.6 × 980 dyne cm^-2

The dimensional formula of pressure P is [ML^-1T^-2]

(ii) To check the dimensional correctness of a given physical equation: 

Example: The equation \(\frac{1}{2}mv^2\) = mgh can be checked by using this method as follows. 

Solution: Dimensional formula for

Dimensional formula for

Both sides are dimensionally the same, hence the equations \(\frac{1}{2} mv^2 = mgh\) is dimensionally correct

(iii) To establish the relation among various physical quantities: 

Example: An expression for the time period T of a simple pendulum can be obtained by using this method as follows. 

Let true period T depend upon 

(i) mass m of the bob 

(ii) length l of the pendulum and 

(iii) acceleration due to gravity g at the place where the pendulum is suspended. Let the constant involved is K = 2π.

Solution:

Here k is the dimensionless constant. Rewriting the above equation with dimensions.

Comparing the powers of M, L and T on both sides, 

a – 0, b + c = 0, -2c = 1 

Solving for a, b and c a = 0, b = 1/2, and c = -1/2 

From the above equation