Question

state the principal of homageneity and hence describe
its applications to derive a formula
and check the correctness of formula​

Answer 1

Answer:

Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another.

Explanation:

We make use of dimensional analysis for three prominent reasons: To check the consistency of a dimensional equation. To derive the relation between physical quantities in physical phenomena. To change units from one system to another.

Answer 2

Answer:

Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another. To better understand the principle, let us consider the following,

Example 1:

 Check the correctness of physical equation s = ut + ½ at2. In the equation, s is the displacement, u is the initial velocity, v is the final velocity, a is the acceleration and t is the time in which change occurs.

Solution:

We know that

L.H.S = s and

R.H.S = ut + 1/2at2

The dimensional formula for the L.H.S can be

written as s = [L1M0T0] ………..(1)

We know that R.H.S is ut + ½ at2 , simplifying we can

write R.H.S as [u][t] + [a] [t]2

[L1M0T-1][L0M0T-1] +[L1M0T-2][L0M0T0]

=[L1M0T0]………..(2)

From (1) and (2),

we have [L.H.S] = [R.H.S]

Hence, by the principle of homogeneity, the given equation is dimensionally correct.

Explanation:

Applications of Dimensional Analysis:-

Dimensional analysis is a fundamental aspect of measurement and is applied in real-life physics. We make use of dimensional analysis for three prominent reasons:

》To check the consistency of a dimensional equation.

》To derive the relation between physical quantities in physical phenomena.

》To change units from one system to another.

Check the correctness of the physical equation

v2 = u2 + 2as2.

Solution:

The computations made on the L.H.S and R.H.S are as follows:

L.H.S: v2 = [v2] = [ L1M0T–1]2 = [ L1M0T–2] ……………(1)

R.H.S: u2 + 2as2

Hence,

[R.H.S] = [u]2 + 2[a][s]2

[R.H.S] = [L1M0T–1]2 + [L1M0T–2][L1M0T0]2

[R.H.S] = [L2M0T–2] + [L1M0T–2][L2M0T0]

[R.H.S] = [L2M0T–2] + [L1M0T–2][L2M0T0]

[R.H.S] = [L2M0T–2] + [L3M0T–2]…………………(2)

From (1) and (2),

we have [L.H.S] ≠ [R.H.S]

Hence, by the principle of homogeneity, the equation is not dimensionally correct.