Dimensional Analysis and Applications

Uses of Dimensional Equations
The dimensional equations have got following three uses:
1. To check the correctness of a physical equation.
2. To derive the relation between different physical quantities involved in a physical phenomenon.
3. To change from one system of units to another.
1. To check the correctness of a physical relation and the principle of homogeneity of dimensions:
Checking the correctness of a physical relation (or equation) is based on the principle of homogeneity of dimensions.

According to this principle, the dimensions of the fundamental quantities (mass, length and time) are same in each and every term on either side of the physical relation.

To check the correctness of a given physical equation, the physical quantities on the two sides of the equation are expressed in terms of fundamental units of mass, length and time.

If the powers of M, L and T on two sides of the equation are same, then the physical equation is correct and otherwise not.

Problem-1

Check the dimensional consistency of the equation:
FS = ,
Where S is the distance moved, u and v are the initial and final velocities of a body of mass m and F is the force acting on it.

Solution
We have,
FS =

L. H. S. Dimensional formula of term FS = [MLT-2] × [L] = [ML2T-2]
R. H. S. Dimensional formula of term = [M] ×[LT-1]2 = [ML2T-2]
Dimensional formula of term = [M] ×[LT-1]2 = [ML2T-2]
As the dimensional formulae of all the terms in the given equation are same, the equation is a correct one.

Problem-2

In Vander Waals’s equation:
= RT.
What are the dimensions of constants ‘a’ and ‘b’?

Solution
Here, = RT
According to principle of homogeneity of dimensions, should have dimensions of P and b should have the dimensions of v.
The dimension of = dimension of pressure.
or dimension of a = pressure × V2 = [ML-1T2][L3]2 = [M L5 T2]
and dimension of b = volume = [L3] = [M0L3T0]

Problem-3

Test by using dimensional analysis, the accuracy of the relation:
=
Where the letters have their usual meanings.

Solution
Here, =
Dimensional formula of wavelength, = [L]
Dimensional formula of L. H. S = [L] = [M0LT0]
Dimensional formula of Planck’s constant, h = [ML2T-1]
Dimensional formula of mass, m = [M]
Dimensional formula of velocity, v = [LT-1]
Dimensional formula of R. H. S = = [M0LT0]
As dimensional formula of L. H. S. is same as that of R. H. S., the given relation is correct.

Problem-4

Check the correctness of the relation τ = I α.

Solution
Now, τ = torque = force × distance
= [M L T-1] [L] = [ML2T-2]
I = Moment of inertia = mass × distance2 = [m][L]2 = [M L2]
a = angular acceleration = = [T-2]
I α[M L2] [T-2] = [M L2 T-2]
Since dimensional formulae of L. H. S. and R. H. S. are the same, the given relation is correct.

Problem-5

Test by using dimensional analysis, the accuracy of the relation:
=

Solution
Here, =

where k is a dimensionless constant.

Dimensional formula of critical velocity, vc = [L T-1]

Dimensional formula of coefficient of viscosity,

η = [ML-1 T-1]

Dimensional formula of radius, R = [L]

Dimensional formula of density, ρ = [ML-3]

Dimensional formula of L. H. S = [LT-1] = [M0 L T-1]

Dimensional formula of R. H. S = = [M0 L T-1]

As dimensional formula of L. H. S. is same as that of R. H. S. the given relation is correct.

In deriving the relation between the various physical quantities, the principle of homogeneity of dimensional equation is used.

To derive a physical relation the final dimensional equation is written in terms of fundamental units of M, L and T.

The powers of M, L and T are equated on both sides of the dimensional equation.

Three equations are got from which the values of three unknown powers can be calculated.

These values are substituted in the dimensional equation.

Problem-6

The velocity (v) of sound through a medium may be assumed to depend on
1. the density (ρ ) of the medium and
2. modulus of elasticity (E).
If the dimensions for elasticity (ratio of stress to strain) are [ML-1T-2], deduce by the method of dimensions the formula for the velocity of sound.

Solution
Let v
α r a            ------(i)
Let v α Eb            ------(ii)

Combining equation (i) and (ii), we get

v ρ a Eb            ------(iii)

Writing the dimensional formula of v, α and E in equation (iii), we get

[LT-1] [ML-3]a [ML-1T-2]b
or[M0LT-1] [Ma+bL-3a-b T-2]b

Comparing dimensions of M, L and T on the two sides, we get

a + b = 0           ------(iv)
-3a –b = 1
and -2b = -1
or b =

Substituting for b in equation (iv), we get

a + = 0 or a =

Putting values of a and b in equation (iii), we get

V ∝ or v = k

The constant of proportionality k is found to be 1.
v =

Problem-7

Assuming that frequency (ν) of a vibrating string depends upon load applied (F), length of the string (l) and mass per unit length (m), prove that

Solution
Let ν Fa                                                                       ------ (i)

lb                                                                                      ------ (ii)

mc                                                                      ------ (iii)

Combining equation (i), (ii) and (iii), we get ν = kFalbmc  ------ (iv)
Where k is constant of proportionality.
Substituting the dimensions of ν , F, l and m, we get
[T-1] = [MLT-2]a [L]b [ML-1]C
[M0 L0 T-1] = [Ma+c L a+b-c T-2a]
Comparing dimensions of M, L and T on the both sides of the equation, we get
-2a = -1            or                a =
a + c = 0           or                c = -a = -
and           a + b-c = 0           or               b=c-a=--= -1
Substituting the values of a, b and c in equation (iv), we get
ν = k F1/2 l-1 m-1/2

Method of dimensional analysis can be used to obtain the value of a physical quantity in some other system, when its value in one system is given.

As discussed earlier, the measure of a physical quantity is given by

X = n u,

Where u is size of the unit and n is numerical value of the physical quantity for the unit chosen. If u1 and u2 are units for measurement of the physical quantity in two system and n1 and n2 are the numerical values of the physical quantity for the two units, then
n1u1 = n2u2

If a, b and c are dimensions of the physical quantity in mass, length and time, then
n1[M1a L1b T1c] = n2[M2a L2b T2c]

Here, M1, L1, T1 and M2, L2 and T2 are units of mass, length and time in the two systems. Therefore
n2 = n1

The dimensional formula of work is [ML2T-2]. Therefore, a = 1, b = 2 and c = -2.
Hence, number of erg on 1 J is given by
n2 = 1× = = 107
1 J = 107 erg