# Dimensional Analysis and Applications

**Uses of Dimensional Equations**

The dimensional equations have got following three uses:

- To check the correctness of a physical equation.
- To derive the relation between different physical quantities involved in a physical phenomenon.
- To change from one system of units to another.

**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 =

^{-2}] × [L] = [ML

^{2}T

^{-2}]

R. H. S. Dimensional formula of term = [M] ×[LT

^{-1}]

^{2}= [ML

^{2}T

^{-2}]

Dimensional formula of term = [M] ×[LT

^{-1}]

^{2}= [ML

^{2}T

^{-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 × V

^{2}= [ML

^{-1}T

^{2}][L

^{3}]

^{2}= [M L

^{5}T

^{2}]

and dimension of b = volume = [L

^{3}] = [M

^{0}L

^{3}T

^{0}]

# 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] = [M

^{0}LT

^{0}]

Dimensional formula of Planck’s constant, h = [ML

^{2}T

^{-1}]

Dimensional formula of mass, m = [M]

Dimensional formula of velocity, v = [LT

^{-1}]

Dimensional formula of R. H. S = = [M

^{0}LT

^{0}]

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] = [ML

^{2}T

^{-2}]

I = Moment of inertia = mass × distance

^{2}= [m][L]

^{2}= [M L

^{2}]

a = angular acceleration = = [T

^{-2}]

∴ I α[M L

^{2}] [T

^{-2}] = [M L

^{2}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, =

_{c}= [L T

^{-1}]

^{-1}T

^{-1}]

^{-3}]

^{-1}] = [M

^{0}L T

^{-1}]

^{0}L T

^{-1}]

# Problem-6

**The velocity (v) of sound through a medium may be assumed to depend on**

**the density (ρ ) of the medium and****modulus of elasticity (E).**

^{-1}T

^{-2}], deduce by the method of dimensions the formula for the velocity of sound.

**Solution**

Let v

α r

^{a}------(i)

Let v α E

^{b}------(ii)

^{a}E

^{b}------(iii)

^{-1}] ∝ [ML

^{-3}]

^{a}[ML

^{-1}T

^{-2}]

^{b}

or[M

^{0}LT

^{-1}] ∝ [M

^{a+b}L

^{-3a-b}T

^{-2}]

^{b}

-3a –b = 1

and -2b = -1

or b =

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 ∝ F

^{a}------ (i)

∝ l^{b }------ (ii)

∝ m^{c} ------ (iii)

^{a}l

^{b}m

^{c}------ (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}

[M

^{0}L

^{0}T

^{-1}] = [M

^{a+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=--= -1a + c = 0 or c = -a = -

Substituting the values of a, b and c in equation (iv), we get

ν = k F

^{1/2}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

_{1 }and u

_{2 }are units for measurement of the physical quantity in two system and n

_{1}and n

_{2}are the numerical values of the physical quantity for the two units, then

n

_{1}u

_{1}= n

_{2}u

_{2}

**n**

_{1}[M_{1}^{a}L_{1}^{b}T_{1}^{c}] = n_{2}[M_{2}^{a}L_{2}^{b}T_{2}^{c}]_{1,}L

_{1}, T

_{1}and M

_{2}, L

_{2}and T

_{2}are units of mass, length and time in the two systems. Therefore

n

_{2}= n

_{1}

^{2}T

^{-2}]. Therefore, a = 1, b = 2 and c = -2.

Hence, number of erg on 1 J is given by

n

_{2}= 1× = = 10

^{7}

**1 J = 10**

^{7}erg