Applications of Dimensional Analysis - Practice Problems, FAQs

Have you ever wondered how the terms associated with an equation in physics relate to each other ? Can the terms signify different physical quantities when they are added together or subtracted from each other ? Can they be multiplied or divided ? Let’s try to find answers to these questions in dimensional analysis and its applications.

Every physical quantity has a dimension which represents its constituent fundamental quantities. By dimensional analysis we can understand the significance of a physical quantity and what its unit will be. In many cases the dimensional analysis can be used to get relations between different physical quantities.

Table of content:

List of applications of Dimensional Analysis
Checking the correctness of the given physical relation
Principle of homogeneity
Conversion of system of Units
Deriving the relationship between various physical quantities
Finding the dimension and unit of a physical constant
Limitations of dimensional analysis
Practice Problems
FAQs

List of applications of Dimensional Analysis

Checking the dimensional correctness of the given physical relation

conversion of system of Units

Deriving the relationship between various physical quantities

Finding the dimensions of a physical constant or coefficient

Finding the unit of a physical quantity

Checking the dimensional correctness of the given physical relation

For a physical relation to be correct, the dimensions of the fundamental quantities of each term on both sides of the physical relation must be the same.

Principle of homogeneity

The magnitude of physical quantities may be added together or subtracted from one another only if they have the same dimensions.
any physical quantity can be multiplied or divided with the same or different dimensions.

For an example 1 km and 2 s cannot be added or subtracted but they can be multiplied or divided.

Conversion of system of Units

Let n₁ and n₂ be the numerical values of a physical quantity and u₁ and u₂ be their units in two different systems of units. Then,

nu= Constant

⇒ n₁u₁= n₂u₂

$n_{1} [M_{1}^{a} L_{1}^{b} T_{1}^{c}] = n 2 [M 2 a L 2 b T 2 c]$

Deriving the relationship between various physical quantities

Establishing a relationship between different physical quantities
If the various factors on which a physical quantity depends are known then we can find a relation among the factors by using the principle of homogeneity.

Finding the dimension and unit of a physical constant

In many cases the physical equation is derived from the relationships of the variables involved. In those cases a proportionality constant is taken to complete the equation. Thus we get many physical constants like Gravitational constant, Plank’s constant etc. The dimensions of such constants can be derived from the dimensions of the terms involved in that equation. From this dimension we get the unit.

Limitations of dimensional analysis

Dimension does not depend on the magnitude of a physical quantity. Thus, a dimensionally correct equation need not be physically correct (by values).

The constant of proportionality of any equation cannot be deduced by the method of dimensional analysis.

This method is only applicable if the relation is of the multiplication or division type. It fails in the case of addition, subtraction, exponential and trigonometric relations.

Note: Quantities in the powers and inside the argument of trigonometric functions are dimensionless.

Practice Problems:

Question.1 Find the dimensions of A and B in the equation v = A + Bt, where v is the velocity and t is the time.

a. [A] = [LT^-2] and [B] = LT^-1

b. [A] = [LT^-1] and [B] = [LT^-2]

c. [A] = [L²T^-1] and [B] = [LT^-1]

d. [A] = [LT^-1] and [B] = [L²T^-2]

Answer. v = A + Bt

Using the principle of homogeneity,

Dimension of v = Dimension of A = Dimension of Bt

⇒ Dimension of A = Dimension of v

[A] = [LT^-1]

Thus we can say that the term A is of velocity.

Dimension of Bt = Dimension of v

[B ][T] = [LT^-1]

[B ] = [LT^-2]

Thus we can say that the term B is of acceleration.

Question.2 Find the dimensional formula of X in the equation A=A0eXt , where A and A0 are the pressures and t is the time.

a. [X] = [ML²T^-1]

b. [X]= [LT^-1]

c. [X] = [ML²T^-2]

d. [X] = [T^-1]

Answer. $A = A_{0} e^{X t}$

Quantities in the powers are dimensionless.

Dimension of Xt,

$[X t] = [M^{0} L^{0} T^{0}]$

$[X] [T] = [M^{0} L^{0} T^{0}]$

$[X] = [T^{- 1}]$

Question.3 Find an expression for the time period (t) of a simple pendulum. The time period (t) may depend upon mass m of the bob of the pendulum, length l of the pendulum, and acceleration due to gravity g at the place where the bob is suspended. (Take C as the constant of proportionality.)

$t = C \sqrt{l g}$
$t = C l^{2} \sqrt{g}$
$t = C l g^{2}$

4. $t = C \sqrt{\frac{l}{g}}$

Answer. We assume, $t \propto m^{a}, t \propto l^{b}, t \propto g^{c}$

Combining all the three factors,

$t \propto m^{a} l^{b} g^{c}$

$[M^{0} L^{0} T^{1}] \propto [M^{a}] [L b] [g c]$

$[M^{0} L^{0} T^{1}] \propto [M^{a}] [L b] [(L T - 2) c]$

$[M^{0} L^{0} T^{1}] = C [M^{a}] [L b] [(L T - 2) c]$

$∴ 0 = a; 0 = b + c; 1 = - 2 c$

$i . e . a = 0, b = \frac{1}{2}, c = - \frac{1}{2}$

So, $t = C m^{0} l^{\frac{1}{2}} g^{- \frac{1}{2}} i . e . t = C \sqrt{\frac{l}{g}}$

Question.4 Power of a car is 2 kW . Express it in erg s-1. Here, erg s-1 is the CGS unit of power.

$2 \times 10^{9}$
$2 \times 10^{10}$
$2 \times 10^{11}$
$2 \times 10^{12}$

Answer. We know,

the unit of Power is W.

$[W] = [M L^{2} T^{- 3}]$

$n_{2} = n_{1} {[\frac{M_{1}}{M_{2}}]}^{a} {[\frac{L_{1}}{L_{2}}]}^{b} {[\frac{T_{1}}{T_{2}}]}^{c} = 2 \times 10^{3} \times {[\frac{k g}{g}]}^{1} \times {[\frac{m}{c m}]}^{2} \times {[\frac{s}{s}]}^{- 3}$

$= 2 \times 10^{3} \times 10^{3} \times 10^{4} \times 1^{- 3} = 2 \times 10^{10} g c m^{2} s^{- 3} = 2 \times 10^{10} e r g s^{- 1}$

FAQs:

Question.1 What are the limitations of dimensional analysis ?
Answer. The limitations are:

a dimensionally correct equation need not be physically correct (by values).
The proportionality constant of any equation cannot be deduced by the method of dimensional analysis.
It is only applicable if the relation is of the multiplication or division type.

Question.2 What is the principle of homogeneity ?
Answer. The principle of homogeneity tells us that,

The magnitude of physical quantities may be added together or subtracted from one another only if they have the same dimensions.
Any physical quantity can be multiplied or divided with the same or different dimensions.

As an example 1 kg and 2 m cannot be added or subtracted but they can be multiplied or divided.

Question.3 The distance travelled in n th second can be written as sn=u + 12a(2n-1). Is it dimensionally correct? Discuss.
Answer. Here the dimension of the term on the left side is of length.

The terms on the right side also mean length. Here the time term of 1 s is not written that gives rise to the confusion. Actually it $s_{n} = u + \frac{1}{2} a (2 n - 1)$ . So it is correct.