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Dimensional Analysis

Dimensions of a physical quantity are the power to which the fundamental quantities must be raised to represent the given physical quantity.

Dimensional AnalysisDimensional AnalysisFor example, density  = (mass/volume) = mass/(length)3
or density = (mass) (length)-3  
Thus, the dimensions of density are 1 in mass and -3 in length. The dimensions of all other fundamental quantities are zero.
For convenience, the fundamental quantities are represented by one-letter symbols. 

  • Generally, mass is denoted by M, length by L, time by T, and electric current by A.
  • The thermodynamic temperature, the amount of substance, and the luminous intensity are denoted by the symbols of their units K, mol, and cd respectively.
  • The physical quantity that is expressed in terms of the base quantities is enclosed in square brackets.
    [sinθ] = [cosθ] = [tanθ] = [ex] = [M0L0T0]

Dimensional Formula

It is an expression that shows how and which of the fundamental units are required to represent the unit of physical quantity.

For example:

By definition, Velocity is a derived quantity. We can write its dimensional formula using:

Velocity = Displacement/Time

Now, displacement is measured in Length, [L1], and time is measured in [T1] which are both fundamental quantities. Hence, we can represent the dimensional formula of velocity as:

[v] = [s/t] = [L/T] = [LT-1]

Question for Dimensional Analysis & Formulas
Try yourself:What is the dimensional formula for velocity?
View Solution

Similarly, we can evaluate the dimensional formula of all physically derived quantities. The table given below shows some of the most occurring physical quantities and their dimensions.

Different quantities with units, symbols, and dimensional formulas

Dimensional Analysis & Formulas | Physics Class 11 - NEET

Dimensional Analysis & Formulas | Physics Class 11 - NEET

Dimensional Analysis & Formulas | Physics Class 11 - NEET

Dimensional Analysis & Formulas | Physics Class 11 - NEET
Dimensional Analysis & Formulas | Physics Class 11 - NEET


Dimensional Analysis & Formulas | Physics Class 11 - NEET

 Dimensional Analysis & Formulas | Physics Class 11 - NEET

Dimensional Analysis & Formulas | Physics Class 11 - NEET

Dimensional Analysis & Formulas | Physics Class 11 - NEET

Dimensional Analysis & Formulas | Physics Class 11 - NEET


Applications of Dimensional Analysis

Conversion of units

This is based on the fact that the product of the numerical value (n) and its corresponding unit (u) is a constant, i.e.,

n[u] = constant

or n1[u1] = n2 [u2]

Suppose the dimensions of a physical quantity are a in mass, b in length, and c in time. If the fundamental units in one system are M1, L1, and T1 and in the other system are M2, L2, and T2 respectively. Then we can write.

Dimensional Analysis & Formulas | Physics Class 11 - NEETHere n1 and n2 are the numerical values in two systems of units respectively. 

Using Eq. (i), we can convert the numerical value of a physical quantity from one system of units into the other system.

Example 1.  The value of the gravitation constant is G = 6.67 × 10-11 Nm2/kg2 in SI units. Convert it into CGS system of units. 
Solution. The dimensional formula of G is [M-1 L3 T-2].
Using equation number (i), i.e.,
Dimensional Analysis & Formulas | Physics Class 11 - NEET
Here, n1 = 6.67 x 10-11
M1 = 1 kg, M2 = 1 g = 10-3 kg, L1  = 1 m, L2 = 1 cm = 10-2 m, T1 = T2 = 1s
Substituting in the above equation, we get
Dimensional Analysis & Formulas | Physics Class 11 - NEET

or n2 = 6.67 x 10-8
Thus, the value of G in the CGS system of units is 6.67 x 10-8 dyne cm2/g2.

To check the dimensional correctness of a given physical equation

  • Every physical equation should be dimensionally balanced. This is called the 'Principle of Homogeneity'
  • The dimensions of each term on both sides of an equation must be the same. On this basis, we can judge whether a given equation is correct or not. However, a dimensionally correct equation may or may not be physically correct.

Question for Dimensional Analysis & Formulas
Try yourself:
Which principle states that every physical equation should be dimensionally balanced?
View Solution

Example 2. Show that the expression of the time period T of a simple pendulum of length l given by
Dimensional Analysis & Formulas | Physics Class 11 - NEET
is dimensionally correct.  
Solution. 
Dimensional Analysis & Formulas | Physics Class 11 - NEET
Dimensional Analysis & Formulas | Physics Class 11 - NEETAs in the above equation, the dimensions of both sides are the same. The given formula is dimensionally correct.

Principle of Homogeneity of Dimensions

This principle states that the dimensions of all the terms in a physical expression should be the same. For example, in the physical expression s = ut + 1/2 at2,  the dimensions of s, ut, and 1/2 at2 all are the same.

Note: The physical quantities separated by the symbols +, -, =, >, <, etc., have the same dimensions.

Example 3. The velocity v of a particle depends upon the time t according to the equation 
Dimensional Analysis & Formulas | Physics Class 11 - NEET
Write the dimensions of a, b, c, and d. 
Solution.
From the principle of homogeneity
Dimensional Analysis & Formulas | Physics Class 11 - NEET
Dimensional Analysis & Formulas | Physics Class 11 - NEET

To establish the relation among various physical quantities

If we know the factors on which a given physical quantity may depend, we can find a formula relating the quantity to those factors. Let us take an example.

Example.4. The frequency (f) of a stretched string depends upon the tension F (dimensions of force), length l of the string, and the mass per unit length m of string. Derive the formula for frequency. 
Solution. Suppose, that the frequency f depends on the tension raised to the power a, length raised to the power b, and mass per unit length raised to the power c. Then.
f ∝ [F]a[l]b[μ]c
or f = k[F]a[l]b[μ]c
Here, k is a dimensionless constant. Thus,
[f] = [F]0[l]b[μ]c
or [M0L0T-1] = [MLT-2]a[L]b[ML-1]c
or [M0L0T-1] = [Ma+c La+b-c T-2a]
For dimensional balance, the dimensions on both sides should be the same.
Thus,    a + c = 0    ...(ii)
a + b - c = 0    ...(iii)
and    - 2a = - 1    ...(iv)
Solving these three equations, we get
a = 1/2, c = (-1/2) and b = -1
Substituting these values in Eq. (i), we get
f = k(F)1/2(l)-1(μ)-1/2
or
Dimensional Analysis & Formulas | Physics Class 11 - NEETExperimentally, the value of k is found to be 1/2
Dimensional Analysis & Formulas | Physics Class 11 - NEET

Limitations of Dimensional Analysis 

The method of dimensions has the following limitations:

(a) By this method the value of dimensionless constant can not be calculated.

(b) By this method the equation containing trigonometrical, exponential, and logarithmic terms cannot be analyzed.

(c) If a physical quantity depends on more than three factors, then a relation among them cannot be established because we can have only three equations by equalizing the powers of M, L, and T.

The document Dimensional Analysis & Formulas | Physics Class 11 - NEET is a part of the NEET Course Physics Class 11.
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FAQs on Dimensional Analysis & Formulas - Physics Class 11 - NEET

1. What is dimensional analysis and why is it important in the field of science?
Dimensional analysis is a mathematical method used to analyze and solve problems involving physical quantities. It involves studying the dimensions of various physical quantities and their relationships. This approach is important in science because it helps to establish the validity of equations, convert units, and derive new equations. It also aids in understanding the physical significance of different quantities and their interdependencies.
2. What is a dimensional formula and how is it determined?
A dimensional formula represents the dimensions of a physical quantity in terms of fundamental dimensions. These fundamental dimensions are typically mass [M], length [L], and time [T]. The dimensional formula for a quantity is determined by expressing it as a product or a quotient of these fundamental dimensions raised to certain powers. For example, the dimensional formula for velocity is [LT^-1] (length divided by time), indicating that velocity has dimensions of length per unit time.
3. What are some applications of dimensional analysis in practical scenarios?
Dimensional analysis has various applications in practical scenarios, including: - Unit conversion: Dimensional analysis allows for the conversion of units from one system to another. For example, it can be used to convert kilometers to miles or liters to gallons. - Checking equations: Dimensional analysis helps in verifying the correctness of equations. By comparing the dimensions on both sides of an equation, one can ensure that the equation is balanced and physically meaningful. - Predicting relationships: Dimensional analysis can be used to predict the relationships between different physical quantities. By analyzing the dimensions of the quantities involved, one can determine how they are related. This is particularly useful when deriving empirical equations. - Scaling laws: Dimensional analysis aids in understanding the behavior of physical systems at different scales. It helps in formulating scaling laws, which describe how certain properties or behaviors change with size or other relevant parameters.
4. Are there any limitations to dimensional analysis?
Yes, there are a few limitations to dimensional analysis: - It cannot provide numerical values: Dimensional analysis only deals with the dimensions of physical quantities, not their actual values. It cannot provide numerical information such as the magnitude or numerical coefficient in an equation. - It assumes linear relationships: Dimensional analysis assumes that the relationship between physical quantities is linear. However, in reality, many relationships are non-linear, and dimensional analysis may not accurately predict these complex relationships. - It ignores dimensionless quantities: Dimensional analysis focuses on quantities with dimensions and does not consider dimensionless quantities. This limitation restricts its application in some fields where dimensionless numbers play a crucial role, such as fluid dynamics.
5. How can dimensional analysis be used in the NEET exam?
Dimensional analysis is an important concept in physics, which is a part of the NEET exam syllabus. It can be used to solve problems involving unit conversions, verifying equations, and predicting relationships between physical quantities. Questions related to dimensional analysis in the NEET exam may require students to analyze the dimensions of different quantities, determine their relationships, or convert units. Understanding dimensional analysis can help students approach such questions systematically and derive the correct solutions.
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