Dimensional Analysis- Principle, Example, Application, and Limitations

The physical dimension is useful because it allows you to monitor what symbols mean using extensive calculations (a technique known as dimensional analysis). It is not a waste of effort to learn how to incorporate it into your skillset. It assists you in comprehending the formulas you are employing, seeing how different expressions are related to one another, and memorising the most important formulas. You’ll become more efficient at solving problems as well, spending less time on each one than if you hadn’t used this strategy.  For better results, you can also solve MCQ on Work Power Energy for NEET as well. It will help to expertise in this particular topic.

Finally, and most crucially, if you take the time to master dimensional analysis, you will find that you will make fewer mistakes.

The study of a relationship through the lens of its units of dimension is known as dimensional analysis. For illustration, constructing an equation like

M = T,

 where M is measured in kilos and T is measured in time, maybe useless. This type of equation is pertained to as dimensionally inconsistent or dimensionally non-homogeneous. As a result, a dimensionally homogeneous equation has the correct confines (or dimensionally harmonious).

Meaning of Dimension

Dimensional analysis is the process of using known equivalent statements or conversion factors to convert between different units of a specific quantity. The dimensional analysis is also called the unit factor method and factor label method.

For example- 365 days =? seconds

365 days *24 hours/day *60 minute/1 hour *60 seconds/1 minute =31536000 seconds

Physical Quantities – Quantities that could not be quantified by measurements. It can be expressed as the combination of numbers, units, or a combination of units. It can be divided into 2 categories: Fundamental and Derived physical quantities.

Fundamental physical quantities: That can’t be denoted by another unit. For example:length,mass,time etc.

S.No.Fundamental QuantitiesSI UnitsSymbol
1LengthMeterM
2MassKilogramKg
3TimeSecondS
4Electric currentAmpereA
5TemperatureKelvinK
6Amount of substanceMolMol
7Luminous intensityCandelaCd

Derived physical quantities: One which is derived from the base or fundamental quantities. Example: pressure, speed, young modulus, etc.

S.NoDerived QuantitySymbolFormulaDerived unit
1AreaALength *widthm2
2VolumeVLength*Height*widthM3
3DensitypMass/volumeKg/m3
4VelocityvDistance/timem/s
5AccelerationaVelocity change/timem/s2
6WorkWForce* displacementKg m2 /s2

Law of Dimensional Analysis or Principle of Homogeneity

Making the units the same on both sides of the equations is the objective of the homogeneity principle. The principle aims to represent the physical units in which the dimensions are consistent for each equation used. Absence of which cannot be expressed as a physically feasible situation. Let us consider some examples for a better understanding of homogeneity.

Example 1 checking the results for the principle of homogeneity, the physical equation v2 = u2+2as2.

Solution:

The calculations 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 equation  (1) & (2),  [L.H.S] ≠ [R.H.S]

Since LHS≠RHS, by the principle of homogeneity, the equation is not correct dimensionally. 

We can now solve specific issues. Set up each problem by putting a question mark next to what you need to find. Then make it equal to the data you’ve been given. To solve the problem, multiply the given data and units by the relevant unit factors until only the desired units remain.

“We have to convert measures to fractions, create a product of fractions, and cancel units in a dimensional analysis issue. The following are the stages to performing a unit analysis:

  1. Write the last fact using the units specified in the solution.
  2. Write the first fact as a fraction with the units included.
  3. Determine how many conversion facts are required to get from the starting point to the desired result.
  4. Build a fraction product between the beginning and the end-user with the relevant conversion facts. Arrange the fractions in such a way that the units that must be deleted are cancelled.
  5. Unwanted units should be cancelled.
  6. Carry out the mathematical calculations.”

Applicability of dimensional analysis in the real world

In real-life physics, dimensional analysis is a crucial part of the measurement. The following are the uses of dimensional analysis in the real world:

1.     To ensure that a dimensional equation is consistent.

2.     To determine the relationship between physical quantities in physical phenomena.

3.     To switch from one system to another’s units.

4.     To re-introduce a formula that has been forgotten.

Drawbacks of dimensional analysis

Despite multiple usages of dimensional analysis in the real world, the following are the disadvantages of dimensional analysis:

1.     In the case of trigonometric or exponential solutions, data from a large number of experiments may still be inconclusive.

2.     Impractical for more than three parameters to be correlated.

3.     The analysis cannot reflect the addition and removal of parameters.

4.     It cannot give the final word on the validity of a relationship. 

5.     The analysis cannot reflect the addition and removal of parameters.

6.     It cannot give the final word on the validity of a relationship. 

Conclusion

Dimensional analysis is used in all branches of science, maths, and statistics to formulate and derive the formulas and units.

FQAs:

1. You’re preparing a soup that calls for 100 ounces of chicken broth. Unfortunately, chicken broth is only sold by the pint at the grocery store. To create the soup, how many pints of chicken broth is needed? How many pints do you think you’ll need? (Use the 1 pint = 16 fluid ounces conversion factor.)

Solution:

We can get the number of points required by the recipe by multiplying 100 ounces by the conversion factor 1 pint / 16 ounces, which cancels out the unit of “ounces” and leaves us with a pint result. 6.25 pints = (100 ounces) * (1 pint / 16 ounces). Because you’ll need 6.25 pints of chicken broth, you’ll need to buy 7 pints at the shop.

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