The physical dimension is useful because it allows you to monitor what symbols mean using extensive calculations (a technique known as dimensional analysis). Learning how to incorporate it into your skillset is not a waste of effort. It assists you in comprehending the formulas you are employing, seeing how different expressions are related to one another, and memorizing the most important formulas. You’ll also become more efficient at solving problems, spending less time on each one than if you hadn’t used this strategy. For better results, you can also solve the MCQ on Work Power Energy for NEET. It will help me gain 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 dimension units 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, may be 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 uses known equivalent statements or conversion factors to convert between different units of a specific quantity. The dimensional analysis is also called the unit factor and factor label methods.
For example- 365 days =? seconds
365 days *24 hours/day *60 minute/1 hour *60 seconds/1 minute =31536000 seconds
Physical Quantities – Quantities that measurements could not quantify. 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: They can’t be denoted by another unit. For example, length, mass, time, etc.
| S.No. | Fundamental Quantities | SI Units | Symbol |
| 1 | Length | Meter | M |
| 2 | Mass | Kilogram | Kg |
| 3 | Time | Second | S |
| 4 | Electric current | Ampere | A |
| 5 | Temperature | Kelvin | K |
| 6 | Amount of substance | Mol | Mol |
| 7 | Luminous intensity | Candela | Cd |
Derived physical quantities are derived from the base or fundamental quantities. Examples: pressure, speed, young modulus, etc.
| S.No | Derived Quantity | Symbol | Formula | Derived unit |
| 1 | Area | A | Length *width | m2 |
| 2 | Volume | V | Length*Height*width | M3 |
| 3 | Density | p | Mass/volume | Kg/m3 |
| 4 | Velocity | v | Distance/time | m/s |
| 5 | Acceleration | a | Velocity change/time | m/s2 |
| 6 | Work | W | Force* displacement | Kg m2 /s2 |
Law of Dimensional Analysis or Principle of Homogeneity
Making the units the same on both sides of the equation is the objective of the homogeneity principle. The principle aims to represent the physical units in which the dimensions are consistent for each equation. The absence of which cannot be expressed as a physically feasible situation. Let us consider some examples for a better understanding of homogeneity.
Example 1 checks 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 incorrect dimensionally.
We can now solve specific issues. Set up each problem by putting a question mark beside 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 of performing a unit analysis:
- Write the last fact using the units specified in the solution.
- Write the first fact as a fraction with the units included.
- Determine how many conversion facts are required to get from the starting point to the desired result.
- Build a fraction product between the beginning and the end-user with the relevant conversion facts. Arrange the fractions so that the units that must be deleted are canceled.
- Unwanted units should be canceled.
- 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 many 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 science, maths, and statistics branches to formulate and derive formulas and units.
FQAs:
1. You’re preparing a soup 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 are 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.
