Question

Define dimensional formula. Give use of dimensional analysis. Write down the limitations.

Answer 1

Hint – Dimension formula is the expression for the unit of a physical quantity in terms of the fundamental quantities. The fundamental quantities are mass (M), Length (L) and time (T). A dimensional formula is expressed in terms of power of M, L and T.

Complete Step-by-Step solution:
Now as we know that all physical quantities can be expressed in terms of seven fundamental base quantities such as mass, length, time, temperature, electric current, luminous intensity and amount of substance.
These seven quantities are called seven dimensions of the physical world.
We can use symbols instead of the names of the base quantities and they are represented in square brackets.
[M] for mass, [L] for length, [T] for time, [K] for temperature, [I] for current, [cd] for luminous intensity and [mol] for the amount of substance.
These will specify the nature of the unit and not its magnitude.
Uses of dimensional analysis:
$\left( 1 \right)$ It will be used to check the consistency of a dimensional equation.
$\left( 2 \right)$ It will be used to derive the relation between physical quantities in physical phenomena.
$\left( 3 \right)$ It will be used to change units from one system to another.
Limitations of dimensional analysis:
The very most important limitation of dimensional analysis is if the given formula carries the sum of two quantities and if only one of those quantities is provided dimensional analysis cannot be used to verify if the formula is correct or not.
For example:
$v = u$ m/sec and $v = at$ m/sec,
Both are dimensionally correct but the formula is not correct.
As the correct formula is $v = u + at$ m/sec (first law of motion).
So this is the required answer.

Note – The concept of dimensional formula will be clearer by taking into consideration of an example. Let’s compute the dimensional formula for the electric field.
As we know that the electric field (E) is the ratio of force (F) to charge (q)
Therefore, E = (F/q)
Now as we know force is the product of mass (M) to square of acceleration (a)
Therefore, F = (M. $a^2$)
Now as we know that charge (q) is the product of current (I) and time (t).
Therefore, q = (I. t)
Now as we know that the dimension of current (I) is ($A^1$) and the dimension of time (t) is ($T^1$).
So the dimension of charge (q) is ($A^1. T^1$).
Now as we know that the dimension of mass (M) is ($M^1$).
And we know the S.I unit of acceleration (a) is $m/s^2$.
The dimension of meter is ($L^1$) and the dimension of second (s) is ($T^1$).
So the dimension of acceleration is $(L^1T^{-2})$.
Therefore, the dimension of force (F) is $[M^1 L^1T^{-2}]$.
Therefore, the dimension of the electric field is $[M^1 L^1T^{-2}]/[A^1T^1]$.
So on simplifying we get,
Dimension of electric field (E) = $[M^1 L^1 A^{-1} T^{-3}]$.
That is how a dimensional formula is deduced for a quantity.