Unit & Dimension, Measurement
Uses of Dimensional equation
A. To change the value of a physical quantity from one system to another :
Example : Convert 1 newton into dyne.
Force = mass x acceleration
In this way the conversion factor for any derived physical quantity can be calculated if the dimensional formula of the derived quantity is known.
B. Homogenity of Dimensions in an equation : On the basis of this principle the accuracy of an equation can be checked. In a physical equation every term should have the same dimensions.
Example : The given relation is v = u + at
Dimensionally:
As in the above equation dimensions of both sides are the same. Hence, this formula is correct dimensionally.
Example : Consider the formula.
As in the above equation dimensions of both sides are not same, this formula is not correct dimensionally.
Example : In van-der Wall’s equation 
= RT, obtain the dimension of the constant a and b, where P is pressure, V is volume, T is temperature and R is gas constant.
Solution : The given equation is 
As pressure can be added only to pressure, therefore, 
represents pressure P.
Again, from volume V, one can subtract only the volume. Therefore, ‘b’ must represent volume i.e., 
C. Deducing Relation Among the Physical Quantities : The relation between different physical quantities can be established with the help of dimensions. It will be possible if we know the quantities on which a particular physical quantity depends.
Example : Einstein Mass-Energy Relation : When mass is converted into energy, let the energy produced depends on the mass (m) and speed of light (c) i.e.,
Solution: 
where K is a dimensionless quantity.
Equating the exponents of similar quantities a = 1 and b = 2
So the required physical relation becomes, E = mc2
The value of dimensionless constant is found unity through experiment.
E = mc2
Example : To find the expression for the time period of a simple pendulum which depends on the its length, mass of the bob and acceleration due to gravity.
Solution : We know
where K is a dimensionless constant and a, b and c are exponents which we have to evaluate. Taking the dimensions of both sides.
Putting these values in the above given equation: 
Example : Assuming that the critical velocity of flow of a liquid through a narrow tube depends on the radius of the tube, density of the liquid and viscosity of the liquid, find an expression for critical velocity.
Solution : Suppose, 
where r = radius of the tube,
ρ = density of the liquid,
η = co-efficient of viscosity of the liquid
and a, b and c are the unknown powers to be determined.
Equating the powers of M, L and T, we have
