DIMENSIONS

We may define the dimensions of a physical quantity as the powers to which the
fundamental units of mass, length and time have to be raised to represent a derived unit
of the quantity.
For example,  For example,
 velocity = [L1 T-1] = [M0L1 T-1]
Hence the dimensions of velocity are: zero in mass, +1 in length and -1 in time. Hence the dimensions of velocity are: zero in mass, +1 in length and -1 in time.

DIMENSIONAL FORMULA
The expression which shows how and which of the base quantities represent the
dimensions of a physical quantity is called the dimensional formula of the given physical
quantity.
For example, the dimensional formula of the volume is [M° L3 T°],
and dimensional formula of speed or velocity is [M° L T-1].
Similarly, [M° L T–2] is the dimensional formula of acceleration and [M L–3 T°] that of
mass density.
DIMENSIONAL EQUATION:
An equation obtained by equating a physical quantity with its dimensional formula is
called the dimensional equation of the physical quantity.
For example, the dimensional equations of volume [V], speed [v], force [F] and mass
density [ρ] may be expressed as
[V] = [M0 L3 T0]
[v] = [M0 L T–1]
[F] = [M L T–2]
[ρ] = [M L–3 T0]

Application of Dimensional Analysis:
1. To check the correctness or consistency of a given formula
2. To derive relationship among various physical quantities.
3. To convert one system of unit into other system.

• To check the correctness or consistency of a given formula    
• The correctness or consistency or accuracy of a formula can be checked by
  applying the principle of homogeneity of dimensions. According to this principle
  dimensions of each term on the both sides of the formula is always same.
  Example: to check the accuracy of v2 –u2 = 2as
  Solution:
  The dimensions of LHS = [LT-1]2 - [LT-1]2 = [L2 T-2] - [L2 T-2] = [L2 T-2]
  The dimensions of RHS = 2aS = [L T-2] [L] = [L2 T-2]
  The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.
•  
•  
• 2. To derive relationship among various physical quantities.
  Let a physical quantity Q depends upon the quantities q1 , q2, and q3 such that:
  Q ∝ Š VŠ Š W
  Or Q = k Š VŠ Š W ← (1)
  Where k is dimensionless constant.
  Now, we will apply the following steps -
  i) Write dimensional formula of each quantity on both sides of equation (1)
  ii) Equate the powers of M, L and T and solve three equations so obtained
  and find a, b and c.
  iii) Now we will put the values of a, b and c in equation (1), hence we get a
  new physical relation(formula)
   
• 3. To convert one system of unit into other system.
  Let M1 ,L1 and T1 are fundamental units of mass, length and time in one system of unit.
  & M2 ,L2 and T2 are fundamental units of mass, length and time in another system.
  Let a, b and c be the dimensions of the given quantity in mass, length and time
  respectively.
  If n1 is the numerical value of the quantity in one system then that n2 in other system is
  given by the formula:
  n2 = n1’“
  “
   
•  
• Limitations of Dimensional Analysis:
  1. Using dimensional analysis we cannot find the value of dimensionless constant.
  2. We cannot derive the relation containing exponential and trigonometric
  functions.
  3. It cannot inform that whether a quantity is scalar or vector.
  4. It cannot find the exact nature of plus or minus, connecting two or more terms in
  formula.
  5. The relation containing more than three physical quantities cannot be derived
  using dimensional analysis.