Dimension

 Dimension

Dimension :The dimensions of a physical quantity are the powers to which the concerned fundamental units must be raised in order to obtain the unit of the given physical quantity

Dimensional Formula :When we represent any derived quantity with appropriate powers of symbols of the fundamental quantities, then such an expression is called dimensional formula.

This dimensional formula is expressed by square bracket and no comma is written in between any of the symbols.

   Dimensional formula of speed:

               

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Let us see some Examples

Dimension of Temperature Gradient:

  Temperature Gradient = Temperature/Distance  = ([ L0  M0 T0  K1 ]) / ([ L1   M0  T ]) = ([ K1 ]) / ([ L1 ]) = [ L-1 K1 ]

Dimension of Velocity:

      Velocity = Displacement/Time = ([ L1  M0 T0   ]) / ([ L0   M0  T1 ])   = [L1  M0 T-1]

Dimension of Acceleration :

      Acceleartion = Velocity/Time  = ([ L1  M0 T-1   ]) / ([ L0   M0  T1 ])  = [L1  M0 T-2]

Dimension of Force  :

     Force = Mass x  Acceleration 

           = [L0 M1 T0] x [L1 M0 T-2]  =  [L1 M1 T-2]

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Uses of Dimensional Analysis

• To check correctness of a Physical Quantities

In any equation relating different physical quantities, if the dimensions of all the terms on both the sides are the same then that equation is said to be dimensionally correct. This is called the principle of homogeneity of dimensions. Consider the first equation of motion.

Let Us Understand:

Consider an Equation : S= ut + 1/2  at2

Writing Dimension For each quantity

S =  [ L1  M0 T0 ]  --------- LHS

ut = [ L1  M0 T0 ] ----------RHS

at2=[ L1  M0 T0] ----------RHS

[L.H.S] = [R.H.S]

As the dimensions of L.H.S and R.H.S are the same, the given equation is dimensionally correct.

• To establish the relationship between related physical quantities

  The period T of oscillation of a simple pendulum depends on length l and acceleration due to gravity g. Let us derive the relation between T, l, g

  Suppose T ∝ la ------ (1)

                  T ∝ gb ------ (2)

            T ∝ la gb  ------ From (1) and (2)

            T = K la gb  

  where k is constant of proportionality and it is a dimensionless quantity and a and b are rational numbers. Equating dimensions on both sides  we get,

  [M0 L0 T1 ] = k [L1 ]a [LT-2]b

               = k [La+b T-2b]

  [L0 T1 ] = k [La+b T-2b] 

  Comparing the dimensions of the corresponding quantities on both the sides we get, a + b = 0 ∴ a = -b   and -2b=1 , b = -1/2 , ∴a = -b = -(-1/2) , a = ½

  T = k l1/2 g -1/2   T = k√(l/g) ,   T= 2π √(l/g) .The value of k is determined experimentally and is found to be 2π

  
  

• To find the conversion factor between the units of the same physical quantity in two different systems of units:

  
  

  
  

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