Units Dimensions Error 4

Physics Helpline
L K Satapathy
Dimensional Analysis
Units Dimensions Error 4
1 1 1
2 1
2 2 2
a b c
M L T
n n
M L T
     
       
     

Physics Helpline
L K Satapathy Units Dimensions Error 4
To convert the value of a Physical quantity from one system of units to another :
1 1 11 , 1 , 1secM kg L m T  System – 1 (SI units)
2 2 21 , 1 , 1secM g L cm T  System – 2 (CGS units)
Uses of Dimensional Analysis
Consider a Physical quantity having dimensional formula [ ]a b c
M L T
Let its magnitude in SI units = n1  its value in SI units 1 1 1 1[ ]a b c
n M L T 
& its magnitude in CGS units = n2  its value in CGS units 2 2 2 2[ ]a b c
n M L T 
 Equating the values we get 1 1 1 1 2 2 2 2[ ] [ ]a b c a b c
n M L T n M L T  
1 1 1
2 1 1
2 2 2
1 1 1sec
1 1 1sec
a b c a b c
M L T kg m
n n n
M L T g cm
           
                         
3 2
2 1 10 a b
n n 
   [ No contribution from T ]

Physics Helpline
Example : The value of gravitational constant in SI units is 6.67 x 10 –11 N.m2.Kg –2.
Express it in CGS units.
Answer :
Gravitational force is given by
2
2
GMm FRF G
MmR
  
2 1 1 2 2
1 3 2
2 2
[ ][ ] [ ][ ]
[ ] [ ]
[ ] [ ]
F R M LT L
G M L T
M M

 
   
11 1 1
1 2 1
2 2
: 6.67 10 & ( 1 , 3)
a b
M L
Given n n n a b
M L
    
          
   
3 2 11 3 6 8
2 1 10 6.67 10 10 6.67 10a b
n n     
       
 The value of G in CGS units 8 2 2
6.67 10 . [ ].dyn cm g Ans 
 

Physics Helpline
Example : In the Van der Waals equation of state
find the dimensions of a and b .
Answer : Concept : Two physical quantities can be added or
subtracted only when they have the same dimensions
2
( )aP V b RT
V
    
 
2
2
[ ] [ ] [ ][ ]aP a P V
V
      
1 1 2 6 1 5 2
[ ] [ ][ ] [ ][ ]a M L T L M L sT An  
  
0 3 0
[ ] [ ] [ ] [ ]Also b V M L T Ans 
1 1 2 3
[ ] [ ] & [ ] [ ]We have P M L T V L 
 
Determination of the dimensions of a Physical Quantity :

Physics Helpline
Example : De Broglie wave length associated with a particle of mass m , moving
with velocity v is given by , where h is the Planck’s constant. Determine the
dimensions of h using dimension analysis.
Answer :
Given :
h
mv
 
1 1 1 1
[ ] [ ] , [ ] [ ] & [ ] [ ]L m M v LT 
  
h h mv
mv
   
We have
[ ] [ ][ ][ ] . . . (1)h m v 
1 1 1 1
(1) [ ] [ ][ ][ ]h L M LT
 
1 2 1
[ ][ [] ]h M L T Ans
 

Physics Helpline
Example : The volume of liquid (V) flowing per second in a tube of length (l) &
radius of cross-section (r) when pressure difference across it is (P), is given by
Poiseuille’s equation . Determine the dimensions of  .
Answer :
4
8
Pr
V
l



We have
1 1 2 1 3 1 1
[ ] [ ] , [ ] [ ] , [ ] [ ] & [ ] [ ]P M L T r L V L T l L  
   
4 4
8 8
Pr Pr
V
l V l
 


  
4
[ ] . . . (1)
P r
V l

 
   
 
[ Since 8 and  are dimensionless ]
1 1 2 4
3 1 1
[ ][ ]
(1) [ ]
[ ][ ]
M L T L
L T L

 

 
1 1 1
[] ][ [ ]M L T Ans  
 

Physics Helpline
Example : Using dimensional analysis check the correctness of the equation  = I ,
where  is the torque acting on a body of moment of inertia I and  is its angular
acceleration .
Answer :
Given that I 
Dimensionally correct  Dimensions of LHS = Dimensions of RHS
Dimensions of LHS = 1 2 2
[ ] [ . ] [ ] . . . (1)F r M L T 
 
Dimensions of moment of inertia 2 1 2
[ ] [ ] [ ]I MR M L 
Dimensions of angular acceleration 2
[ ] [ ]T 

 Dimensions of RHS = 1 2 2 1 2 2
[ ] [ ][ ] [ ] . . . (2)I M L T M L T  
 
(1) & (2)  The given equation is Dimensionally correct
Checking the correctness of a Physical equation

Physics Helpline
Example : Frequency ( f ) of vibrations of a stretched string is found to depend on its
length ( l ) , tension ( T ) and mass per unit length ( m ). Derive an equation of its
frequency using dimensional analysis.
Answer :
Frequency depends on l a
n l 
Derivation of Physical equation using dimensional analysis :
Frequency depends on T b
n T 
Frequency depends on m c
n m 
Combining , we get a b c
n l T m
. . . (1)a b c
n k l T m 
Where k is a dimensionless constant

Physics Helpline
Dimensions of frequency :
1 1
[ ] [ ]m M L

0 0 1 1 1 2 1 1
(1) [ ] [ ] [ ] [ ]a b c
M L T L M LT M L  
 
11[ ] [ ]n T
T
  
  
Dimensions of length : 1
[ ] [ ]l L
Dimensions of tension : 1 1 2
[ ] [ ]T M LT

Dimensions of mass/unit length :
0 0 1 2
[ ] [ ]b c a b c b
M L T M L T    
 
Applying the principle of homogeneity of dimensions , we get
0 . . . (2)
0 . . . (3)
2 1 . . . (4)
b c
a b c
b
 
  
  

Physics Helpline
& (3) 1a c b    
1(4)
2
b 
1(2)
2
c b     
1 1
1 2 2
(1) n k l T m

  
. . . (5)k Tn
l m
 
The value of the constant k cannot be determined by dimensional analysis
The value of k was experimentally found to be ½ .
1 . . . (6)
2
Tn
l m
 The required equation is

Physics Helpline
L K Satapathy
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Units Dimensions Error 4

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