Dimensions & Dimensional Formula

 Dimension : Dimension of a physical quantity is defined as the power to which the fundamental units have to be raised to represent the derived unit of that quantity.

Dimensional Formula of Some Physical Quantities:

Quantity                       Dimensions Quantity                    Dimensions
Acceleration                         $ LT^{-2}$

Angular acceleration          T-2

Angular frequency/ speed   T-1

Angular Momentum       $M L^2 T^{-1} $

Angular velocity              T-1

Area                                    L2

Displacement                   L

Energy                      $M L^2 T^{-2}$

(Total /Kinetic /potential/ Internal)

Force                             $ MLT^{-2}$

Frequency                    T1

Gravitational Field strength   LT2

Gravitational potential      L2T2

Length                        L

Mass                           M

Density                     ML3

Momentum            $M L T^{-1}$

Power                      $ML^2 T^{-3}$

Pressure                 $ML^{-1}T^{-2}$

Rotational Inertia     ML2

Time                       T

Torque                 $ML^2T^{-2}$

Velocity                 LT1

Heat                       $ML^2T^{-2}$

Capacitance      $M^{-1}L^{-2}T^4 I^2 $

Charge                        IT

Conductivity      $M^{-1}L^{-3}T^3I^2$

Current                      IT°

Current Density      L2T°I1

Electric dipole moment     LIT

Electric field Strength       MLT3I1

Electric Flux              $ML^3T^{-3}I^{-1}$

Electric Potential      $ML^2T^{-3}I^{-1}$

Electromotive force  $ML^2T^{-3}I^{-1}$

Inductance        $ML^2T{-2}I{-2}$

Magnetic dipole moment     L2T0I

Magnetic field Strength     MT0I1

Magnetic Flux             $M L^2 T^{-2}I^{-1}$

Magnetic Induction     $M T^{-2}I^{-1}$

Permeability            $M L T^{-2} I^{-2}$

Permittivity            $M^{-1}L^{-3} T^4 I^2$

Resistance               $M L^2 T^{-3}I^{-2}$

Resistivity              $M L^3 T^{-3}I^{-2}$

Voltage                 $M L^2 T^{-3}I^{-1}$

Volume                  L3

Wavelength         L

Work /Energy      $M L^2 T^{-2}$

Dimensional Equation :

Whenever the dimension of a Physical quantity is equated with its dimensional formula , we get a dimensional Equation .

Principle of Homogeneity :

According to this , we can multiply Physical quantity with same or different dimensional formula , however in addition & subtraction only like quantities can be added or subtracted .
e.g. If P + Q ⇒ P & Q both represent same Physical quantity .

Solved Example :  Calculate the dimensional formula of Energy from the equation $E = \frac{1}{2}mv^2$

Solution : Dimensionally , E = mass × (velocity)2

Since 1/2 is a number and has no dimension .

$\large [E] = M \times (\frac{L}{T})^2$

$\large [E] = M L^2 T^{-2}$

Solved Example : The kinetic Energy of a particle moving along elliptical trajectory is given by $K = \alpha s^2 $ ; Where s is the distance travelled by the particle . Determine dimension of α .

Solution : $\large K = \alpha s^2 $

$\Large \alpha = \frac{K}{s^2}  $

$\large  [\alpha] = \frac{M L^2 T^{-2}}{L^2} $

$\large [\alpha] = M L^0 T^{-2} $

Solved Example : The distance covered by the particle is given by $\large x = a + bt + ct^2 + dt^3 $ . Find the dimension of a , b , c & d .

Solution : According to principle of homogeneity , all the five quantities should have same dimension . Since [x] = [L]

[a] = [L]

[bt] = [L] ⇒ [b] = [LT-1]

[ct2] = [L] ⇒ [c] = [LT-2]

[dt3] ⇒ [L] = [d] = [LT-3]

What are the Uses of Dimensional equations ?

(i) Conversion of one system of units into another.

(ii) Checking the accuracy of various formulae.

(iii) Derivation of formula

(i) Conversion of one system of units into another.

It is based on fact that ;

Numerical value × Unit = Constant

So on changing unit the numerical value also gets changed . If n1 and n2 are the numerical values of a given physical quantity and u1 & u2 be the units respectively in two different system of units , then

n1 u1 = n2 u2

$\large n_2 = n_1 [\frac{M_1}{M_2}]^a [\frac{L_1}{L_2}]^b [\frac{T_1}{T_2}]^c $

Solved Example : Convert 1 N into dyne.

Solution: Dimensional formula of force is

F = [M1L1T-2]

Now we have to convert M.K.S system into C.G.S system

M = 1kg    L = 1m    T = 1s (M.K.S)

M = 1g    L = 1cm    T = 1s (C.G.S)

F = M1L1T-2

F (SI) = [1 kg] [1m] [s-2]

= [103g] [102 cm] [s-2]

= 105 dyne

Solved Example : Check the accuracy of the relation

$ \displaystyle \nu = \frac{1}{2l}\sqrt\frac{T}{m}$

where ν is the frequency, l is length , T is tension and m is mass per unit length of the string.

Solution : The given relation is

$ \displaystyle \nu = \frac{1}{2l}\sqrt\frac{T}{m}$

Writing the dimensions on either side, we get

LHS = ν = [T-1] = [M0L0T-1]

RHS = $ \displaystyle \nu = \frac{1}{2l}\sqrt\frac{T}{m}$

$ \displaystyle = \frac{1}{L}\sqrt{\frac{MLT^-2}{ML^-1}}$

= [T-1]

As LHS = RHS

Dimensionally the formula is correct

Solved Example :    If force, length and time would have been the fundamental units what would have been the dimensional formula for mass ?

Solution:Let   M = K FaLbTc

[M1L0T0] =  [MLT-2]a [Lb] Tc

[M1L0T0]= [MaL(a+b)T(-2a+c)]

a = 1 , a + b = 0     &  – 2a + c = 0

=> a = 1 , b = – 1 ,  c = 2

Dimensional formula for mass : [FL1T]

Solved Example : In the equation y = A sin(ωt – kx) obtain the dimensional formula of ω and k. Given x is distance and t is time.

Solution:     The given equation is

y = A sin(ωt – kx)

The argument of a trigonometrical function i.e. angle is dimensionless

i.e.   ωt = θ

$ \displaystyle \omega = \frac{\theta}{t} = \frac{1}{T}$

= [T-1]

= [M0L0T-1]

Also      kx = θ

$ \displaystyle k = \frac{\theta}{x} = \frac{1}{L}$

= [ L-1 ]

= [M0L-1T0]

What are the Limitations of Dimensional Analysis:

1. Dimension does not depend on the magnitude of the quantity involved. Therefore, a dimensionally correct equation need not be actually correct.

e.g. : dimension of 1/T and 2π/T are same

2. Dimensional method cannot be used to derive relations other than those involving products of physical parameters.

e.g. : y = a cos(ωt − kx) can not be derived using this method.

3. This method cannot be applied to derive formula if in mechanics a physical quantity depends on more than three physical quantities (mass, length, time). e.g. : $ \displaystyle T = 2\pi \sqrt{\frac{I}{m g L}} $ cannot be derived by using dimensions .

Exercise :   “The dimension of torque is equal to the dimension of work”. Yet, the two quantities are different. Explain.

Also Read :

→  Fundamental Units & Derived Units
→  Significant figures , Rules of counting significant figure
→  Errors in Measurement
→  Vernier Callipers & Screw Gauge
→  Solved Problems : Physical World & Measurement

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