# Dimensions and uses of dimensions

# Dimensions and uses of dimensions

In this topic we will discuss about what is Dimensions and uses of dimensions ?Also what are the limitations of dimensions ? This topic is easy to understand and video related to the topic is also linked with the given notes of Dimensions and uses of dimensions .

Dimensions and uses of dimensions-

**DIMENSIONS**

When a physical quantity expressed In terms of fundamental units , it is written as a product of different powers of the fundamental quantity is known as the dimension of the physical quantity .

Some fundamental units are – Length – [L] , Mass – [M] , Time – [T] , Ampere-[A] etc.

Here – [L] , [M] , [T] , [A] are called dimensions .

Dimension of some physical quantity may be written as –

(i) speed – distance/time =[L]/[T] = [LT^{-1}] , Or , [M^{0}L^{1}T^{-1}] ;

(II) Acceleration- Change in velocity/time = dimension of velocity/dimension of time = [LT^{-1}]/[T] =[LT^{-2}]

(III) Force – Mass x acceleration = [M] [LT^{-2}]= [MLT^{-2}] Similarly we can find the dimension of all other physical quantity .

**Uses of dimensions – **

**(i) To check the correctness of the formula** – [ Principle of Homogeneity – according to this principle for a given formula , the dimension of every term of L.H.S. = dimensions of every term of R.H.S. ]

{** we must keep in mind that when two physical quantity is added or subtracted there dimensions will be same }

Lets check the correctness of given formula S= ut + ½ (at^{2} )

Here L.H.S. dimension of s= [L]

Dimension of ut = [LT^{-1}] X [T] = [L]

Dimension of next term of R.H.S ½ at^{2} = [LT^{-2}][T]^{2} = [L]

So dimension of R.H.S = [L] + [L] = [L]

So we can say dimension of LHS = Dimension of RHS , it follows the principle of homogeneity and hence the given formula is correct .

[ note- using this process we can find the dimensions of unknown physical quantity ]

**(ii) Conversion of unit of a physical quantity** – Dimension also help us to convert the unit of a physical quantity from one form to another Ex- MKS to CGS or vice-versa .

As we know , when we change the unit then the magnitude of the physical quantity changes but it express the same physical quantity . example – 1 km = 1000m ; here when we changes the unit km to m its numeric value changes but it express the same physical quantity which is distance/displacement .

So we can say for a given physical quantity n_{1}u_{1}=n_{2}u_{2} ;

Using this we can write-

n_{1}[M_{1} ^{a} L_{1} ^{b} T_{1 }^{c} ] = n_{2}[M_{2} ^{a} L_{2} ^{b} T_{2 }^{c} ]

So ,

n_{2} = n_{1} [M_{1} ^{a} L_{1} ^{b} T_{1 }^{c} ]/ [M_{2} ^{a} L_{2} ^{b} T_{2 }^{c} ]

= [M_{1}/M_{2}]^{a} [L_{1}/L_{2}]^{b} [T_{1}/T_{2}]^{C} ………..Eq(I)

Suppose we have to convert 10 Newton( MKS) force into Dyne (CGS) . At first we have to write the dimension of the physical quantity which is force ( since Newton is the unit of force ) .

Dimension of force is [MLT^{-2}] i.e. a=1 , b=1, c=-2 and n_{1} = 10 ;

So applying the eq. (i) n_{2} = n_{1} [M_{1}/M_{2}]^{a} [L_{1}/L_{2}]^{b} [T_{1}/T_{2}]^{C}

n_{2} = 10 x [ 1kg/1gm] [1m/1cm] [1s/1s]

n_{2}= 10 x [1000gm/1gm] [ 100cm/1cm ] [1s/1s] = 10 x 10^{5}

10 Newton = 10 x 10^{5} Dyne ;

Similarly we can convert a physical quantity from one unit to another .

**(iii) Deducing relation among the physical quantity **– Dimension is also used to find the actual relationship among the given physical quantities or in another words we can say it is used to derive the formula with the help of given physical quantities .

Suppose if it is given the pressure (p) acting on a body depends on force applied (F) and area of cross-section A ,

We can write P α F^{a} A^{b}

Or, P = K F^{a} A^{b} ( Where K is a constant ) ………………………..(I)

Dimension of, pressure P = [ ML^{-1} T^{-2}]

Dimension of , force F = [MLT^{-2}]

Dimension of area A = [L^{2}]

Putting all these values in equation (i) we can write

[ ML^{-1}T^{-2}] = [MLT^{-2}]^{a} [L^{2}]^{b}

[ ML^{-1}T^{-2}] = [M^{a} L^{a+2b} T^{-2a}] , on comparing the power of both sides we get ;

a=1 , a+2b=-1 and -2a =- 2

so a=1 and b=-1 . putting these values in eq. 1 we get

pressure p = F/A ;

again , suppose time period (t) of a pendulum depends on mass of the bob ’m’ length of the pendulum ‘l’ and acceleration of the gravity ‘g’. we have to find the formula of the time period .

we can write

t α m^{a} l^{b} g^{c}

or, t = K m^{a} l^{b} g^{c} ……………..(i) where K is a constant .

[M^{0}L^{0}T^{1}] = [ M]^{a} [L]^{b} [LT^{-2}]^{c} = [M^{a }L^{b+c} T^{-2c}]

On comparing we get , a=0 ,b+c=0 and -2c=1

On solving all the above relations we get , a=0 . b=1/2 and c=-1/2 ; putting all these value in the above eq (i) we get,

t= Kl^{1/2}g^{-1/2} here K= 2∏ √l/g .

**limitations of dimensions** –

(i) The dimensional method works only when the dependence is of the product time . it does not show the relation in addition or subtraction .

(ii) The numerical constant having no dimensions cannot be deduced by the method of dimension.

(iii) the method works only when a physical quantity depends on only three physical quantity . This method is not valid for four or more physical quantity dependence.

Assignment related to the unit and dimension is also available for the students , which will help the student to solve questions related to these topics .

For objective type questions for unit and dimension-

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