11.1 Measurements-01

Find the dimensions of

a.      Power

b.     Force

c.      Permittivity of free space

Solution-

b. ‘ Force’ -

According to Newton’s second law of motion

F = m a

Therefore,

Dimensions of ‘F’ = Dimensions of ‘m’ X Dimensions of ‘a’

Dimensions of ‘ F’ = [M¹,L^o,S^o] X [M⁰,L¹, S^-2]

Dimensions of ‘ F’= [M¹,L¹,S^-2]

_ _ _ _ _ _ _ _ _ _ (1)

a.     Power –

Power can be defined as the work done per unit time.

i.e.

P = w/t

and work done on particle is nothing but product of force applied on that particle and displacement done by the same particle.

i.e.

w = F s

Therefore,

P = ( F s ) / t

Dimensions of ‘P’={ Dimensions of ‘F’ X Dimensions of ‘ s ’} / Dimensions of ‘ t’

Dimensions of ‘ P ’= {[M¹,L¹,S^-2] X [M⁰,L¹,S^o]}/ [M⁰,L^o,S¹]

Dimensions of ‘ P ’= [M¹,L²,S^-3]

c. Permittivity of free space

 Let’s consider Coulomb’s law of electrostatic force of attraction.

F = {1/(4πε_o)} {(q₁q₂)/r² }

Therefore,

ε_o = {1/(4πF)} {(q₁q₂)/r² }

Dimensions of ‘ε_o’ = Dimensions of {1/(4πF)}  X Dimensions of {(q₁q₂)/r² }

_ _ _ _ _ _ _ _ _ _ (2)

Here,

4 and π are numbers without any dimension

And

Dimensions of ‘ F’= [M¹,L¹,S^-2]

Dimensions of ‘ q’ =

We know that current is rate of flow of charge .

Therefore,

I = q/t

i.e.

 q = I t

Therefore,

Dimensions of ‘ q’ = Dimensions of ‘ I ’ X Dimensions of ‘ t ’

Dimensions of ‘ q’ =   [M⁰,L⁰,S^-1 ,A¹]

And r is radius

Dimensions of ‘ r’ = [M⁰,L¹,S⁰ ,A⁰]

Putting this value in equation ( 2 ) we get,

Dimensions of ‘ε_o’ = Dimensions of {1/(F)}  X Dimensions of {(q₁q₂)/r² }

Dimensions of ‘ε_o’={1/[M¹,L¹,S^-2]}  X [M⁰,L⁰,S^-1 ,A¹] [M⁰,L⁰,S^-1 ,A¹] / [M⁰,L²,S⁰ ,A⁰] }

Dimensions of ‘ε_o’=[M^-1,L^-1,S²]  X {[M⁰,L⁰,S^-2 ,A²] / [M⁰,L²,S⁰ ,A⁰] }

Dimensions of ‘ε_o’= [M^-1,L^-3,S⁴, A² ]