Question

Taking force, length and time to be the fundamental quantities find the dimensions of (a) density, (b) pressure, (c) momentum and (d) energy.

Concept of Physics - 1 , HC VERMA , Chapter "Introduction to Physics".

Answer 1

Hello Dear.


Let us taking the Dimension of the Force as F.


(a) ∵ Density = Mass/Volume.
Since, as per as the Question, Fundamental Quantities are Length, force and acceleration.(but not mass)

  ∴ Density = (Force/Acceleration)/Length³

∵ Acceleration = Velocity/Time.
∴ Acceleration = Velocity/Time².
∴ Dimension of Acceleration =  L T⁻²

∴ Dimension of Density [D] =(F/LT⁻²) ÷ L³
   [D]= F L⁻⁴ T²

Hence, Dimensions of Density is F L⁻⁴ T².

_____________________

(b) ∵ Pressure = Thrust(or Force) ÷ Area.
     ∴ Dimension of Pressure = F/L²
       [P] = F L⁻².

Hence, the Dimension of the Pressure is F L⁻².

______________________

(c) ∵ Energy = Force × Displacement.
∴ Dimension of the Energy = F × L
 [E] = F L

Hence, the Dimensions of the Energy is F L.

________________________

Note ⇒ Force is the Derived units, thus it depends on the other units. But in the Question, it is to be considered that it is the Fundamental unit, therefore I have taken its Dimension as F.


Hope it helps.

Answer 2

Force=F,Length=L,Time=T

Therefore,

1) Density=Mass/Volume

=Force/Acceleration/volume

=[F/LT^-2]/[L^3]

=[FL^-1T^2][L^-3]

=[FL^-4T^2]

2) Pressure=Force/Area

=[F]/[L^2]

=[F][L^-2]

=[FL^-2]

3) Momentum=Mass×Velocity

=[Force/Acceleration]×Velocity

=[F/L^1T^-2][LT^-1]

=[FL^-1T^2][LT^-1]

=[FL^0T^1]

=[FT]

4) Energy=Work done

=Force×Displacement

=[F][L]

=[FL]