Question

Find the dimension formula of momentum if frequency,density and velocity are taken as fundamental units.

Answer 1

Answer:

$P=[F^{-3}DV^4]$

Explanation:

Let the relation between Momentum P, frequency F, density D and velocity V be

$P=[F^aD^bV^c]$    ............(a)

We know that

Dimensions of Frequency F = [T⁻¹]

Dimensions of Density D = [ML⁻³]

Dimensions of Velocity V = [LT⁻¹]

Dimensions of Momentum P = [MLT⁻¹]

Putting the dimensions in eq (a)

$[MLT^{-1}]=[T^{-1}]^a[ML^{-3}]^b[LT^{-1}]^c$

$\implies [MLT^{-1}]=[M^bL^{c-3b}T^{-a-c}]$

comparing the powers on both sides we get

b = 1

c - 3b = 1 ⇒c = 1+3b = 1+3 = 4

-a-c=-1

⇒ -a = -1+c = -1+4 = 3

⇒ a = -3

Thus, the momentum P in terms of frequency, density and velocity can be written as

$P=[F^{-3}D^1V^4]$

or, $P=[F^{-3}DV^4]$