Question

in a new system of units energy, density and power are taken as fundamental units, then the dimensional formula of momentum will be

Answer 1

the dimesion of the fundamental units is

E is energy -------------> ML²T⁻²

d is density -------------> ML⁻³

P is power --------------> ML²T⁻³

G is grav constant ----> M⁻¹L³T⁻²

using the dimensional analysis we'll get

G = k • (E)^α • (d)^β • (P)^γ ................ k = constant

⇔ M⁻¹L³T⁻² = (ML²T⁻²)^α • (ML⁻³)^β • (ML²T⁻³)^γ

⇔ M⁻¹L³T⁻² = M^(α + β + γ) • L^(2α - 3β + 2γ) • T^(-2α - 3γ) .................(1)

based on the eqn (1) we get

α + β + γ = -1

2α - 3β + 2γ = 3

-2α - 3γ = -2

solve for α, β, and γ until you get value of

α = -2

β = -1

γ = 2

then, the dimensional formula of universal gravitational constant is

#nancy1D

G = k(E⁻² d⁻¹ P²) .

I hope this will help

Answer 2

Answer:

G = E⁻²  D⁻¹  P²

Explanation:

E = [ M  L²  T⁻² ]

P = [ M  L²  T⁻³ ]

D = [ M  L⁻³  T⁰ ]

G = Nm²/Kg²  => [ M⁻¹  L³  T⁻² ]

G = Eᵃ  Dᵇ  Pˣ

Put values of E, D, P

G = [ M  L²  T⁻² ]ᵃ  [ M  L⁻³ ]ᵇ  [ M  L²  T⁻³ ]ˣ

G = [ Mᵃ⁺ᵇ⁺ˣ  L²ᵃ⁻³ᵇ⁺²ˣ  T⁻²ᵃ⁻³ˣ ]

COMPARE

a + b + x = -1

2a - 3b + 2x = 3

-2a -3x = -2

a = -2

b = -1

x = 2

G = [ E⁻²  D⁻¹  P² ]