Question

in a new system of units energy, density and power are taken as fundamental units, then the dimensional formula of universal gravitational constant G 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

To find: gravitational constant G

Explanation:

here

energy  E  = $[ML^2T^-^2]$

Density  d = $[ML^-^3]$

power P = $[ML^2T^-^3]$

Gravitational constant G will be

$G =\frac{Nm^2}{kg^2}= [M^-^1L^3T^-^2]$

$G = E^ad^bP^c$

$[M^-^1L^3T^-^2]= [ML^2T^-^2]^a[ML^-^3]^b[ML^2T^-^3]^c$

$[M^a^+^b^+^cL^2^a^-^3^b^+2^cT^-^2^a^-^3^c]$

$\text{on comparing}\\\\a+b+c = -1 \\\\2a-3b+2c=3 \\\\-2a-3c=-2\\\\thus \\\\a = -2\\\b = -1\\\c =2 \\\\G = E^-^2d^-^1P^2$

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