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(B) The dimensional formula of the universal gravitational constant $G$ is $[M^{-1} L^3 T^{-2}]$.Let $G = [E^a d^b P^c]$.The dimensions of the fundame

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$[E^{-2} d^{-1} P^2]$

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(B) The dimensional formula of the universal gravitational constant $G$ is $[M^{-1} L^3 T^{-2}]$.Let $G = [E^a d^b P^c]$.The dimensions of the fundamental units are:$E = [M L^2 T^{-2}]$$d = [M L^{-3}]$$P = [M L^2 T^{-3}]$Substituting these into the equation:$[M^{-1} L^3 T^{-2}] = [M L^2 T^{-2}]^a [M L^{-3}]^b [M L^2 T^{-3}]^c$$[M^{-1} L^3 T^{-2}] = [M^{a+b+c} L^{2a-3b+2c} T^{-2a-3c}]$Comparing the powers of $M, L,$ and $T$ on both sides:$1) a + b + c = -1$$2) 2a - 3b + 2c = 3$$3) -2a - 3c = -2 \Rightarrow 2a + 3c = 2$From $(3)$,$c = (2 - 2a)/3$. Substituting $c$ and $b = -1 - a - c$ into $(2)$:Solving the system of linear equations yields $a = -2, b = -1, c = 2$.Therefore,$G = [E^{-2} d^{-1} P^2]$.

In a new system of units,energy $(E)$,density $(d)$,and power $(P)$ are taken as fundamental units. Then the dimensional formula of the universal gravitational constant $G$ will be .......

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