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(B) Let the density $\rho$ be expressed as $\rho = k C^a G^b h^c$.Dimensional formulas are:$[\rho] = M L^{-3}$$[C] = L T^{-1}$$[G] = M^{-1} L^3 T^{-2}

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$C^5 G^{-2} h^{-1}$

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(B) Let the density $\rho$ be expressed as $\rho = k C^a G^b h^c$.Dimensional formulas are:$[\rho] = M L^{-3}$$[C] = L T^{-1}$$[G] = M^{-1} L^3 T^{-2}$$[h] = M L^2 T^{-1}$Substituting these into the equation:$M^1 L^{-3} T^0 = (L T^{-1})^a (M^{-1} L^3 T^{-2})^b (M L^2 T^{-1})^c$$M^1 L^{-3} T^0 = M^{-b+c} L^{a+3b+2c} T^{-a-2b-c}$Comparing the exponents of $M, L,$ and $T$:$1$) $-b + c = 1 \implies c = 1 + b$$2$) $a + 3b + 2c = -3$$3$) $-a - 2b - c = 0 \implies a = -2b - c$Substitute $c = 1 + b$ into $(3)$: $a = -2b - (1 + b) = -3b - 1$Substitute $a$ and $c$ into $(2)$: $(-3b - 1) + 3b + 2(1 + b) = -3$$-3b - 1 + 3b + 2 + 2b = -3$$2b + 1 = -3 \implies 2b = -4 \implies b = -2$Now find $c$: $c = 1 + (-2) = -1$Now find $a$: $a = -3(-2) - 1 = 6 - 1 = 5$Thus, the dimension of density is $[\rho] = C^5 G^{-2} h^{-1}$.

If the velocity of light $C$, the gravitational constant $G$, and Planck's constant $h$ are chosen as the fundamental units, the dimension of density in the new system is

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