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(D) Let $\frac{Q}{A} = \eta^a \left( \frac{S\Delta 	heta}{h} ight)^b \left( \frac{1}{ho g} ight)^c$. Dimensions of $\frac{Q}{A}$ (Heat flux) ar

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$\eta \frac{S\Delta 	heta}{h}$

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(D) Let $\frac{Q}{A} = \eta^a \left( \frac{S\Delta 	heta}{h} ight)^b \left( \frac{1}{ho g} ight)^c$. Dimensions of $\frac{Q}{A}$ (Heat flux) are $[M T^{-3}]$. Dimensions of $\eta$ are $[M L^{-1} T^{-1}]$. Dimensions of $\frac{S\Delta 	heta}{h}$ are $[L^2 T^{-2} K^{-1} \cdot K \cdot L^{-1}] = [L T^{-2}]$. Dimensions of $\frac{1}{ho g}$ are $[(M L^{-3})^{-1} (L T^{-2})^{-1}] = [M^{-1} L^3 \cdot L^{-1} T^2] = [M^{-1} L^2 T^2]$. Equating dimensions: $[M T^{-3}] = [M L^{-1} T^{-1}]^a [L T^{-2}]^b [M^{-1} L^2 T^2]^c$. $[M T^{-3}] = [M^{a-c} L^{-a+b+2c} T^{-a-2b+2c}]$. Comparing powers: $a - c = 1$ $-a + b + 2c = 0$ $-a - 2b + 2c = -3$ Solving these equations: From the first,$a = 1 + c$. Substituting into the third: $-(1+c) - 2b + 2c = -3 \Rightarrow -1 + c - 2b = -3 \Rightarrow c - 2b = -2$. Substituting into the second: $-(1+c) + b + 2c = 0 \Rightarrow c + b = 1$. Adding $c - 2b = -2$ and $2(c + b) = 2$ gives $3c = 0$,so $c = 0$. Then $b = 1$ and $a = 1$. Therefore,$\frac{Q}{A} = \eta \frac{S\Delta 	heta}{h}$.

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