d
\(\begin{array}{l}
Let\,\frac{Q}{A} = {\eta ^a}{\left( {\frac{{S\Delta \theta }}{h}} \right)^b}{\left( {\frac{1}{{\rho g}}} \right)^c}\\
{\rm{Using}}\,{\rm{dimensional}}\,method\\
M{T^{ - 3}} = {\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]^a}{\left[ {L{T^{ - 2}}} \right]^b}{\left[ {{M^{ - 1}}{L^2}{T^2}} \right]^c}\\
or,\,M{T^{ - 3}} = \left[ {{M^{a - c}}{L^{ - a + b + 2c}}{T^{ - a - 2b + 2c}}} \right]\\
Equating\,powers\,and\,solving\\
we\,get,a = 1,b = 1,c = 0\\
\therefore \frac{Q}{A} = \eta \frac{{S\Delta \theta }}{h}
\end{array}\)