Question
Pressure gradient has the same dimension as that of
Potential gradient$ = \frac{V}{x} = \frac{{[M{L^2}{T^{ - 3}}{A^{ - 1}}]}}{{[L]}}$$ = [ML{T^{ - 3}}{A^{ - 1}}]$
Energy gradient $ = \frac{E}{x} = \frac{{[M{L^2}{T^2}]}}{{[L]}} = [ML{T^{ - 2}}]$
and pressure gradient$ = \frac{P}{x} = \frac{{[M{L^{ - 1}}{T^{ - 2}}]}}{{[L]}} = [M{L^{ - 2}}{T^{ - 2}}]$
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[Area of cross section of wire $=0.005 \mathrm{~cm}^2$, $\mathrm{Y}=2 \times 10^{11}\ \mathrm{Nm}^{-2}$ and $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$
