Results from the permeability test of rock samples, E, F, and G are shown. The correct order of hydraulic conductivity (k) of the rock samples is Note: Q is water flow rate, A is area of cross-section of the sample, h is water head, L is length of the sample.
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The slope of the Q versus Ah/L line for each sample equals its hydraulic conductivity by Darcy's law.
This question checks whether we can read hydraulic conductivity straight off a Q versus $Ah/L$ plot without doing any extra maths, using Darcy's law $Q = k(Ah/L)$.
Rearranging Darcy's law for $k$ gives $k = \dfrac{Q}{Ah/L}$, which is just the slope of each line on the graph. So instead of comparing lines by eye as steep or flat, we can compare $k$ by picking the same value of $Ah/L$ on the x-axis and reading off which sample gives the largest $Q$ there.
$k_E > k_F > k_G$: at any fixed $Ah/L$, line E sits highest, meaning it produces the largest flow $Q$ for that gradient, so it has the largest $k$. Line F sits below E but above G, so its $k$ is in between. Line G sits lowest, so it has the smallest $k$. This matches the graph exactly.
$k_E < k_F < k_G$: this reverses the order completely, it would need G to sit above E on the graph, which it does not.
$k_E = k_F = k_G$: this would only hold if all three lines coincided, but the graph clearly shows three separate slopes.
$k_F > k_G > k_E$: this puts E last, but E is the steepest line on the graph, so E cannot have the smallest $k$.
Let's summarize:
On a Q versus $Ah/L$ plot, the slope of each straight line is numerically equal to the hydraulic conductivity $k$ of that sample.
The steeper the line, the more water it lets pass for the same head and geometry, so the higher its $k$.
Since line E is the steepest, F is next, and G is the flattest, the order is $k_E > k_F > k_G$.
