Question

Among the given layered models, labelled as P, Q, R and S, which of the following pairs is/are NOT possible to distinguish according to the principle of equivalence?

Model parameters read off the figure -- (P): \(\rho_1 = 1\ \Omega\text{m}, h_1 = 10\ \text{m}\); \(\rho_2 = 100\ \Omega\text{m}, h_2 = 20\ \text{m}\); \(\rho_3 = 0\ \Omega\text{m}\).
(Q): \(\rho_1 = 5\ \Omega\text{m}, h_1 = 10\ \text{m}\); \(\rho_2 = 400\ \Omega\text{m}, h_2 = 5\ \text{m}\); \(\rho_3 = 0\ \Omega\text{m}\).
(R): \(\rho_1 = 40\ \Omega\text{m}, h_1 = 20\ \text{m}\); \(\rho_2 = 2\ \Omega\text{m}, h_2 = 10\ \text{m}\); \(\rho_3 = 100\ \Omega\text{m}\).
(S): \(\rho_1 = 30\ \Omega\text{m}\); \(\rho_2 = 3\ \Omega\text{m}, h_2 = 10\ \text{m}\), with \(h_1 = 5\ \text{m}\); \(\rho_3 = 100\ \Omega\text{m}\).

• A. P, R
• B. P, S
• C. P, Q
• D. R, S

Hint:

For a resistive middle layer (K-type), curves are equivalent when rho2*h2 (transverse resistance) matches; for a conductive middle layer (H-type), they are equivalent when h2/rho2 (conductance) matches. Compute both for all four models.

Answer 1

The Correct Option is C

The principle of equivalence in DC resistivity sounding says that when a middle layer is thin compared to its depth, a range of (thickness, resistivity) combinations for that middle layer produce essentially indistinguishable apparent-resistivity sounding curves, PROVIDED a particular combined parameter is held constant:

• If the middle layer is resistive relative to its neighbours (a K-type sequence, \(\rho_1 < \rho_2 > \rho_3\)), the curves are equivalent when the transverse resistance \(T = \rho_2 h_2\) is the same (current is forced to flow mostly across/through the resistive layer, so only the total resistance it presents matters).
• If the middle layer is conductive relative to its neighbours (an H-type sequence, \(\rho_1 > \rho_2 < \rho_3\)), the curves are equivalent when the longitudinal conductance \(S = h_2/\rho_2\) is the same (current channels along the conductive layer, so only its total conductance matters).

Classify and compute for each model:

P: \(\rho_1=1 < \rho_2=100 > \rho_3=0\) -- resistive middle layer (K-type). \(T_P = \rho_2 h_2 = 100 \times 20 = 2000\ \Omega\text{m}^2\).

Q: \(\rho_1=5 < \rho_2=400 > \rho_3=0\) -- resistive middle layer (K-type). \(T_Q = \rho_2 h_2 = 400 \times 5 = 2000\ \Omega\text{m}^2\).

R: \(\rho_1=40 > \rho_2=2 < \rho_3=100\) -- conductive middle layer (H-type). \(S_R = h_2/\rho_2 = 10/2 = 5\ \text{S}\).

S: \(\rho_1=30 > \rho_2=3 < \rho_3=100\) -- conductive middle layer (H-type). \(S_S = h_2/\rho_2 = 10/3 \approx 3.33\ \text{S}\).

P and Q are both K-type with IDENTICAL transverse resistance (\(T = 2000\ \Omega\text{m}^2\) for both), so their sounding curves are equivalent and CANNOT be distinguished from one another. R and S are both H-type but their conductances differ (5 S versus 3.33 S), so R and S ARE distinguishable from each other -- ruling out option (D). P and R (option A) and P and S (option B) are of different curve types (K versus H) entirely and are trivially distinguishable, not a matched equivalence pair.

\(\boxed{\text{P and Q cannot be distinguished -- option (C)}}\)