GATE-2018 (24) :Wyllie – Time average equation

CSIR-NET 2015

113. The following table shows the values of resistance R measured in a Wenner sounding for different electrode spacing, a at a location covered by clay.

a(m)          =        5                      10                 15               20            25

R(ohms)    =       1.0                    0.8                0.6              0.4           0.2

Which of the following conclusions can be made?

A) clay extends to a depth of at least 25m

B) clay is underlain by a thick basement of high resistivity 

C) clay is underlain by water bearing sands overlying a thick basement

D) clay is underlain by water bearing gravel overlying a clay formation

Answer: D

Solution:

In Wenner configuration the current and potential 

electrode pairs have a common mid-point and the

 distance between adjacent electrodes are equal,

apparent resistivity ρ_a = 2Пa ΔV /I  ---------(1)

  a= electrode spacing

ΔV / I = R , Resistance

therefore ρ_a = 2ΠaR

1. apparent resistivity at a=5m , R= 1.0 Ω is

ρ₁ = 2*3.14*5*1.0

    =31.40 Ωm

2. apparent resistivity at a = 10m R = 0.8Ω is

ρ₂ = 2*3.14*10*0.8

    = 50.24Ωm

3. apparant resistivity at a=15m R=0.6Ω is

ρ₃ = 2*3.14*15*0.6

    = 94.20 Ωm

4. apparent resistivity at a = 20m R = 0.4Ω is

ρ₄ = 2*3.14*20*0.4

    = 50.24 Ωm

5. apparent resistivity at a=25m R=0.2Ω is

ρ₅ = 2*3.14*25*0.2

    =31.40Ωm

apparent resistivity values → ρ₁ < ρ₂ < ρ₃ > ρ₄ > ρ₅

Therefore , Clay is presented upper layer and inner 

layer, In between water bearing gravel present ( 

because gravel is high resistivity).