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

CSIR NET- JUNE 2016

104) In a electrical logging a current of 1 A is passed through the electrodes C₁ and C_2, and a voltage of 1.4V is measured between the potential electrodes P₁ and P₂. The electrode P₂ & C₂ being grounded at the surface and P₁ & C₁ being lowered into the borehole separated by a distance of 40cm. The apparent resistivity is 

A)35 Ωm 

B) 70 Ωm

C) 105 Ωm 

D) 120 Ωm 

 (Thank to Chandrasekhar, ANU)

Solution:

Where M,N are the potential electrodes and A,B are the current electrodes

$Apparent resistivity(\rho_{a}) = 4\pi(\frac{\triangledown V}{I})AM$

where,

   spacing(AM) = 40cm = 0.4m

   $\triangledown V= 1.4 V$   

  
   I= 1A

$Apparent resistivity(\rho_{a}) = 4\pi(\frac{\triangledown V}{I})AM$

                                                      =4*3.14*1.4*0.4

$\rho_{a}$ =7.03 Ωm

(I am getting 7.03 ohm m, if anyone get the answer please share in this blog)