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

Gate_2020

23) The mean resistivity of a horizontally stratified cuboid rock sample is 100 Ωm and the co-efficient of anisotropy is 1.15. the transverse resistivity of the rock sample is _________Ωm

(Thanks to Chandrasekhar, ANU)

Solution:

Given that 

 

Coefficient of anisotropy ($\lambda$) = 1.15

Mean resistivity of rock($\rho_{m}$) = 100 Ωm

Transverse resistivity of rock ($\rho_{t}$) =?

We know that 

 

Coefficient of anisotropy($\lambda) =\sqrt{\frac{\rho_{t} }{\rho_{s}}}$--------(1)

Mean resistivity of rock($\rho_{m}$) = $\sqrt{\rho_{t}\rho_{s}}$---------------(2)

 

 $\rho_{t}$ Transverse resistivity 

$\rho_{s}$ longitudinal resistivity

Dividing the both the equations we get

$\frac{\lambda}{\rho_{m}}=\sqrt{\frac{\rho_{t} }{\rho_{s}}\times \frac{1}{\rho_{t}\rho_{s}}}$

$\frac{\lambda}{\rho_{m}}=\sqrt{\frac{1}{\rho_{s}^2}}$

$\frac{\lambda}{\rho_{m}}={\frac{1}{\rho_{s}}}$

$\rho_{s}=\frac{\rho_{m}}{\lambda}$

$\rho_{s}=\frac{100}{1.15}$

$\rho_{s}=86.95 Ωm $

By substituting this value in any one of the above two equations

$\rho_{m} = \sqrt{\rho_{t}\rho_{s}}$

$\rho_{m}^2=\rho_{t}\rho_{s}$

$\rho_{t}=\frac{\rho_{m}^2}{\rho_{s}}$

$\rho_{t}=\frac{100\times100}{86.95}$

$\rho_{t}=115.88 Ωm$