Calculation of true thickness of the bed

 GATE-2018

78. The apparent resistivities obtained at 0.1 Hz and 10 Hz in the frequency domain I.P. measurement are 100 ohm m and 80 ohm m, respectively. The Percentage Frequency Effect is_________.

Solution:

f and F are the frequencies (F>f) and f=0.05 to 0.5 Hz and F=1-10 Hz. Then $\rho_f$ >$\rho_F$ then, frequency effect is

$FE = \frac{\rho_f-\rho_F}{\rho_F}---(1)$

The ratio FE is often multiplied by 100 to express it as a precentage frequecy effect (PFE)

$PFE = 100*\frac{\rho_f-\rho_F}{\rho_F}---(2)$

$PFE = 100*\frac{\rho_{0.1}-\rho_{10}}{\rho_{10}}----(3)$

$\rho_f$ - 100 ohm m

$\rho_F$ - 80 ohm m

Substitute above values in eq (3)

$PFE = 100*\frac{100-80}{80}$

$PFE = 100*\frac{20}{80}$

$PFE = 100*0.25$

$PFE = 25$

Reference : Fundamental of Geophysics, William Lowrie.