CSIR_NET_JUNE_2019
101)
An MT survey was conducted in two areas having a half space
resistivity of 100 Ωm and 25 Ωm respectively. In the first area a
100 s period signal is required to investigate up to 50km depth. The
signal period in seconds required to investigate up to the same depth
in second area is
A)
100
B) 50
C) 200
D) 400
(Thanks to Chandrasekhar, ANU)
Solution:
The
relation between half-space resistivity. Time period (inverse
frequency) and Skin depth($\delta$) is
$\delta$=
$503.8\sqrt{\frac{\rho}{f}}$-------------(1)
=$503.8\sqrt{{\rho}{T}}$--------------------(2)
case(1):
$\rho_{1}$=100Ωm, T 1=
100s,$\delta_{1}$ = 50km
For
checking
$\delta_{1}
= 503.8\sqrt{{\rho_{1}}{T_{1}}}$
=$503.8
\sqrt{{100\times100}}$
=503.8*100
=50380m
=50.38km
Therefore
$\delta_{1}$ = 50km our estimation is correct.
Case(2):
$\rho_{2}$=25Ωm,
T2 = ?, $\delta_{2}$ = 50km
$\delta_{2}
= 503.8\sqrt{{\rho_{2}}{T_{2}}}$
$50380=
503.8\sqrt{{25}{T_{2}}}$
$50380/503.8
= \sqrt{{25}{T_{2}}}$
$100
= 5\sqrt{{T_{2}}}$
$\sqrt{{T_{2}}}
= 20$
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