91. The gravity anomaly at a point P distant 1.0 km from the position of
the maximum anomaly, along a profile across a horizontal circular cylinder, is
twice the anomaly at a point 1.0 km farther away from P. The depth of the
cylinder is
(Thanks to Nanajee form AU)
A. 1.0 km
B.√2 km
C. 2 km
D. 2 √2 km
Solution:
The gravity anomaly of a horizontal circular cylinder is
$ \triangle g= \frac{2\pi R^2 \rho z }{x^2+z^2}$----(1)
Radius of the circular cylinder = R
Density of the cylinder = ρ
Depth to the cylinder = Z = ?
at X= 1 km the anomaly of the horizontal circular cylinder is
$\triangle g= \frac{2\pi R^2 \rho z }{1+z^2}$---(2)
At X= 2 km the twice the anomaly of the horizontal circular cylinder
is
$2\triangle g= \frac{2\pi R^2 \rho z }{4+z^2}$---(3)
Therefore Δg =2Δg
$ \frac{2\pi R^2 \rho z }{1+z^2}=\frac{2\pi R^2 \rho z }{4+z^2}$
$ \frac{1 }{1+z^2}=\frac{2 }{4+z^2}$
$ {4+z^2}={2+2z^2}$
$z^2=2$
Z =√2 km
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Thank you Sir. Please help me to solve the rest of the anomaly related problems which i had already posted here.
ReplyDeleteNET 2016 DECEMBER:113. A potential difference of 10 mV is measured in an electrical survey with Wenner configuration using an electrode spacing of 10 m and 1.0 amp current. What would be the potential difference if the measurements are made with a Schlumberger configuration employing current electrode spacing of AB =
ReplyDelete30 m, potential electrode spacing of MN = 4 m and the same current?
1. 27.6 mV 2. 2.76 mV
3. 3.62 mV 4. 36.2 mV
3.3.62
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DeleteNET 2016 DECEMBER . 114. Self potential (SP) logs from two shaly sandstone reservoirs A and B record their pseudo-static self potential (PSP) and static self potential (SSP). The ratio of PSP to SSP in A and B are 0.4 and 0.7, respectively. For both A and B, the mud filtrate resistivity is more than the formation water resistivity (Rmf > Rw). This implies that the shale volume of A is
ReplyDelete1. half of that in B
2. twice that in B
3. one-fourth of that in B
4. four times of that in B
Vsh=(1-PSP/SSP) use this relation for both the reservoir A and B, you will find out Vsh of A is Twice than in B.
DeleteNice explanation sujit...
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NET 2016 DECEMBER:
ReplyDeleteIn the above figure, the MT apparent resistivity indicates a vertical contact
1. away from the sounding point, overlain by
thick overburden.
2. away from the sounding point, overlain by thin overburden.
3. at the sounding point, overlain by thick overburden.
4. at the sounding point, overlain by thin overburden.
116. Over a perfect electrical conductor, the elliptically polarized electromagnetic field and quadrature component will, respectively, be
ReplyDelete1. a straight line and zero
2. a straight line and /2
3. a circle and zero
4. a circle and /2
For perfect electrical conductors, the elliptically polarized electromagnetic field and quadrature component will be Straight line and 90 degrees (pi/2)
DeleteFor perfect insulators (non conductive bodies ), the elliptically polarized electromagnetic field and quadrature component will be Circle and 0 degrees (zero)
NET 2015 DECEMBER:93. A spherical body of 1.0 km radius buried at a depth of 2.0 km produces gravity and magnetic anomalies and , respectively at a horizontal distance of 2.0 km from the position of the centre of the sphere. Another spherical body of identical physical properties and of radius 1.0 mile and depth 2.0 miles produces gravity and magnetic anomalies and , respectively at a horizontal distance of 2.0 miles from the position of its centre. Then
ReplyDelete1.
2.
3.
4.
94. Two sedimentary layers of infinite horizontal
extension sit one over the other. The layer
parameters are
Layer 1: depth to top = 2000 m; thickness =
1000 m; density contrast = 0.4gm/cc
Layer 2: depth to top = 3000m; thickness = 1000m; density contrast = 0.4gm/cc.
The total gravity anomaly produced by these two layers on the surface is
1. 84 mGal 2. 42 mGal
3. 0 mGal 4. 21 mGal
93) The gravity anomaly in miles is 1.6 times greater than anomaly in km and there is no change in magnetic anomaly value. we can find these using respective anomaly formulae and substituting 1mile=1.6km
Delete94) By using infinite horizontal thin sheet formula i.e 2piG(rho)t
Delete