Gravity Methods Questions and Answers

Q: In Bouguer gravity anomaly map, which of the following statements are/is correct. Select appropriate answer :

The answer is All of the above are correct.A Bouguer anomaly map shows the effect of density variations beneath the surface. Broad, long-wavelength anomalies come from deep density contrasts, such as crustal thickness changes or major basins, and reveal regional structure. Short-wavelength anomalies come from shallow sources, such as ore bodies, salt domes or buried channels, and are the targets of exploration.The residual anomaly, left after the regional trend is removed, has no unique interpretation: many different bodies can produce the same anomaly. Geological knowledge of the area is needed to constrain it. Since each of the three statements is true, choosing any single one is incomplete.

Q: In gravity surveys the readings are function of :

Gravity readings are a function of the density and mass of the rocks.A gravimeter measures the vertical attraction of the earth at a point. By Newton's law of gravitation, this depends on the masses beneath the station and their distances, and so variations in reading reflect lateral differences in rock density, the density contrast between bodies and their surroundings.Rigidity and bulk modulus are elastic properties that control seismic velocities, not gravity. Hardness has no effect on gravity, and specific gravity alone, paired with hardness, misses the dependence on the mass and geometry of the bodies.

Q: The value of 'g' (acceleration due to gravity) at a place varies with time due to which of the following :

The value of g at a place varies with time because of isostatic rebound.Where an ice sheet has melted, as in Scandinavia and Canada, the crust is still rising slowly to regain isostatic equilibrium, and mantle material flows in beneath it. The station moves away from the Earth's centre and the mass distribution beneath it changes, so the measured gravity there changes measurably over years and decades.The topography of the continents and lateral differences in rock composition make g vary from place to place, not with time at one place. The Earth's rotation lowers g by a fixed amount that depends on latitude, so it too gives no change over time at a given station.

Q: Bouguer correction is applied to correct for the gravity anomaly due to mass between station location and .......................... .

The key completes the sentence with local datum plane.After the free-air correction has allowed for the height of the station, the Bouguer correction removes the attraction of the slab of rock between the station and the reference level to which the survey is reduced: \(0.0419\,\rho h\) mGal for density \(\rho\) in g/cm\(^{3}\) and height h in metres. That reference is the survey's datum plane, which in regional work is usually mean sea level.The Moho and the base of the upper crust are deep boundaries within the Earth and are not reference levels for gravity reduction.

Q: At a point on the Earth at height 1000 m, the observed value of gravity is 979700 mGal. The value at sea level is 980000 mGal. Calculate the Bouguer anomalies :

The Bouguer anomaly is about -1046 gu (gravity units; 1 mGal = 10 gu).Apply the free-air correction (0.3086 mGal/m) and the Bouguer correction \(0.0419\,\rho\) mGal/m, the stem leaving the density to the standard value. With \(\rho = 2.7\) g/cm\(^{3}\), the Bouguer factor is 0.1132 mGal/m:\[ \begin{aligned} FA &= 0.3086 \times 1000 = 308.6 \\ BC &= 0.1132 \times 1000 = 113.2 \\ \Delta g_B &= 979700 + 308.6 - 113.2 - 980000 \\ &= -104.6\ \text{mGal} = -1046\ \text{gu} \end{aligned} \]The density 2.67 g/cm\(^{3}\) gives -1033 gu, still nearest to -1046. Forgetting the free-air correction gives about -4130 gu, and the remaining options do not follow from any standard reduction.

Q: For a series of points in a line and at zero height which are affected by the gravitational attraction exerted a buried sphere of density contrast 1.5 gm/cm3, the anomaly versus horizontal distance curve has a maximum of 4.526 mGal and a point of inflexion at 250 m from maximum. Calculate the depth of the point :

The depth of the buried sphere is 500 m.For a sphere of radius \(R\) whose centre is at depth \(z\), the vertical attraction at horizontal distance \(x\) from the point above its centre is:\[ \begin{aligned} g(x) &= \frac{4}{3}\pi G\,\Delta\rho\,R^{3} \\ &\quad \times \frac{z}{(x^{2}+z^{2})^{3/2}} \end{aligned} \]Put \(u = x/z\), so that \(g \propto (1+u^{2})^{-3/2}\). The curve changes curvature where its second derivative is zero:\[ \begin{aligned} g' &\propto -3u\,(1+u^{2})^{-5/2} \\ g'' &\propto (1+u^{2})^{-7/2}\,(12u^{2} - 3) \\ g'' = 0 &\Rightarrow u = \tfrac{1}{2} \end{aligned} \]So the point of inflexion lies at \(x = z/2\) from the maximum. With the inflexion at 250 m:\[ z = 2 \times 250 = 500\ \text{m} \]Check with the other data: \(\tfrac{4}{3}\pi G\,\Delta\rho = 4.19\times10^{-7}\) in SI units for a density contrast of 1500 kg/m³, and a maximum of 4.526 mGal (\(4.526\times10^{-5}\) m/s²) then gives \(R^{3}/z^{2} \approx 108\) m, hence \(R \approx 300\) m. A sphere of radius 300 m centred at 500 m depth lies wholly below the surface, so the data are consistent. The other options (700, 400 and 800 m) do not follow from \(x = z/2\).

Q: At a point on the Earth at height 1000 m, the observed value of gravity is 979700 mGal. At sea level it is 980000 mGal. Calculate the free-air anomaly.

The free-air anomaly is 86 gu (8.6 mGal).The free-air correction adds back the decrease of gravity with height, 0.3086 mGal per metre:\[ \begin{aligned} \Delta g_{FA} &= 0.3086 \times 1000 \\ &= 308.6\ \text{mGal} \end{aligned} \]\[ \begin{aligned} g_{FA} &= g_{obs} + \Delta g_{FA} - g_0 \\ &= 979700 + 308.6 - 980000 \\ &= 8.6\ \text{mGal} \end{aligned} \]One mGal is 10 gravity units (gu, micrometres per second squared), so the anomaly is 86 gu. Forgetting the correction gives a large negative anomaly of -300 mGal, and none of the other printed values comes from the standard correction.

Q: The negative Bouger anomaly across the Mid-Oceanic ridges is due to :

The key attributes the anomaly to Hot mantle surrounded by cold mantle.The Bouguer correction removes the effect of the ridge's topography, so what remains reflects density below. Beneath the ridge axis lies hot, partly molten, low-density mantle, while the older lithosphere on either side has cooled and become denser. The mass deficit of this hot upwelling compared with the surrounding colder mantle produces the negative Bouguer anomaly over the ridge.Ridge topography is exactly what the Bouguer correction strips out. Mantle convergence would occur at subduction zones, not ridges, and ridge push is a force driving plates, not a density structure.

Q: The gravity measurements over the oceans indicate that generally the free-air anomaly is nearly zero. This is because :

The free-air anomaly is near zero over the oceans because ocean basins are in isostatic equilibrium.The low density of seawater compared with rock is compensated at depth by denser mantle rising closer to the surface beneath the thin oceanic crust. Because the mass deficit of the water is balanced by the mass excess below, the total mass in each column is about the same as under the continents, and the free-air anomaly averages close to zero over broad areas.The amount of sediment does not cause this balance, option C restates part of the idea vaguely without the compensation of the water, and the free-air anomaly is perfectly valid at sea.

Q: The magnitude of gravity on the earth's surface depends on :

Gravity depends on latitude, elevation, topography, earth tides and density variations in the subsurface.Each factor changes measured gravity and must be corrected for in a gravity survey:Latitude: rotation and the Earth's flattening make gravity about 5 Gal higher at the poles than at the equator.Elevation: free-air and Bouguer corrections for height and the rock in between.Topography: terrain corrections for nearby hills and valleys.Earth tides: changes of up to about 0.3 mGal through the day.Subsurface density: the anomaly that a survey looks for.The shorter options each leave out tides or subsurface density, both real influences on gravity.

In Bouguer gravity anomaly map, which of the following statements are/is correct. Select appropriate answer :

A.Long wavelength anomalies suggest deep level density contrast and help in understanding large-scale or regional structures of the Earth's crust
B.Short wavelength anomalies are due to density contrast at shallower levels and are useful in understanding shallower structures and in exploration.
C.Geological knowledge of the area is essential in understanding the residual anomalies in the given area.
D.All of the above are correct.

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: D

The answer is All of the above are correct.

A Bouguer anomaly map shows the effect of density variations beneath the surface. Broad, long-wavelength anomalies come from deep density contrasts, such as crustal thickness changes or major basins, and reveal regional structure. Short-wavelength anomalies come from shallow sources, such as ore bodies, salt domes or buried channels, and are the targets of exploration.

The residual anomaly, left after the regional trend is removed, has no unique interpretation: many different bodies can produce the same anomaly. Geological knowledge of the area is needed to constrain it. Since each of the three statements is true, choosing any single one is incomplete.

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In gravity surveys the readings are function of :

A.Rigidity of the rocks
B.Bulk's modulus of the rocks
C.Density and mass of the rocks
D.Hardness and specific gravity of rocks

Exam Year(s)

MH SET - 2018, SET Exam - 2018,

Answer: C

Gravity readings are a function of the density and mass of the rocks.

A gravimeter measures the vertical attraction of the earth at a point. By Newton's law of gravitation, this depends on the masses beneath the station and their distances, and so variations in reading reflect lateral differences in rock density, the density contrast between bodies and their surroundings.

Rigidity and bulk modulus are elastic properties that control seismic velocities, not gravity. Hardness has no effect on gravity, and specific gravity alone, paired with hardness, misses the dependence on the mass and geometry of the bodies.

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The value of 'g' (acceleration due to gravity) at a place varies with time due to which of the following :

A.Isostatic rebound
B.The topography of the continents
C.Lateral differences in the compositions of rocks
D.The Earth's rotation

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The value of g at a place varies with time because of isostatic rebound.

Where an ice sheet has melted, as in Scandinavia and Canada, the crust is still rising slowly to regain isostatic equilibrium, and mantle material flows in beneath it. The station moves away from the Earth's centre and the mass distribution beneath it changes, so the measured gravity there changes measurably over years and decades.

The topography of the continents and lateral differences in rock composition make g vary from place to place, not with time at one place. The Earth's rotation lowers g by a fixed amount that depends on latitude, so it too gives no change over time at a given station.

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Bouguer correction is applied to correct for the gravity anomaly due to mass between station location and .......................... .

A.Mohorovicic discontinuity
B.Base of upper crust
C.Mean sea level
D.Local datum plane

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: D

The key completes the sentence with local datum plane.

After the free-air correction has allowed for the height of the station, the Bouguer correction removes the attraction of the slab of rock between the station and the reference level to which the survey is reduced: \(0.0419\,\rho h\) mGal for density \(\rho\) in g/cm\(^{3}\) and height h in metres. That reference is the survey's datum plane, which in regional work is usually mean sea level.

The Moho and the base of the upper crust are deep boundaries within the Earth and are not reference levels for gravity reduction.

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At a point on the Earth at height 1000 m, the observed value of gravity is 979700 mGal. The value at sea level is 980000 mGal. Calculate the Bouguer anomalies :

A.-1046 gu
B.-4443 gu
C.-5246 gu
D.-7013 gu

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The Bouguer anomaly is about -1046 gu (gravity units; 1 mGal = 10 gu).

Apply the free-air correction (0.3086 mGal/m) and the Bouguer correction \(0.0419\,\rho\) mGal/m, the stem leaving the density to the standard value. With \(\rho = 2.7\) g/cm\(^{3}\), the Bouguer factor is 0.1132 mGal/m:

\[ \begin{aligned} FA &= 0.3086 \times 1000 = 308.6 \\ BC &= 0.1132 \times 1000 = 113.2 \\ \Delta g_B &= 979700 + 308.6 - 113.2 - 980000 \\ &= -104.6\ \text{mGal} = -1046\ \text{gu} \end{aligned} \]

The density 2.67 g/cm\(^{3}\) gives -1033 gu, still nearest to -1046. Forgetting the free-air correction gives about -4130 gu, and the remaining options do not follow from any standard reduction.

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For a series of points in a line and at zero height which are affected by the gravitational attraction exerted a buried sphere of density contrast 1.5 gm/cm³, the anomaly versus horizontal distance curve has a maximum of 4.526 mGal and a point of inflexion at 250 m from maximum. Calculate the depth of the point :

A.500 meters
B.700 meters
C.400 meters
D.800 meters

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The depth of the buried sphere is 500 m.

For a sphere of radius \(R\) whose centre is at depth \(z\), the vertical attraction at horizontal distance \(x\) from the point above its centre is:

\[ \begin{aligned} g(x) &= \frac{4}{3}\pi G\,\Delta\rho\,R^{3} \\ &\quad \times \frac{z}{(x^{2}+z^{2})^{3/2}} \end{aligned} \]

Put \(u = x/z\), so that \(g \propto (1+u^{2})^{-3/2}\). The curve changes curvature where its second derivative is zero:

\[ \begin{aligned} g' &\propto -3u\,(1+u^{2})^{-5/2} \\ g'' &\propto (1+u^{2})^{-7/2}\,(12u^{2} - 3) \\ g'' = 0 &\Rightarrow u = \tfrac{1}{2} \end{aligned} \]

So the point of inflexion lies at \(x = z/2\) from the maximum. With the inflexion at 250 m:

\[ z = 2 \times 250 = 500\ \text{m} \]

Check with the other data: \(\tfrac{4}{3}\pi G\,\Delta\rho = 4.19\times10^{-7}\) in SI units for a density contrast of 1500 kg/m³, and a maximum of 4.526 mGal (\(4.526\times10^{-5}\) m/s²) then gives \(R^{3}/z^{2} \approx 108\) m, hence \(R \approx 300\) m. A sphere of radius 300 m centred at 500 m depth lies wholly below the surface, so the data are consistent. The other options (700, 400 and 800 m) do not follow from \(x = z/2\).

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At a point on the Earth at height 1000 m, the observed value of gravity is 979700 mGal. At sea level it is 980000 mGal. Calculate the free-air anomaly.

A.78 gu
B.38 gu
C.52 gu
D.86 gu

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: D

The free-air anomaly is 86 gu (8.6 mGal).

The free-air correction adds back the decrease of gravity with height, 0.3086 mGal per metre:

\[ \begin{aligned} \Delta g_{FA} &= 0.3086 \times 1000 \\ &= 308.6\ \text{mGal} \end{aligned} \]

\[ \begin{aligned} g_{FA} &= g_{obs} + \Delta g_{FA} - g_0 \\ &= 979700 + 308.6 - 980000 \\ &= 8.6\ \text{mGal} \end{aligned} \]

One mGal is 10 gravity units (gu, micrometres per second squared), so the anomaly is 86 gu. Forgetting the correction gives a large negative anomaly of -300 mGal, and none of the other printed values comes from the standard correction.

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The negative Bouger anomaly across the Mid-Oceanic ridges is due to :

A.Ridge Topography
B.Hot mantle convergence
C.Hot mantle surrounded by cold mantle
D.Ridge push mechanism

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The key attributes the anomaly to Hot mantle surrounded by cold mantle.

The Bouguer correction removes the effect of the ridge's topography, so what remains reflects density below. Beneath the ridge axis lies hot, partly molten, low-density mantle, while the older lithosphere on either side has cooled and become denser. The mass deficit of this hot upwelling compared with the surrounding colder mantle produces the negative Bouguer anomaly over the ridge.

Ridge topography is exactly what the Bouguer correction strips out. Mantle convergence would occur at subduction zones, not ridges, and ridge push is a force driving plates, not a density structure.

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The gravity measurements over the oceans indicate that generally the free-air anomaly is nearly zero. This is because :

A.The ocean basins are poorly sedimented
B.Ocean basins are in isostatic equilibrium
C.Mantle below oceans are isostatic
D.Free-air anomaly is invalid due to ocean waters

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The free-air anomaly is near zero over the oceans because ocean basins are in isostatic equilibrium.

The low density of seawater compared with rock is compensated at depth by denser mantle rising closer to the surface beneath the thin oceanic crust. Because the mass deficit of the water is balanced by the mass excess below, the total mass in each column is about the same as under the continents, and the free-air anomaly averages close to zero over broad areas.

The amount of sediment does not cause this balance, option C restates part of the idea vaguely without the compensation of the water, and the free-air anomaly is perfectly valid at sea.

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10.

The magnitude of gravity on the earth's surface depends on :

A.Latitude, elevation and topography
B.Latitutde, elevation, topography and density variations in the subsurface
C.Latitude, elevation, topography and earth tides
D.Latitude, elevation, topography, earth tides and density variations in the subsurface

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

Gravity depends on latitude, elevation, topography, earth tides and density variations in the subsurface.

Each factor changes measured gravity and must be corrected for in a gravity survey:

Latitude: rotation and the Earth's flattening make gravity about 5 Gal higher at the poles than at the equator.
Elevation: free-air and Bouguer corrections for height and the rock in between.
Topography: terrain corrections for nearby hills and valleys.
Earth tides: changes of up to about 0.3 mGal through the day.
Subsurface density: the anomaly that a survey looks for.

The shorter options each leave out tides or subsurface density, both real influences on gravity.

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