Elements of Hydrogeology and the Principles of Gravity, Magnetic, Electrical, Electromagnetic, Seismic and Radiometric Prospecting

This chapter answers the Part A heading of the Geology and Geophysics (GG) paper that reads “Elements of hydrogeology: Principles and applications of gravity, magnetic, electrical, electromagnetic, seismic, and radiometric methods of prospecting for oil, mineral, and groundwater”. It teaches the water table, aquifers, porosity, permeability and Darcy’s law, and then, for each of the six prospecting methods, the principle, the property it senses, the quantity measured and the usual target, with the depth rules and formulas that GATE asks. The numericals are Darcy discharge, depth from a gravity half-width rule and a magnetic zero-crossing rule, Wenner apparent resistivity, skin depth and seismic depth from travel times, with every rule stated in the question.

1. Elements of hydrogeology: water table, aquifers and Darcy's law

Groundwater fills the pores and fractures below the water table, the surface in the saturated zone where the water pressure equals atmospheric pressure. Above it lies the vadose (unsaturated) zone, with a capillary fringe at its base. An aquifer is a body of rock or sediment that stores and yields useful water (sand, gravel, fractured or weathered rock, vesicular basalt); an aquitard transmits little (silty clay) and an aquiclude almost none. An unconfined aquifer has the water table as its top. A confined aquifer lies under a low-permeability layer, its water is under pressure, and a well in it rises to the potentiometric surface, which is artesian if it stands above the ground.

Porosity n = V_v/V is the share of the volume that is void, and measures storage. The specific yield S_y is the part that drains by gravity and the specific retention S_r the part that clings to the grains, so n = S_y + S_r. Permeability is the ability to transmit water and depends on the size and connection of the pores; the hydraulic conductivity K (m/day or m/s) combines rock and fluid, K = kρg/μ. Transmissivity T = Kb for an aquifer of thickness b. Clay has high porosity but very low K, whereas a well-sorted gravel has both high. Fresh and saline pore water differ strongly in electrical resistivity, which is why resistivity surveys are used to find and grade groundwater.

Darcy's law states that the specific discharge (flux) is proportional to the hydraulic gradient: q = Q/A = K i, with i = Δh/L the head loss per unit flow length, and the discharge is Q = K i A, with A the whole cross-section of the flow; water moves from high to low head. The flux q is not the speed of the water, because water moves only through the pores: the seepage velocity is v = q/n_e, with n_e the effective porosity, and is larger than q. The law holds for laminar flow, which is normal in groundwater. Worked example: K = 10 m/day, a head drop of 2 m over 500 m (i = 0.004), A = 1000 m² and n_e = 0.20 give q = 0.04 m/day, Q = 40 m³/day and v = 0.04/0.20 = 0.2 m/day.

⚠️ Flux, velocity, porosity and permeability are four different things
Use the whole area A in Q = KiA, not the pore area, and divide q by the effective porosity only to get the seepage velocity. A rock with high porosity can still be a poor aquifer if its pores are not connected, as in clay or a pumice-like rock. Peninsular India is largely hard-rock terrain, where water lies in the weathered layer and in fractures, so locating fractures is the main groundwater task there.

2. The six methods at a glance: property, measurement, target

Each method senses one physical property that differs between the target and its surroundings, and it works only where that contrast exists and is large enough, at a depth that is not too great, to be measured at the surface.

Prospecting methods: property, measured quantity and use
MethodProperty that must contrastQuantity measuredTypical target and use
GravityDensityGravitational acceleration g, in mGal (1 mGal = 10⁻⁵ m/s²)Dense ore bodies such as chromite; salt domes (lighter); basin depth and faults for oil; buried valleys
MagneticMagnetic susceptibility and remanenceTotal magnetic field anomaly, in nTMagnetite-bearing iron ore, basic dykes, kimberlite pipes; depth to magnetic basement under sediments
Electrical (resistivity, SP, IP)Resistivity, natural potential, chargeabilityPotential difference ΔV for injected current I, giving apparent resistivity ρa in Ω·mGroundwater (depth, fresh against saline, fractures), bedrock depth, sulphide ore
ElectromagneticElectrical conductivityIn-phase and quadrature parts of the secondary field, or the decay of a transientConductive sulphide and graphite bodies, saline water, conductive overburden; deep crust with natural fields
SeismicElastic properties: velocity and acoustic impedance ρvTravel times of reflected or refracted wavesLayers and traps for oil; bedrock depth and water table at shallow depth
RadiometricNatural radioelement content (K, U, Th)Gamma-ray counts and energiesUranium and thorium ores, lithological mapping, potassic alteration, phosphate

A survey normally moves from regional to local: airborne magnetic and gravity surveys outline basins and belts, then electromagnetic, resistivity or seismic surveys refine targets, and a drill hole tests the best one. Several different bodies can give the same anomaly (non-uniqueness), so geology and a second method are needed to choose between them.

3. Gravity and magnetic methods

Gravity. Newton's law makes the vertical attraction g depend on the mass below, so a body of density contrast Δρ = ρ_body − ρ_host gives a positive anomaly if denser (a chromite or sulphide body) and a negative one if lighter (a salt dome, with salt of about 2.2 g/cm³ against sediments of about 2.5 g/cm³, or a sediment-filled basin). Readings are first reduced: latitude correction, the free-air correction (0.3086 mGal per metre of elevation, g falls with height), the Bouguer correction for the slab of rock between station and datum (0.0419ρ mGal per metre with ρ in g/cm³, 0.112 mGal/m for 2.67 g/cm³), terrain, and tide and drift. The result is the Bouguer anomaly. For a buried sphere the depth to its centre is z = 1.305 x½, and for a horizontal cylinder z = x½, where x½ is the horizontal distance from the peak to where the anomaly falls to half its peak value. Worked example: x½ = 120 m over a sphere gives z = 1.305 × 120 ≈ 157 m.

Magnetic. Rocks acquire magnetisation in the Earth's field, induced J = χF (χ the susceptibility, large for magnetite and pyrrhotite) plus any remanent part, and the instrument measures the total-field anomaly ΔT in nT after the diurnal variation and the regional field (IGRF) are removed. A magnetised body acts as a dipole, so at mid-latitudes its anomaly is a positive peak paired with a negative trough, and at the magnetic pole it is a symmetric positive peak with weak negative flanks. A compact body's gravity anomaly falls as 1/z² but its magnetic anomaly as 1/z³, so magnetic anomalies are sharper and favour shallow sources. For a sphere magnetised vertically (at the pole) the anomaly along a profile through its centre varies as (2z² − x²)/(x² + z²)^(5/2), changing sign at x₀ = √2 z, so z = x₀/√2. Worked example: x₀ = 141 m gives z = 141/1.414 ≈ 100 m. Magnetite loses its ferromagnetism above its Curie temperature (about 580 °C), which limits magnetic sources to the shallower crust.

⚠️ Contrast, not density or susceptibility itself
A body gives an anomaly only through its contrast with the host, so a body as dense as its surroundings is invisible to gravity however dense it is. Gravity is always one-signed for a given contrast, whereas a magnetic anomaly depends on the field direction and on remanence. The free-air correction is added when the station is above the datum, and the Bouguer correction is subtracted.

4. Electrical and electromagnetic methods

Resistivity. Most rocks conduct current by ions in pore water, so resistivity falls as porosity, saturation and salinity rise (Archie: the ratio of rock to water resistivity is about φ⁻ᵐ with m near 2 for clean sands). Clay and metallic sulphides conduct well, while dry crystalline rock resists. Current I is passed through two electrodes and the potential difference ΔV is read across two others; the apparent resistivity is ρa = K ΔV/I, where K is the geometric factor. For the Wenner array with equal spacing a, K = 2πa; for the Schlumberger array with half-spread L and potential electrodes 2l apart, K = π(L² − l²)/(2l). Worked example: a = 6 m, ΔV = 24 mV and I = 80 mA give ρa = 2π × 6 × 0.024/0.080 = 11.3 Ω·m.

In a vertical electrical sounding the electrodes are expanded about a fixed centre so that the current reaches greater depth, and ρa is plotted against spacing on log axes and interpreted as layers (a groundwater sounding reads depth to water, thickness of a weathered layer and depth to bedrock). A curve that rises to a maximum and then falls shows a resistive middle layer. In profiling a fixed spacing is moved along a line to find lateral change such as fractures, dykes and ore. Self-potential (SP) measures natural potentials, negative by up to a few hundred millivolts over sulphide bodies, and induced polarisation (IP) measures the chargeability of disseminated sulphides.

Electromagnetic. An alternating current in a transmitter coil makes a primary field, which induces eddy currents in any conductor; these make a secondary field that the receiver separates into in-phase and quadrature parts. No ground contact is needed, so surveys can be airborne. The fields decay with depth over the skin depth δ = 503 √(ρ/f) metres (ρ in Ω·m, f in Hz): a lower frequency or a more resistive ground penetrates deeper. Worked example: ρ = 25 Ω·m and f = 10 kHz give δ = 503 × √(25/10⁴) = 25 m. Variants are ground-loop systems, airborne EM, transient (TEM) systems that read the decay after the current is switched off, the radio-transmitter VLF method, natural-field magnetotellurics for deep structure, and ground-penetrating radar (GPR), whose pulses travel at v = c/√ε_r and give depth = v t/2. Conductive overburden, such as wet clay or saline ground, shields the target and limits all of these.

5. Seismic reflection and refraction

A source (dynamite, vibrator, air gun or hammer) sends elastic waves into the ground, and geophones record when they return. The velocities are v_P = √((K + 4μ/3)/ρ) and v_S = √(μ/ρ), and they rise with compaction and cementation (limestone and sandstone are faster than shale, saturated sediment faster than dry). At an interface part of the energy is reflected and part refracted, with Snell's law sin i/v₁ = sin r/v₂, and the reflected fraction depends on the contrast in acoustic impedance Z = ρv: the reflection coefficient is R = (Z₂ − Z₁)/(Z₂ + Z₁).

Reflection. A wave sent down to a reflector at depth h and back takes the two-way time t₀ = 2h/v, so h = v t₀/2; at offset x the time is t² = t₀² + x²/v² (normal moveout). Worked example: v = 2400 m/s and t₀ = 0.5 s give h = 2400 × 0.5/2 = 600 m. Reflection maps layers, folds, faults and salt for oil traps.

Refraction. If v₂ > v₁, a ray striking the interface at the critical angle sin i_c = v₁/v₂ travels along the top of the faster layer as a head wave and returns to the surface. On a plot of first arrivals the direct wave has slope 1/v₁ and the refracted wave slope 1/v₂; the lines cross at the crossover distance x_c = 2z √((v₂ + v₁)/(v₂ − v₁)), and the intercept time is t_i = 2z cos i_c/v₁, so z = t_i v₁v₂/(2√(v₂² − v₁²)). Refraction gives depth to bedrock or the water table at shallow depth, but it fails to see a low-velocity layer beneath a faster one, since no head wave forms.

⚠️ Reflection time is two-way
A record shows the time down and back, so divide v t by 2 for depth; forgetting the halving doubles the depth. In refraction work the velocity must increase with depth, so a layer slower than the one above it is invisible and every layer below it comes out at the wrong depth.

6. Radiometric prospecting and choosing methods for oil, minerals and groundwater

The radiometric method records the natural gamma rays of K-40 (1.46 MeV), of the uranium series through Bi-214 (1.76 MeV) and of the thorium series through Tl-208 (2.61 MeV), with a scintillometer or a spectrometer on the ground or in the air, and reports K (per cent), eU and eTh (ppm). The “e” means equivalent: uranium and thorium are inferred from their daughters on the assumption of radioactive equilibrium, so loss of radon gas makes eU read low. Gamma rays are absorbed within a few tens of centimetres of rock or soil, so the method maps the surface only. It finds uranium and thorium ores and monazite sands, maps lithology (granites are high in K, U and Th; basalt and limestone are low) and outlines potassic alteration around ore.

Which methods for which target
TargetUsual methods, in order of useReason
Oil and gasGravity and magnetic, then seismic reflection, then wellsPotential fields give basin shape and basement depth; reflection images the traps
Metallic mineralsAirborne magnetic, radiometric and EM, then ground IP, resistivity and gravity, then drillingMagnetite, conductors and radioactive minerals have strong contrasts; IP finds disseminated sulphide
GroundwaterResistivity sounding and profiling, EM, seismic refraction, then boreholesResistivity separates clay, fresh and saline water and fractured rock; refraction gives bedrock depth

Because one anomaly can come from a small shallow body or a large deep one, interpretation needs constraints from geology, drilling and a second method, such as a magnetic and a gravity map of the same area.

Key takeaways

  • The water table is the surface where pore pressure equals atmospheric; confined aquifers rise to the potentiometric surface. n = S_y + S_r, and clay is porous but not permeable. Darcy: q = Ki, Q = KiA, seepage velocity v = q/n_e.
  • Each method needs a contrast: gravity in density (mGal), magnetic in susceptibility (nT), electrical in resistivity, EM in conductivity, seismic in velocity and impedance ρv, radiometric in K, U and Th (eU assumes equilibrium and radon loss lowers it). Oil suits gravity, magnetic and seismic, minerals suit magnetic, EM and IP, groundwater suits resistivity. Surveys go from regional to local and are non-unique.
  • Gravity: z = 1.305 x½ for a sphere and z = x½ for a horizontal cylinder; free-air 0.3086 mGal/m; a salt dome gives a negative anomaly. Magnetic: dipole anomalies fall as 1/z³ against 1/z² for gravity, and a vertically magnetised sphere gives z = x₀/√2.
  • Resistivity: ρa = KΔV/I with K = 2πa for Wenner; soundings expand the spacing for depth and profiling moves it for lateral change. EM skin depth δ = 503√(ρ/f) m, so lower frequency or higher resistivity goes deeper; conductive overburden screens targets.
  • Seismic: depth h = v t₀/2 for reflection (two-way time), crossover distance x_c = 2z√((v₂ + v₁)/(v₂ − v₁)) for refraction, R = (Z₂ − Z₁)/(Z₂ + Z₁). Refraction needs velocity to increase with depth and misses low-velocity layers.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Which description correctly defines the water table of an unconfined aquifer?

    1. The surface in the saturated zone where the water pressure equals atmospheric pressure
    2. The upper edge of the capillary fringe, above which the ground is completely dry of water
    3. The level to which water rises in a well drilled into a confined aquifer under pressure
    4. The lower boundary of the aquifer, where it rests on an impermeable bedrock layer below it
    Show answer

    Answer: A — The surface in the saturated zone where the water pressure equals atmospheric pressure

    The water table is where the pore pressure equals atmospheric pressure, and below it the ground is saturated. The capillary fringe lies above it and is still wet. The level to which water rises in a well in a confined aquifer is the potentiometric surface, a different thing, and the base of the aquifer is its lower boundary.

  2. Groundwater flows through a confined aquifer with hydraulic conductivity K = 2.0 m/day. The head falls by 3.0 m over a flow length of 150 m, and the cross-sectional area of the aquifer normal to the flow is 500 m². The discharge, in m³/day, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 20

    The hydraulic gradient is i = Δh/L = 3.0/150 = 0.02. By Darcy's law Q = K i A = 2.0 × 0.02 × 500 = 20 m³/day. Dividing by the porosity would give a velocity, not a discharge, and is not required here.

  3. A clay layer has a porosity of 0.50 and a clean gravel layer has a porosity of 0.30. Which statement about them as aquifers is correct?

    1. The clay is the better aquifer, because its higher porosity means that it can store and also yield more water
    2. The gravel is the far better aquifer, because its large connected pores give a much higher hydraulic conductivity
    3. The two are equally good as aquifers, because the hydraulic conductivity of a material depends only on its porosity
    4. The clay is the better aquifer, because very small pores hold water more firmly and so let it flow more readily
    Show answer

    Answer: B — The gravel is the far better aquifer, because its large connected pores give a much higher hydraulic conductivity

    An aquifer must yield water, which needs high hydraulic conductivity as well as storage. Clay has the higher porosity but its tiny pores hold water by surface tension and transmit it extremely slowly, so its K and specific yield are very low. Gravel has the larger, well-connected pores and a high K. Porosity alone does not fix K.

  4. Which of the following statements about groundwater are correct?

    1. The seepage velocity v = q/n_e is larger than the Darcy flux q through the same section
    2. Specific yield is the part of the water that is held on the grains against gravity
    3. Water in a well in a confined aquifer rises to a potentiometric surface above the top of the aquifer
    4. Hydraulic conductivity always increases whenever the porosity of the rock or sediment increases
    Show answer

    Answer: A — The seepage velocity v = q/n_e is larger than the Darcy flux q through the same section; C — Water in a well in a confined aquifer rises to a potentiometric surface above the top of the aquifer

    Water moves only through the pores, so its actual speed q/n_e exceeds the flux q, and in a confined aquifer the pressure raises the water above the aquifer top to the potentiometric surface. Specific yield is the part that drains under gravity, and the part held on the grains is the specific retention. Clay shows that K can fall as porosity rises.

  5. A survey must map the depth of the boundary between fresh water and saline water in a sandy coastal aquifer. Which method is the most suitable, and why?

    1. Gravity, because saline water is much denser than fresh water and gives a large density contrast across the boundary
    2. Seismic refraction, because saline water makes the P-wave velocity of the saturated sand much higher than fresh water does
    3. Electrical resistivity sounding, because saline-saturated sand is far more conductive than fresh-saturated sand
    4. Magnetic survey, because the dissolved salts change the magnetic susceptibility of the saturated sand strongly
    Show answer

    Answer: C — Electrical resistivity sounding, because saline-saturated sand is far more conductive than fresh-saturated sand

    Dissolved salts raise the conductivity of the pore water, so saline-saturated sand has a resistivity many times lower than fresh-saturated sand, and a sounding shows the interface as a drop in resistivity. Gravity and seismic velocity depend on the bulk density and the elastic moduli of the saturated sand, which hardly change between fresh and saline pore water, and dissolved salts do not alter the susceptibility of the sand.

  6. Which of the following statements about the prospecting methods are correct?

    1. The gravity method responds to a density contrast between the body and its surroundings
    2. A magnetic anomaly can include a remanent part in addition to the induced magnetisation
    3. Seismic reflection strength depends on the contrast in acoustic impedance across an interface
    4. The radiometric method can detect a uranium ore body buried several hundred metres below the surface
    Show answer

    Answer: A — The gravity method responds to a density contrast between the body and its surroundings; B — A magnetic anomaly can include a remanent part in addition to the induced magnetisation; C — Seismic reflection strength depends on the contrast in acoustic impedance across an interface

    Gravity senses density contrast, magnetics senses induced plus remanent magnetisation, and the reflection coefficient depends on the impedance contrast ρv. Gamma rays are absorbed within a few tens of centimetres of rock or soil, so the radiometric method sees only the surface and cannot detect deeply buried ore directly.

  7. A salt dome of density 2.2 g/cm³ rises through sediments of density 2.5 g/cm³. What gravity anomaly, after the usual reductions, is expected over the centre of the dome?

    1. Positive, because a large body of evaporite represents an extra mass of rock below the station
    2. Negative, because salt is lighter than the sediments it replaces, a density contrast of −0.3 g/cm³
    3. Zero, because gravity responds to magnetic rather than density contrasts between the rock bodies
    4. Positive, because a body that rises towards the surface always brings denser rock closer to the station
    Show answer

    Answer: B — Negative, because salt is lighter than the sediments it replaces, a density contrast of −0.3 g/cm³

    The anomaly follows the density contrast Δρ = ρ_body − ρ_host = 2.2 − 2.5 = −0.3 g/cm³, so the salt dome gives a negative anomaly. Its size or its rise toward the surface does not change the sign, and gravity responds to density, not to magnetic properties. Salt domes therefore appear as gravity lows, which is why gravity is used in oil exploration for them.

  8. A gravity profile across a buried sphere shows that the anomaly falls to half its peak value at a horizontal distance of 200 m from the peak. Use the rule z = 1.305 x½ for the depth z to the centre of a sphere. The depth, in m to the nearest metre, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 261

    With x½ = 200 m the rule gives z = 1.305 × 200 = 261 m. The factor comes from solving (1 + x²/z²)^(−3/2) = 1/2, which gives x/z = 0.766 and z = 1.305 x. For a horizontal cylinder the factor would be 1, giving 200 m, which is the wrong body for this stem.

  9. A compact body, treated as a sphere or dipole, is moved from 50 m to 150 m depth with all other properties unchanged. Take the peak gravity anomaly ∝ 1/z² and the peak magnetic anomaly of a dipole ∝ 1/z³. By what factors do the two anomalies fall?

    1. Gravity falls by a factor of 3 and the magnetic anomaly by a factor of 9
    2. Gravity falls by a factor of 9 and the magnetic anomaly by a factor of 27
    3. Gravity and the magnetic anomaly both fall by a factor of 9 for the same body
    4. Gravity falls by a factor of 27 and the magnetic anomaly by a factor of 9
    Show answer

    Answer: B — Gravity falls by a factor of 9 and the magnetic anomaly by a factor of 27

    The depth is tripled, so a 1/z² anomaly falls by 3² = 9 and a 1/z³ anomaly by 3³ = 27. The magnetic anomaly of a dipole falls faster with depth than the gravity anomaly, which is why magnetic anomalies are sharper and favour shallow sources. The pair with the factors reversed is the usual slip.

  10. Over a sphere magnetised vertically, the total-field anomaly along a profile through its centre varies as (2z² − x²)/(x² + z²)^(5/2), where z is the depth to the centre. The anomaly changes sign at a horizontal distance of 212 m from the peak. The depth z, in m to the nearest metre, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 150

    The anomaly is zero when 2z² − x² = 0, that is x = √2 z. So z = x/√2 = 212/1.4142 = 149.9, which is 150 m to the nearest metre. Taking z = x would give 212 m, and dividing by 2 instead of √2 gives 106 m.

  11. In a Wenner array with an electrode spacing a = 10 m, a current of 100 mA produces a potential difference of 50 mV between the inner electrodes. Take π = 3.1416. The apparent resistivity, in Ω·m to one decimal place, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 31.4

    For the Wenner array K = 2πa = 2 × 3.1416 × 10 = 62.83 m. Then ρa = K ΔV/I = 62.83 × 0.050/0.100 = 31.4 Ω·m. Leaving the factor 2 out of K, or using the volts and amperes in millivolts and milliamperes inconsistently, are the common slips.

  12. A vertical electrical sounding over a three-layer ground gives an apparent-resistivity curve that rises to a maximum and then falls at the largest spacings. Which layering is indicated?

    1. Three layers of steadily increasing resistivity with depth, such as soil over weathered rock over fresh rock
    2. A conductive middle layer between a more resistive top layer and a more resistive bottom layer
    3. A uniform half-space, because the curve of a homogeneous ground rises and falls with spacing
    4. A resistive middle layer between a more conductive top layer and a more conductive bottom layer
    Show answer

    Answer: D — A resistive middle layer between a more conductive top layer and a more conductive bottom layer

    At small spacings ρa reflects the top layer. As the spacing grows the current reaches the resistive middle layer, and ρa rises toward its value, and at the largest spacings the conductive bottom layer pulls ρa down, giving a maximum. A conductive middle layer gives a minimum instead, steadily increasing resistivity gives a steadily rising curve, and a homogeneous ground gives a flat curve at its true resistivity.

  13. In an electromagnetic survey over ground of fixed resistivity, the transmitter frequency is raised from 1 kHz to 4 kHz. By what factor does the skin depth change?

    1. It doubles, because a higher frequency drives the fields further into the ground
    2. It falls to a quarter, because the skin depth is inversely proportional to the frequency
    3. It is halved, because the skin depth varies as the inverse square root of the frequency
    4. It is unchanged, because the skin depth depends only on the resistivity of the ground
    Show answer

    Answer: C — It is halved, because the skin depth varies as the inverse square root of the frequency

    δ = 503 √(ρ/f), so δ varies as f^(−1/2). Raising f by a factor of 4 divides δ by √4 = 2, so the skin depth is halved. It does not quarter, because the dependence is on the square root, and it does not rise, because higher frequencies penetrate less. The skin depth depends on both ρ and f.

  14. A seismic reflection record shows a reflector at a two-way time of 0.8 s. The average velocity above it is 2000 m/s. What is the depth of the reflector?

    1. 1600 m, from the product of the velocity and the two-way time
    2. 400 m, from halving the product of the velocity and the two-way time twice
    3. 800 m, from half the product of the velocity and the two-way time
    4. 2500 m, from dividing the velocity by the two-way time
    Show answer

    Answer: C — 800 m, from half the product of the velocity and the two-way time

    The wave travels down and back, so the depth is h = v t₀/2 = 2000 × 0.8/2 = 800 m. The value 1600 m forgets that t₀ is a two-way time and is the usual error. Halving twice and dividing the velocity by the time have no basis in the geometry of the path.

  15. In a seismic refraction survey the upper layer has v₁ = 600 m/s and the lower layer v₂ = 1000 m/s. The first-arrival lines cross at a distance x_c = 80 m. Use x_c = 2z √((v₂ + v₁)/(v₂ − v₁)) for the thickness z of the upper layer. The thickness, in m, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 20

    The ratio (v₂ + v₁)/(v₂ − v₁) = 1600/400 = 4, whose square root is 2. So x_c = 2z × 2 = 4z and z = 80/4 = 20 m. Using v₂/v₁ in place of the ratio of sum to difference gives a wrong factor, and treating x_c as 2z gives 40 m.

  16. Which of the following statements about seismic reflection and refraction are correct?

    1. The depth of a flat reflector is the product v t₀, where t₀ is the two-way travel time
    2. A refraction survey cannot detect a layer whose velocity is lower than that of the layer above it
    3. The normal-incidence reflection coefficient depends on the contrast in acoustic impedance ρv
    4. A head wave forms along the interface when the lower layer is slower than the layer above it
    Show answer

    Answer: B — A refraction survey cannot detect a layer whose velocity is lower than that of the layer above it; C — The normal-incidence reflection coefficient depends on the contrast in acoustic impedance ρv

    A head wave needs v₂ > v₁, so a slower layer sends none back and is hidden, and R = (Z₂ − Z₁)/(Z₂ + Z₁) depends on the impedance contrast. The depth of a reflector is v t₀/2, because t₀ includes the path down and up. A head wave does not form when the lower layer is slower.

  17. A ground gamma-ray spectrometer over a granite reads eK and eTh in agreement with chemical analyses, but eU is clearly lower than the uranium content found by chemical analysis. Which explanation is the most likely?

    1. Thick soil and moisture have absorbed the gamma rays, because they penetrate only a few tens of centimetres
    2. The uranium series is in transient equilibrium, so the Bi-214 activity is always lower than that of U-238
    3. Weathering has increased the decay constant of U-238, so that the nuclide gives out fewer gamma rays
    4. Radon has escaped from the rock, so the counted daughter Bi-214 is below its equilibrium amount
    Show answer

    Answer: D — Radon has escaped from the rock, so the counted daughter Bi-214 is below its equilibrium amount

    eU is inferred from Bi-214 gamma rays on the assumption of secular equilibrium. If radon (Rn-222, a gas) escapes, the Bi-214 that follows it in the chain is depleted and eU underestimates the uranium. Absorption by soil or moisture would lower eK and eTh too, which the stem excludes; the uranium series is in secular, not transient, equilibrium; and decay constants are not altered by weathering.

  18. Which ground condition most strongly limits the depth of penetration of ground-penetrating radar?

    1. A thick layer of dry, clean sand with a very low electrical conductivity and little moisture
    2. A conductive ground such as wet clay or saline soil, which strongly attenuates the radar waves
    3. The use of a lower-frequency antenna, which gives a longer wavelength and a coarser resolution
    4. A deep water table in massive resistive granite with very few fractures or clay-filled joints
    Show answer

    Answer: B — A conductive ground such as wet clay or saline soil, which strongly attenuates the radar waves

    Radar waves are attenuated by conduction currents, so a conductive ground, such as wet clay or saline soil, stops them within a short distance. Dry sand and resistive granite are among the best media for GPR, and a lower frequency penetrates deeper, at the cost of resolution.