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Earth Sciences: Second-round sample problems Part 1. Choose one correct answer (2 points each) 1.1 In which geological era did dinosaurs live?

A - Paleozoic; B - Cenozoic; C - Mesozoic; D - Mesoproterozoic; E - Mesoarchean. Answer: C

1.2 Which depositional environments are associated with rocks formed in river deltas? A - continental; B - transitional; C - marine; D - lacustrine; E - eolian. Answer: B

1.3 Which of the stratigraphic units of the International Chronostratigraphic Chart DOES NOT belong to the stages of the Paleozoic Era?

A - Dapingian; B - Danian; C - Drumian; D - Darriwilian; E - Gorstian. Answer: B

1.4 Which of these rocks CANNOT form on the ocean floor? A - chert; B - siltstone; C - limestone; D - sandstone; E - tuff. Answer: C

1.5 Which wave type is NOT considered in the seismic survey of the Earth? A - longitudinal waves; B - anisotropic waves; C - transverse waves; D - surface Rayleigh waves;

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E - surface Love waves. Answer: B

1.6 The direct problem of geophysics is: A – geophysical field measurements; B - finding the contour of the causative body, using the geophysical field measurements; C - calculating the geophysical field from the known parameters of the causative body; D - determining the parameters of the causative body, using the geophysical field measurements; E – transformations of the measured geophysical field. Answer: C

1.7 Which ‘father of geochemistry’ gave his name to the value indicating the average content of a chemical element in the Earth’s crust?

A – Vladimir Vernadsky; B - Frank Wigglesworth Clarke; C – Alexander Fersman; D - Victor Moritz Goldschmidt; E – Alexander Vinogradov. Answer: B

1.8 Which continent is the smallest by area? A - South America; B - Antarctica; C - Australia; D - North America; E - Eurasia. Answer: C

1.9 The presence of ground and surface water during mining, strip-mining and deposit development is called:

A - water encroachment; B - water influx; C - storativity; D - hydrogeological conditions; E - settling velocity. Answer: A

1.10 The angle between the direction of true north and a given line is called: A - magnetic heading; B - directional angle; C - true azimuth;

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D - magnetic azimuth; E - conditional angle. Answer: C

1.11 An inclined mine shaft with direct access to the Earth’s surface and a system for hoisting people, material and the mineral is called:

A - auxiliary adit; B – auxiliary shaft; C - main tunnel; D - main haulage gate; E - main pitshaft. Answer: B

1.12 The ratio of stress to permanent longitudinal strain in a rock is: A – post-peak modulus B - Young’s modulus; C - shear modulus; D - deformation modulus; E - none of the above. Answer: A

Part 2. Multiple-choice questions

2.1 Which TWO principles of stratigraphy describe the patterns of facies variability and changes in the age of boundaries between sedimentary bodies?

A - Stenon’s; B - Gressly-Renevier’s; C - Huxley’s; D - Smith’s; E - Lyell’s; F - Golovkinsky-Walther’s; G - Hutton’s. Answer: B, F 2 points for one correct answer 4 points for two correct answers

2.2 Which THREE folded dislocations are common in platform conditions? A - uniform; B - plicative; C - disjunctive;

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D - radial; E - landslide; F - salt; G - glacial; H - tangential. Answer: B, D, F 1 point for one correct answer 3 points for two correct answers 5 points for three right answers

2.3 Which THREE minerals are diamagnets? A - quartz; B - magnetite; C - native gold; D - hematite; E - biotite; F - pyrite; G - anhydrite. Answer: A, C, G 1 point for one correct answer 2 points for two correct answers 4 points for three correct answers

2.4 Choose THREE elements that are most abundant by mass in the Earth’s crust. A - copper; B - titanium; C - gold; D - sodium; E - zinc; F - nickel; G - iron; H - chlorine. Answer: B, D, G 1 point for one correct answer 3 points for two correct answers 5 points for three correct answers

2.5 The THREE geospheres in which living organisms comprising the biosphere can develop are:

A - cryosphere;

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B - mantle; C - magnetosphere; D - troposphere; E - hydrosphere; F - core; G - lithosphere; H - stratosphere. Answer: D, E, G 1 point for one correct answer 3 points for two correct answers 5 points for three correct answers

2.6 Choose THREE main purposes of frothing agents during flotation: A - regulation of the pH of the medium; B – increased hydrophobicity of the surface of the mineral; C - hardening of air bubbles; D - coagulation of particles; E - grinding particles in the effluent; F - prevention of bubble coalescence; G - lowering the surface tension at the water-air interface H - dissolution of silicates and carbonates. Answer: C, F, G 1 point for one correct answer 2 points for two correct answers 4 points for three correct answers

2.7 Choose THREE operations that are mandatory when building a block model of an ore mineral deposit in modern mining and geological information systems:

A - mine surveying of the area; B – creating a database based on earlier geological explorations; C – decomposing the ore body into separate units; D - naming each block; E - selecting a coordinate system; F - compressive ore strength determination; G - ore body contouring; H - reserves calculation. Answer: B, C, G 1 point for one correct answer 3 points for two correct answers

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5 points for three correct answers Part 3. Open-ended questions (11 points each) 3.1 The figure contains two sections with information on the lithology and the organic remains.

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Task:

a) identify local stratigraphic units;

b) determine the age of sections of one large sedimentary paleobasin;

c) correlate these sections with each other;

d) perform facies analysis and establish the paleogeographic position of each section (as well as

its parts) in the sedimentary paleobasin;

e) determine the latitudinal position of the sedimentary paleobasin.

Solution:

a)

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b) according to the combination of organic remains, it is Ordovician, Silurian or Devonian.

c)

d) both sections are located in the same zone of the marine paleobasin at a medium distance

from the coast. At depths below the storm level (20-25 m), they nevertheless receive a large

amount of light (corals grow there – tabulates and rugoses), i.e. the depth is no more than 30-

50 m.

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e) the marine paleobasin containing the two sections (see above) of sedimentary formations

was located in low paleolatitudes (equatorial and tropical belts), i.e. 0-30 degrees of the

northern or southern hemispheres; this is evidenced by the presence of carbonates and corals –

tabulates and rugoses.

Points awarded:

a) is worth 2 points

b) is worth 2 points

c) is worth 2 points

d) is worth 3 points

e) is worth 2 points

TOTAL: 11 points

3.2 The table shows a three-layer seismic model of a geological environment: the density and velocity distributions. There is a normal incident P-wave.

Layer number Density (g/cm3) Velocity (m/s) 1 2.1 1600 2 2.8 2900 3 2.5 2600

Task: 1. Calculate the acoustic stiffness (seismic impedance) Z for each of the three layers,

specifying the unit of measurement. 2. Calculate the reflection coefficient R1 (from the boundary between layers 1 and 2) and

the reflection coefficient R2 (from the boundary between layers 2 and 3). 3. Is the polarity of the wave reflected from the boundary between layers 1 and 2 normal

or reverse? Why? Is the polarity of the wave reflected from the boundary between layers 2 and 3 normal or reverse? Why?

Solution 1. The seismic impedance of the layer is calculated according to the formula: Z = ρ·V,

where ρ is the density in the layer, V is the wave velocity in the layer. The values of density [g/cm3] and velocity [m/s] have to be converted to the same system of units (SI):

2.1g/cm3 = 2,100kg/m3; 2.8g/cm 3 = 2,800kg/m3; 2.5g/cm 3 = 2,500kg/m3. Seismic impedance for layer 1:

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Z1 = 2,100kg/m3 · 1600 m/s = 3,360,000kg/(m2·s) Seismic impedance for layer 2: Z2 = 2,800kg/m3 · 2,900 m/s = 8,120,000kg/(m2·s) Seismic impedance for layer 3: Z3 = 2,500kg/m3 · 2,600 m/s = 6,500,000kg/(m2·s) 2. The reflection coefficient at normal wave incidence is determined through the seismic

impedances of the layers below and above the reflecting boundary. Thus, the reflection coefficient R1 is: R1 = (Z2-Z1)/(Z2+Z1) = (8 120 000-3 360 000)/(8 120 000+3 360 000) =

4 760 000/11 480 000 = 0.415. The reflection coefficient R2 is: R2 = (Z3-Z2)/(Z3+Z2) = (6 500 000-8 120 000)/(6 500 000+8 120 000) =

-1 620 000/14 620 000 = -0.111. 3. The polarity of the reflected wave has the same sign as the reflection coefficient. Thus,

the polarity of the wave reflected from the boundary between layers 1 and 2 is normal (R1 = 0.415); the polarity of the wave reflected from the boundary between layers 2 and 3 is reverse (R2 = -0.111).

Points awarded: 1. The seismic impedances of each of the three layers have been calculated correctly, the

unit of measurement specified – 4 points. 2. The reflection coefficients R1 and R2 have been calculated correctly – 4 points. 3. The polarities of the waves reflected from the boundaries between layers 1 and 2 and

between layers 2 and 3 have been determined correctly, an explanation provided – 3 points. TOTAL: 11 points.

3.3 Overburden rocks are represented by five lithological varieties in the following proportions by volume: Ⅰ- 27%, Ⅱ- 29%, Ⅲ- 15%, Ⅳ- 13%, Ⅴ- 16. The density of the rocks is as follows:Ⅰ- 2500kg/m3, Ⅱ- 2100kg/m3, Ⅲ- 2400kg/m3, Ⅳ- 2200kg/m3, Ⅴ- 2600kg/m3. Calculate the average overburden density (round your answer to the nearest integer and enter it in SI units) and the volume of 1 metric ton of the overburden that consists of only the first two lithological varieties in equal proportions (round the answer to the nearest thousandth and give it in SI units). Solution: 1. Let us calculate the mass of each lithological variety in 1m3 of the overburden: I - 0.27 * 2500 = 675kg II - 0.29 * 2100 = 609kg III - 0.15 * 2400 = 360kg

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IV - 0.13 * 2200 = 286kg V - 0.16 * 2600 = 416kg The mass of one cubic meter of the averaged overburden is 2346kg. Thus, the average density of the overburden rocks is 2346kg/m3. 2. The density of the overburden consisting of only the first two types of lithological variety is calculated the same way. It is 2300kg/m3. The volume of 1 metric ton of such lithological variety mixture is: 1000/2300 = 0.435m3. Points awarded: Correctly calculated average overburden density - 6 points The overburden density consisting of the first two lithological varieties has been calculated correctly - 2 points; The volume of 1 metric ton of the overburden is calculated correctly – 3 points. 3.4 An exploration company has drilled 16 wells. They are located in two sections located at a distance of 120 meters. The table shows the absolute marks of the top and soil of the ore body established by the initial analysis of the obtained core material.

Section (cross-section) I-I

Well number 11 12 13 14 15 16 17 18

Distance along the section from well 11

0 50 75 125 165 200 225 275

Absolute elevation of the ore body top

124 128 130 139 148 151 153 157

Absolute elevation of the ore body soil

122 120 119 120 118 128 134 148

Section (cross-section) II –II

Well number 21 22 23 24 25 26 27 28

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Distance along the section from well 11

0 50 75 125 175 200 225 275

Absolute elevation of the ore body top

134 138 140 149 158 161 163 167

Absolute elevation of the ore body soil

133 129 127 130 128 138 144 158

Task:

Estimate the volume of the ore body between the two sections. Additional conditions needed to reach the correct solution: · Calculate the pinch-out points of the ore body by linear extrapolation of the two absolute extreme elevations of the top and soil of the ore body. · Assume that the shape of the ore body is a truncated pyramid.

Attach schemes of sections to your solution.

Solution:

Well number

Left interface point

11 12 13 14 15 16 17 18 Right interface point

Distance along the section from well 11

-16.67 0 50 75 125 165 200 225 275 320

Absolute elevation of the ore body top

122.67 124 128 130 139 148 151 153 157 160.6

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Absolute elevation of the ore body soil

122.67 122 120 119 120 118 128 134 148 160.6

Area between the wells

16.67 250 237.5 750 980 927.5 525 700 202.5

Section area

S1= 4589.17

Well number

Left interface point

21 22 23 24 25 26 27 28 Right interface point

Distance along the section from well 11

-6.25 0 50 75 125 175 200 225 275 320

Absolute elevation of the ore body top

133.5 134 138 140 149 158 161 163 167 170.6

Absolute elevation of the ore body soil

133.5 133 129 127 130 128 138 144 158 170.6

Area between the wells

3.125 250 275 800 1225 662.5 525 700 202.5

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Section area

S2 = 4643.125

Schematic cross-section along line I-I

Schematic cross-section along line II -II

The ore body volume is: V = (S1+S2+)/3*L = 553,934m3.

L = 120m (distance between the sections).

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Points awarded:

Correctly calculated pinch-out points - 3 points

Cross-sections - 1 point each

Correctly calculated cross-section areas - 2 points each

The volume of the ore body has been calculated correctly using the formula for a truncated pyramid - 2 points

TOTAL: 11 points