Gate Geology and Geophysics 2024 Question paper with solution

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Question: 11

The Earth’s magnetic field originates from convection in which one of the following layers?

• (A) Inner core
• (B) Outer core
• (C) Lithosphere
• (D) Asthenosphere

Detailed Answer:

The Earth’s magnetic field is primarily generated by the movement of molten iron and nickel in the outer core, through a process called the geodynamo. Let’s break down the layers and their roles:

1. Inner Core (A):
   • The inner core is solid, composed mainly of iron and nickel. While it plays a role in Earth’s overall magnetic behavior, it doesn’t directly contribute to the generation of the magnetic field because it is solid and lacks the fluid motion needed for the dynamo effect.
2. Outer Core (B):
   • The outer core is composed of liquid iron and nickel. It’s about 2,200 kilometers thick and surrounds the inner core. The convection currents in this molten metal, along with Earth’s rotation, create electric currents. These currents generate Earth’s magnetic field through a phenomenon called the dynamo effect.
   • The heat from the inner core drives the convection in the outer core, and as the molten iron moves, it creates and maintains the Earth’s magnetic field.
3. Lithosphere (C):
   • The lithosphere is the rigid outermost shell of the Earth, consisting of the crust and the uppermost part of the mantle. It’s involved in tectonic activity but does not generate the magnetic field.
4. Asthenosphere (D):
   • The asthenosphere lies beneath the lithosphere and is part of the upper mantle. It’s a semi-fluid layer that allows tectonic plates to move, but like the lithosphere, it does not play a role in generating the magnetic field.

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Question:12

Which one of the following logging tools is used to measure the diameter of a borehole?

• (A) Sonic
• (B) Density
• (C) Neutron
• (D) Caliper

Detailed Answer:

In borehole logging, various tools are used to measure different physical and chemical properties of the subsurface formations. The diameter of a borehole is crucial for understanding the borehole’s integrity and stability. The tool specifically designed for measuring the diameter is called a caliper.

Here’s a breakdown of the options:

1. Sonic (A):
   • Sonic logging tools are used to measure the velocity of sound waves traveling through the formation. This data helps determine porosity and mechanical properties of the formation but does not measure the borehole diameter.
2. Density (B):
   • Density logging tools measure the density of the surrounding rock formation by using gamma radiation. It helps assess the porosity of the rock but is not used for measuring borehole diameter.
3. Neutron (C):
   • Neutron logging tools emit neutrons to measure the hydrogen content of the formation, which is related to the water and hydrocarbon content. Like the density log, it helps with porosity and fluid saturation analysis but does not measure borehole diameter.
4. Caliper (D):
   • A caliper log is a mechanical tool specifically designed to measure the diameter of the borehole. It consists of arms or sensors that extend to the borehole walls, providing an accurate measurement of the borehole diameter as the tool is lowered into the well. It helps detect any washouts, cave-ins, or irregularities in the borehole shape.

Correct Answer is:

To measure the diameter of a borehole, the caliper logging tool is used. It provides a precise profile of the borehole diameter, which is essential for wellbore stability analysis and drilling operations.

Correct Answer: (D) Caliper

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Question: 13

The given figure depicts an array used in DC resistivity surveys, where the current electrodes are denoted by C1 and C2, and potential electrodes by P1 and P2. If all the electrodes are equally spaced, then the given array corresponds to which one of the following configurations?

• (A) Wenner
• (B) Schlumberger
• (C) Dipole–Dipole
• (D) Pole–Pole

Detailed Answer:

In DC resistivity surveys, several electrode configurations are used to measure the resistivity of subsurface materials. The type of array depends on how the current (C1, C2) and potential (P1, P2) electrodes are arranged. Each configuration has different characteristics in terms of resolution and depth penetration. Here’s an analysis of the configurations:

1. Wenner Array (A):
   • In the Wenner array, all four electrodes (C1, C2, P1, P2) are equally spaced along a straight line. The spacing between the electrodes is the same, which makes this array highly symmetrical. This array is known for its simplicity in field setup and is commonly used for shallow subsurface investigations.
   • Given that the question states that “all the electrodes are equally spaced,” this configuration fits the Wenner array setup.
2. Schlumberger Array (B):
   • In the Schlumberger array, the current electrodes (C1 and C2) are placed farther apart than the potential electrodes (P1 and P2). The potential electrodes remain close to each other, and the spacing between the current electrodes is usually varied during the survey. This array provides a good balance between depth of investigation and resolution but does not have equally spaced electrodes.
3. Dipole–Dipole Array (C):
   • In the Dipole–Dipole array, two pairs of electrodes are used: one pair for current (C1, C2) and another pair for potential (P1, P2). The distance between each pair is usually greater than the spacing within the pairs. This array is often used for detailed imaging and identifying horizontal changes in resistivity, but the electrodes are not equally spaced.
4. Pole–Pole Array (D):
   • In the Pole–Pole array, one of the current electrodes and one of the potential electrodes are placed at a great distance (in theory, at infinity). This setup simplifies the calculations but also does not involve equally spaced electrodes.

Correct Answer is:

Since the question specifies that all electrodes are equally spaced, the array described corresponds to the Wenner array, where C1, C2, P1, and P2 are arranged with equal spacing between them.

Correct Answer: (A) Wenner

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Question:14

Which one of the following is an ultramafic rock?

• (A) Granite
• (B) Gabbro
• (C) Dunite
• (D) Basalt

Detailed Answer:

Ultramafic rocks are igneous rocks that have very low silica content and are composed mostly of mafic minerals such as olivine and pyroxene. These rocks originate from the Earth’s mantle and are characterized by high magnesium and iron content.

Let’s look at the options:

1. Granite (A):
   • Granite is a felsic rock with high silica content (around 70%) and is composed mostly of quartz, feldspar, and mica. It is not ultramafic but rather felsic.
2. Gabbro (B):
   • Gabbro is a mafic rock, composed mostly of pyroxene, plagioclase, and sometimes olivine. It has a lower silica content than granite but still does not classify as ultramafic.
3. Dunite (C):
   • Dunite is a type of ultramafic rock, composed almost entirely of olivine (more than 90%), with very little silica content. It originates from the Earth’s mantle and is often found in ophiolite complexes. Dunite is a prime example of an ultramafic rock.
4. Basalt (D):
   • Basalt is a mafic rock that has a higher silica content compared to ultramafic rocks. It is primarily composed of pyroxene and plagioclase, with some olivine, but it is not ultramafic.

Correct Answer is:

Among the options, Dunite is the ultramafic rock, as it is composed almost entirely of olivine and has very low silica content.

Correct Answer: (C) Dunite

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Question: 15

Gold is being produced from which one of the following mines in India?

• (A) Baula
• (B) Hutti
• (C) Dariba
• (D) Jaduguda

Detailed Answer:

India has a few prominent gold-producing mines, and the Hutti mine stands out as one of the most significant.

1. Baula (A):
   • The Baula mine is known for producing chromite, a type of ore used primarily for chromium production, not gold.
2. Hutti (B):
   • The Hutti Gold Mine is located in the Raichur district of Karnataka and is the only active primary gold mine in India. The Hutti Gold Mines Company Limited (HGML) operates this mine, and it is the leading producer of gold in India.
   • Hutti has a long history of gold mining and continues to produce a significant amount of gold today.
3. Dariba (C):
   • The Dariba mine is known for the production of lead, zinc, and silver but not gold.
4. Jaduguda (D):
   • Jaduguda is primarily known for uranium mining. Located in Jharkhand, the Jaduguda mines supply uranium for India’s nuclear power sector, not gold.

Correct Answer is:

Gold is actively produced from the Hutti Gold Mine in Karnataka, India.

Correct Answer: (B) Hutti

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Question: 16       

Which of the following hydrocarbon fields is/are located in the western offshore of India?

• (A) Tapti
• (B) Lakwa
• (C) Ravva
• (D) Panna

Detailed Answer:

India has several important hydrocarbon fields, both onshore and offshore. Let’s look at where these fields are located:

1. Tapti (A):
   • Tapti is located in the western offshore region of India, specifically in the Gulf of Khambhat (Cambay Basin) off the coast of Gujarat. It is part of the Western Offshore Basin, which is a significant hydrocarbon-producing region in India.
2. Lakwa (B):
   • The Lakwa oil and gas field is located in Assam, which is in the northeastern part of India. This is an onshore field and not part of the western offshore region.
3. Ravva (C):
   • Ravva is an offshore oil and gas field located in the Krishna-Godavari Basin off the eastern coast of India (Andhra Pradesh). This field is part of the eastern offshore region, not the western offshore.
4. Panna (D):
   • Panna is located in the western offshore region, north of Mumbai, in the Arabian Sea. It is part of the Mumbai Offshore Basin and is known for both oil and gas production.

Correct Answer is:

The hydrocarbon fields located in the western offshore of India are Tapti (A) and Panna (D).

Correct Answer:

• (A) Tapti
• (D) Panna

Q.17: A cylindrical sample of granite (diameter = 54.7 mm; length = 137 mm) shows a linear relationship between axial stress and axial strain under uniaxial compression up to the peak stress level at which the specimen fails. If the uniaxial compressive strength of this sample is 200 MPa and the axial strain corresponding to this peak stress is 0.005, the Young’s modulus of the sample in GPa is _______ (in integer).

Answer: To calculate the Young’s modulus of the granite sample, follow these steps:

1. Identify the given values:
• Uniaxial compressive strength, σ = 200 MPa
• Axial strain, ε = 0.005
2. Convert the uniaxial compressive strength from MPa to GPa:

We convert MPa to GPa by dividing by 1000:

• σ = 200 MPa = 200 / 1000 GPa = 0.2 GPa
3. Calculate the Young’s modulus using the formula:

The formula for Young’s modulus E is:

E = σ / ε

Substitute the values:

E = 0.2 GPa / 0.005 = 40 GPa

Therefore, the Young’s modulus of the granite sample is 40 GPa.

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Q.19

Question: Match the geophysical methods in Group–I with their associated physical properties in Group–II.

Group–I

• P. Magnetic
• Q. Gravity
• R. Magnetotelluric
• S. Induced Polarization

Group–II

Chargeability

Electrical conductivity 

Susceptibility 

Density

Options: (A) P-3, Q-4, R-2, S-1 (B) P-3, Q-4, R-1, S-2 (C) P-4, Q-3, R-2, S-1 (D) P-2, Q-1, R-4, S-3

Answer: To match the geophysical methods with their associated physical properties, let’s analyze each method:

1. Magnetic Method (P):
   • Measures variations in the Earth’s magnetic field, which relate to the magnetic susceptibility of rocks.
   • Thus, Magnetic → Susceptibility (3).
2. Gravity Method (Q):
   • Measures variations in the Earth’s gravitational field, which relate to the density of rocks.
   • Thus, Gravity → Density (4).
3. Magnetotelluric Method (R):
   • Measures variations in the Earth’s electric and magnetic fields, which relate to the electrical conductivity of rocks.
   • Thus, Magnetotelluric → Electrical conductivity (2).
4. Induced Polarization Method (S):
   • Measures the delayed electrical response of rocks, which is related to chargeability.
   • Thus, Induced Polarization → Chargeability (1).

Matching:

• P → 3
• Q → 4
• R → 2
• S → 1

Thus, the correct option is:

(A) P-3, Q-4, R-2, S-1

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Q.20: The number of planes of symmetry in a tetrahedron is

(A) 9
(B) 6
(C) 4
(D) 3

Answer: In a regular tetrahedron, each plane of symmetry divides the shape into two mirror-image halves. There are:

1. 4 planes of symmetry that pass through a vertex and the centroid of the opposite face.
2. 2 additional planes of symmetry that pass through the midpoints of opposite edges.

Therefore, the total number of planes of symmetry in a tetrahedron is 6.

Thus, the correct answer is: 6

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Q.22: Which one or more of the following minerals shows an O ratio of 4:1 in its silicate structure?

• (A) Olivine
• (B) Quartz
• (C) Diopside
• (D) Albite

Answer: To determine which mineral has an O ratio of 4:1, let’s analyze each mineral’s silicate structure and its ratio of oxygen (O) to silicon (Si) atoms:

1. Olivine:
• Olivine has the formula (Mg,Fe)2SiO4.
• In this structure, there are 4 oxygen atoms for every 1 silicon atom.
• Thus, the O ratio for Olivine is 4:1.
2. Quartz:
• Quartz has the formula SiO2.
• In this structure, there are 2 oxygen atoms for every 1 silicon atom.
• Thus, the O ratio for Quartz is 2:1.
3. Diopside:
• Diopside has the formula CaMgSi2O6.
• In this structure, there are 6 oxygen atoms for every 2 silicon atoms.
• Thus, the O ratio for Diopside is 3:1.
4. Albite:
• Albite has the formula NaAlSi3O8.
• In this structure, there are 8 oxygen atoms for every 3 silicon atoms.
• Thus, the O ratio for Albite is approximately 2.67:1, which is not 4:1.

Therefore, the only mineral listed with an exact O ratio of 4:1 is Olivine.

Thus, the correct answer is:

• (A) Olivine

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Q.23: Which of the following rock structures is/are fold(s)?

(A) Antiform
(B) Horst
(C) Syncline
(D) Synform

Answer: Folds are rock structures resulting from deformation:

1. Antiform – A fold where the oldest rocks are in the center, and the limbs dip away from the center.
2. Syncline – A fold where the youngest rocks are in the center, and the limbs dip towards the center.
3. Synform – A fold where the youngest rocks are in the center, similar to a syncline but can be used in different contexts.

Horst is a block of the Earth’s crust uplifted between two faults, not a fold.

Thus, the correct answers are:

(A) Antiform, (C) Syncline, and (D) Synform

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Q.24: Assume heat-producing elements are uniformly distributed within a 16 km thick layer in the crust in a heat flow province. Given that the surface heat flow and reduced heat flow are 54 mW/m² and 22 mW/m², respectively, the radiogenic heat production in the given crustal layer in µW/m³ is ___________ (in integer).

Answer: To determine the radiogenic heat production, follow these detailed steps:

1. Convert heat flow values to the same units:

We are given two heat flow values:

• Surface heat flow: 54 mW/m²
• Reduced heat flow: 22 mW/m²
2. Calculate the difference in heat flow:

The radiogenic heat production is calculated by subtracting the reduced heat flow from the surface heat flow:

Heat Production = Surface Heat Flow – Reduced Heat Flow

Substitute the given values:

Heat Production = 54 mW/m² – 22 mW/m² = 32 mW/m²

3. Convert the thickness of the crustal layer to meters:

The thickness of the crustal layer is given as 16 km. Convert this to meters:

Thickness = 16 km = 16,000 meters

4. Convert heat production from mW/m² to µW/m³:

To find the radiogenic heat production in µW/m³, divide the heat production by the thickness of the layer:

Radiogenic Heat Production = Heat Production / Thickness

Substitute the values:

Radiogenic Heat Production = 32 mW/m² / 16,000 meters

Convert mW/m² to µW/m² by multiplying by 1,000:

Radiogenic Heat Production = (32 × 1,000 µW/m²) / 16,000 meters = 32,000 µW/m² / 16,000 meters = 2 µW/m³

Therefore, the radiogenic heat production in the given crustal layer is 2 µW/m³.

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Q.25: A confined aquifer with a uniform saturated thickness of 10 m has a hydraulic conductivity of \(10^{-2}\) cm/s. Considering a steady flow, the transmissivity of the aquifer in m²/day is _________________ (rounded off to one decimal place).

Answer: To calculate the transmissivity (T), follow these detailed steps:

1. Convert hydraulic conductivity from cm/s to m/s:

Given hydraulic conductivity is \(10^{-2}\) cm/s.

Convert cm/s to m/s by multiplying by \(10^{-2}\):

Hydraulic Conductivity = \(10^{-2}\) cm/s × \(10^{-2}\) = \(10^{-4}\) m/s

2. Calculate transmissivity in m²/s:

Transmissivity (T) is calculated as follows:

T = Hydraulic Conductivity × Saturated Thickness

Substitute the values:

T = \(10^{-4}\) m/s × 10 m = \(10^{-3}\) m²/s

3. Convert transmissivity from m²/s to m²/day:

There are 86,400 seconds in a day:

T = \(10^{-3}\) m²/s × 86,400 s/day = 86.4 m²/day

4. Round the result to one decimal place:

The transmissivity, rounded to one decimal place, is:

T = 86.4 m²/day

Therefore, the transmissivity of the aquifer is 86.4 m²/day.

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Q.26: A current of 2 A passes through a cylindrical rod with a uniform cross-sectional area of 4 m² and resistivity of 100 Ω-m. The magnitude of the electric field (E) measured along the length of the rod in V/m is _______________ (in integer). Answer: The electric field (E) in a conductor is given by Ohm’s Law: \[ E = \rho \times \frac{I}{A} \] where: – \( \rho \) is the resistivity (100 Ω-m), – \( I \) is the current (2 A), – \( A \) is the cross-sectional area (4 m²). 1. Calculate the electric field: \[ E = 100 \, \Omega \text{-m} \times \frac{2 \, \text{A}}{4 \, \text{m}^2} = 100 \times 0.5 = 50 \, \text{V/m} \] Thus, the magnitude of the electric field is 50 V/m.

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Q.27: Which one of the following lineations can be observed on a foliation with an attitude of \( 210^\circ, 40^\circ \) NW? (A) \( 40^\circ \rightarrow 300^\circ \) (B) \( 40^\circ \rightarrow 040^\circ \) (C) \( 40^\circ \rightarrow 220^\circ \) (D) \( 40^\circ \rightarrow 350^\circ \) **Answer:** To determine which lineation can be observed on a foliation with the given attitude: 1. **Understand the Foliation Attitude:** The foliation has a strike of \( 210^\circ \) and a dip of \( 40^\circ \) towards the northwest. 2. **Determine the Lineation Direction:** Lineations are typically perpendicular to the strike direction of the foliation. For a foliation strike of \( 210^\circ \), the lineation direction would be orthogonal to this strike direction. 3. **Calculate the Perpendicular Lineation Direction:** The direction perpendicular to the foliation strike can be calculated as: \[ \text{Lineation Direction} = \text{Strike Direction} + 90^\circ \] Substitute the given strike direction: \[ \text{Lineation Direction} = 210^\circ + 90^\circ = 300^\circ \] 4. **Match with the Given Options:** The correct lineation direction among the options provided is \( 40^\circ \rightarrow 300^\circ \). Thus, the correct answer is: **(A) \( 40^\circ \rightarrow 300^\circ \)**

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Q.28: Match the minerals in Group–I with the corresponding cleavage types in Group–II.

Group–I

P. Diopside
Q. Galena
R. Calcite
S. Fluorite

Group–II

1. Cubic
2. Octahedral
3. Prismatic
4. Rhombohedral

(A) P-3, Q-2, R-4, S-1
(B) P-4, Q-3, R-1, S-2
(C) P-3, Q-1, R-4, S-2
(D) P-4, Q-1, R-2, S-3

Answer:

To match the minerals with their corresponding cleavage types:

1. Diopside:
   • Diopside has prismatic cleavage. It typically cleaves along three directions forming prismatic shapes.
2. Galena:
   • Galena exhibits cubic cleavage. It cleaves in three directions at right angles, forming cubic shapes.
3. Calcite:
   • Calcite has rhombohedral cleavage. It cleaves in three directions not at right angles, forming rhombohedral shapes.
4. Fluorite:
   • Fluorite shows octahedral cleavage. It cleaves in four directions forming octahedral shapes.

Matching these properties with the options:

• Diopside: Prismatic (3)
• Galena: Cubic (1)
• Calcite: Rhombohedral (4)
• Fluorite: Octahedral (2)

Therefore, the correct match is:

(C) P-3, Q-1, R-4, S-2

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Q.29: The composition of which one of the following reservoirs closely matches with that of iron meteorites?

(A) Primitive Mantle
(B) Earth’s Core
(C) Depleted Mantle
(D) Bulk Silicate Earth

Answer:

Iron meteorites are primarily composed of metallic iron-nickel alloys. The composition of the Earth’s core is known to be rich in iron and nickel, making it closely resemble the composition of iron meteorites.

Therefore, the correct answer is:

(B) Earth’s Core

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Q.30: Match the microstructures in Group–I with their characteristics in Group–II.

Group–I

P. Core–mantle
Q. Decussate
R. Spherulite
S. Millipede

Group–II

1. Radiating fibrous aggregate of K-feldspar with or without quartz
2. Large strained mineral grains surrounded by fine-grained, recrystallized grains
3. Inclusion trails in a porphyroblast curve into the matrix foliation by developing concave outward pattern
4. Randomly oriented mineral grains dominated by crystal faces, such as in sheet silicates

(A) P-2, Q-3, R-4, S-1
(B) P-3, Q-4, R-1, S-2
(C) P-2, Q-4, R-1, S-3
(D) P-4, Q-2, R-3, S-1

Answer:

To match the microstructures with their characteristics:

1. Core–mantle:
   • This refers to large strained mineral grains surrounded by fine-grained, recrystallized grains. This description fits the core–mantle structure which shows a distinct boundary between large and fine-grained minerals.
2. Decussate:
   • Randomly oriented mineral grains dominated by crystal faces, such as in sheet silicates, describes the decussate structure. It features a complex pattern of randomly oriented crystals.
3. Spherulite:
   • Radiating fibrous aggregate of K-feldspar with or without quartz describes the spherulite, which is a spherical aggregate of minerals radiating outward.
4. Millipede:
   • Inclusion trails in a porphyroblast curve into the matrix foliation by developing a concave outward pattern describes the millipede microstructure.

Matching these properties with the options:

• Core–mantle: Large strained mineral grains surrounded by fine-grained, recrystallized grains (2)
• Decussate: Randomly oriented mineral grains dominated by crystal faces (4)
• Spherulite: Radiating fibrous aggregate of K-feldspar with or without quartz (1)
• Millipede: Inclusion trails in a porphyroblast curve into the matrix foliation by developing a concave outward pattern (3)

Therefore, the correct match is:

(C) P-2, Q-4, R-1, S-3

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Q.31: Which one among the following is the least abundant sedimentary rock in the stratigraphic record?

(A) Sandstone
(B) Limestone
(C) Conglomerate
(D) Shale

Answer:

Sedimentary rocks differ in their abundance based on their depositional environments:

• Sandstone: Common, forming in a variety of environments from sand deposits.
• Limestone: Also common, forming from carbonate sediments in marine environments.
• Shale: Very common, forming from fine-grained sediments in quiet water environments.
• Conglomerate: Less common compared to sandstone, limestone, and shale because it forms from coarse gravel and cobbles, which require more specific conditions for formation.

Given the conditions for conglomerate formation, it is less frequently found compared to the more abundant sandstone, limestone, and shale.

Correct Answer: (C) Conglomerate

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Q.32: Which one of the following sequences of index minerals correctly represents the order of increasing metamorphic grade during regional metamorphism of siliceous dolomitic limestones?

(A) Tremolite → Diopside → Talc
(B) Diopside → Tremolite → Forsterite
(C) Talc → Tremolite → Diopside
(D) Talc → Forsterite → Tremolite

Answer:

During regional metamorphism of siliceous dolomitic limestones, the order of increasing metamorphic grade is as follows:

1. Talc: Forms at the lowest metamorphic grades.
2. Tremolite: Forms at intermediate grades.
3. Diopside: Forms at higher grades.

The correct sequence of minerals from lowest to highest grade is:

• Talc → Tremolite → Diopside

Correct Answer: (C) Talc → Tremolite → Diopside

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Q.33: Which one among the following is the oldest horse genus?

(A) Orohippus
(B) Mesohippus
(C) Merychippus
(D) Pliohippus

Answer:

To determine the oldest horse genus, we need to look at the evolutionary timeline of horses. Here’s a brief overview:

1. Orohippus: The oldest genus among the options, appearing during the late Eocene epoch (around 30-25 million years ago). This genus represents an early stage in horse evolution.
2. Mesohippus: Evolved after Orohippus, during the Miocene epoch (about 20-10 million years ago). It is a more advanced genus compared to Orohippus.
3. Merychippus: Appeared later, during the Miocene epoch (around 15-5 million years ago). This genus is more evolved compared to Mesohippus.
4. Pliohippus: The most recent genus among the options, appearing in the late Miocene to early Pliocene epoch (around 5-2 million years ago). It is the latest in the evolutionary timeline.

Therefore, the oldest horse genus among the options is:

(A) Orohippus

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Q.34: The measured plate velocity is maximum (in International Terrestrial Reference Frame) at which one of the following locations on the Indian Plate?

(A) Leh
(B) Delhi
(C) Bengaluru
(D) Maldives

Answer:

The Indian Plate is moving towards the northeast. The rate of plate motion varies across different locations on the plate:

• Leh: Located in the northern part of the Indian Plate, it experiences relatively lower velocities compared to the southern regions.
• Delhi: Positioned more centrally, it also experiences moderate plate velocities.
• Bengaluru: Located in the southern part of the Indian Plate, the velocity here is lower compared to locations further north.
• Maldives: Situated towards the southern edge of the Indian Plate, it experiences the highest plate velocity due to its proximity to the divergent boundary between the Indian Plate and the surrounding oceanic plates.

Thus, the measured plate velocity is maximum at:

(D) Maldives

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Q.35: Which one of the following textures is called the chalcopyrite disease?

(A) Chalcopyrite blebs in sphalerite
(B) Sphalerite stars in chalcopyrite
(C) Chalcopyrite lamellae in bornite
(D) Bornite lamellae in chalcopyrite

Detailed Answer:

The term “chalcopyrite disease” is used in mineralogy to describe a specific textural feature observed in ore minerals. This term is associated with chalcopyrite, a copper iron sulfide mineral, and its interaction with other minerals, particularly sphalerite, a zinc sulfide mineral.

Textural Features Explained

1. Chalcopyrite blebs in sphalerite (Option A):
1. Chalcopyrite blebs: Small, rounded inclusions or blebs of chalcopyrite within sphalerite. These blebs can appear as irregular, blob-like features scattered throughout the sphalerite matrix.
2. Sphalerite: A common zinc sulfide mineral often found in association with chalcopyrite in ore deposits.
3. Chalcopyrite Disease: This term specifically refers to the presence of chalcopyrite blebs within sphalerite. The term highlights a characteristic texture where chalcopyrite forms discrete, small inclusions within sphalerite, giving the impression of a “disease” affecting the sphalerite.
4. Sphalerite stars in chalcopyrite (Option B):
5. This refers to the formation of star-like patterns of sphalerite within chalcopyrite. However, this texture is not referred to as chalcopyrite disease.
6. Chalcopyrite lamellae in bornite (Option C):
7. Chalcopyrite lamellae: Thin, layered forms of chalcopyrite within bornite, another copper iron sulfide mineral. While this texture is significant in ore mineralogy, it is not termed chalcopyrite disease.
8. Bornite lamellae in chalcopyrite (Option D):
9. Bornite lamellae: Thin layers of bornite within chalcopyrite. Similar to chalcopyrite lamellae in bornite, this texture does not correspond to chalcopyrite disease.

Conclusion

The texture specifically called chalcopyrite disease is the presence of chalcopyrite as small, rounded blebs within sphalerite. This distinctive texture is used to describe how chalcopyrite inclusions are distributed within the sphalerite mineral matrix.

Therefore, the correct answer is:

(A) Chalcopyrite blebs in sphalerite

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Q.38: Which of the following sedimentary basins is/are producing hydrocarbon commercially?

(A) Ganga
(B) Krishna–Godavari
(C) Kerala–Konkan
(D) Cauvery

Detailed Answer:

Hydrocarbon production is associated with sedimentary basins that have the right geological conditions for the formation and accumulation of oil and gas. Here’s a breakdown of each basin:

1. Ganga Basin (Option A):
2. Hydrocarbon Production: The Ganga Basin is not known for significant commercial hydrocarbon production. It is primarily known for its sedimentary sequences and other geological features.
3. Krishna–Godavari Basin (Option B):
4. Hydrocarbon Production: This basin is one of the most prolific hydrocarbon-producing regions in India. It is known for significant oil and gas fields, including the discoveries of major reserves by companies like Reliance Industries and ONGC.
5. Kerala–Konkan Basin (Option C):
6. Hydrocarbon Production: The Kerala–Konkan Basin has limited hydrocarbon production compared to other basins. It is not a major hydrocarbon-producing region.
7. Cauvery Basin (Option D):
8. Hydrocarbon Production: The Cauvery Basin is another significant hydrocarbon-producing region in India. It has multiple oil and gas fields, including those operated by ONGC and other companies.

Conclusion

The sedimentary basins in India that are known for commercial hydrocarbon production are:

(B) Krishna–Godavari and (D) Cauvery

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Q.39: Which of the following bivalves is/are swimmers?

(A) Aspergillum
(B) Lima
(C) Tellina
(D) Pecten

Detailed Answer:

In bivalves, some species are adapted for a swimming lifestyle, while others are more sedentary or burrowing. The bivalves that are known for their swimming abilities are typically equipped with adaptations for active movement in the water.

1. Aspergillum (Option A):
2. Aspergillum: This genus of bivalves is not known for swimming. It is typically a burrowing bivalve.
3. Lima (Option B):
4. Lima: Commonly known as the Lima or Fan Shell, this bivalve has the ability to swim by clapping its valves together. It uses this movement to escape from predators and relocate.
5. Tellina (Option C):
6. Tellina: Generally, these bivalves are burrowers rather than swimmers. They live in sandy or muddy substrates.
7. Pecten (Option D):
8. Pecten: Known as Scallops, these bivalves are active swimmers. They move by rapidly opening and closing their valves, which expels water and propels them through the water.

Conclusion

The bivalves known for swimming are:

(B) Lima and (D) Pecten

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Q.40: Which of the following structures is/are associated with duplexes in fold–thrust belts?

(A) Roof thrust
(B) Floor thrust
(C) Imbricate fan
(D) Horses

Detailed Answer:

In fold-thrust belts, duplex structures are characterized by a stack of imbricate thrust sheets, and several specific structural features are associated with them:

1. Roof Thrust (Option A):
2. Roof Thrust: This is a major thrust fault that forms the upper boundary of a duplex system. It is a significant feature in duplex structures, providing the upper limit of the stacked imbricate sheets.
3. Floor Thrust (Option B):
4. Floor Thrust: This is a major thrust fault that forms the lower boundary of a duplex system. It acts as the base for the stacked thrust sheets in a duplex.
5. Imbricate Fan (Option C):
6. Imbricate Fan: This structure is a series of overlapping thrust faults that fan out from a common point. It is not exclusive to duplex structures but is often associated with them.
7. Horses (Option D):
8. Horses: These are blocks of rock trapped between two thrust faults within a duplex. They are integral components of duplex structures, representing the various slices of rock between the faults.

correct Answer

The structures commonly associated with duplexes in fold-thrust belts are:

(A) Roof thrust, (B) Floor thrust, and (D) Horses

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Q.41: Which of the following statements is/are CORRECT?

(A) Karst topography is formed in limestone terrains
(B) Fjords are formed by aeolian activities
(C) Oxbow lakes are formed in fluvial environments
(D) Ventifacts are formed by glaciers

Detailed Answer:

1. Karst topography is formed in limestone terrains (Option A):
2. Correct: Karst landscapes develop in regions with soluble rock such as limestone, resulting in features like caves, sinkholes, and karst towers.
3. Fjords are formed by aeolian activities (Option B):
4. Incorrect: Fjords are formed by glacial activity, not by wind (aeolian) processes. They are U-shaped valleys carved out by glaciers.
5. Oxbow lakes are formed in fluvial environments (Option C):
6. Correct: Oxbow lakes are crescent-shaped lakes formed by the meandering of rivers and are typical features in fluvial (river) environments.
7. Ventifacts are formed by glaciers (Option D):
8. Incorrect: Ventifacts are rocks shaped by windblown sand, not by glaciers. They are features of aeolian (wind) processes.

correct Answer

The correct statements are:

(A) Karst topography is formed in limestone terrains and (C) Oxbow lakes are formed in fluvial environments

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Q.43: The support pressure of 20 kPa is required to stabilize the loose blocks of the Excavation Disturbed Zone (EDZ) at the crown of a circular tunnel with a horizontal axis. The EDZ is to be stabilized by inserting rock bolts vertically into the roof. If the working capacity of a bolt is 160 kN, the area of the roof supported by a single bolt in m² is ____________ (in integer). **Detailed Answer:** To determine the area of the roof supported by a single bolt, use the formula: \[ \text{Area} = \frac{\text{Capacity of a Bolt}}{\text{Support Pressure}} \] Given: – Support Pressure = 20 kPa – Capacity of a Bolt = 160 kN = 160 kPa (since 1 kN/m² = 1 kPa) Now, substitute the values: \[ \text{Area} = \frac{160 \text{ kPa}}{20 \text{ kPa}} \] Simplifying the expression: \[ \text{Area} = \frac{160}{20} = 8 \text{ m}^2 \] **Conclusion:** The area of the roof supported by a single bolt is \( 8 \text{ m}^2 \).

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Q.44: The areas of drainage basins A and B are 25 km² and 50 km², respectively. The total length of drainages of all orders in basin A is 20 km. If both basins have the same drainage density, the total length of drainages of all orders in basin B in km is _____________ (in integer). **Detailed Answer:** Drainage density is given by the formula: \[ \text{Drainage Density} = \frac{\text{Total Length of Drainages}}{\text{Area}} \] For **Basin A**: \[ \text{Drainage Density} = \frac{20 \text{ km}}{25 \text{ km}^2} \] Simplifying: \[ \text{Drainage Density} = 0.8 \text{ km/km}^2 \] For **Basin B**, using the same drainage density: \[ \text{Total Length of Drainages in Basin B} = \text{Drainage Density} \times \text{Area} \] Substitute the values: \[ \text{Total Length of Drainages in Basin B} = 0.8 \text{ km/km}^2 \times 50 \text{ km}^2 \] Calculate the result: \[ \text{Total Length of Drainages in Basin B} = 40 \text{ km} \] **Conclusion:** The total length of drainages of all orders in Basin B is \( 40 \text{ km} \).

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Q.45: Match the stratigraphic units in Group–I with the sedimentary basins in Group–II.

Group–I:

P. Ramgundam Sandstone
Q. Raipur Formation
R. Bagalkot Group
S. Sonia Sandstone

Group–II:

1. Chhattisgarh
2. Kaladgi
3. Marwar
4. Godavari

Detailed Answer:

1. Ramgundam Sandstone (P):
1. Correct Basin: Godavari (4)
2. The Ramgundam Sandstone is part of the Gondwana sequence found in the Godavari basin. This sandstone is significant in the geological history of the Godavari region and forms a part of the sedimentary record of this basin.
3. Raipur Formation (Q):
2. Correct Basin: Chhattisgarh (1)
3. The Raipur Formation is a geological formation within the Chhattisgarh basin. This formation is known for its sedimentary deposits and contributes to the stratigraphy of the Chhattisgarh basin.
4. Bagalkot Group (R):
3. Correct Basin: Kaladgi (2)
4. The Bagalkot Group is a significant geological unit located in the Kaladgi basin. It is an important component of the sedimentary rock formations in the Kaladgi basin.
5. Sonia Sandstone (S):
4. Correct Basin: Marwar (3)
5. The Sonia Sandstone is found in the Marwar basin. This sandstone is part of the sedimentary sequences in the Marwar region and contributes to the geological framework of the basin.

Correct Answer is:

The correct matching is:

(B) P-4, Q-1, R-2, S-3

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Q.46: Which one of the following openings is a type of decline in underground mines?

(A) Crosscut
(B) Winze
(C) Spiral tunnel
(D) Drift

Detailed Answer:

In underground mining, a decline is an inclined shaft that descends into the mine, allowing access to different levels of the ore body and facilitating the movement of materials and personnel. Among the options provided:

1. Crosscut (A):
2. A crosscut is a horizontal or nearly horizontal tunnel that intersects the ore body and is used to connect different parts of the mine. It is not a type of decline.
3. Winze (B):
4. A winze is a vertical or nearly vertical shaft used for hoisting ore and waste from lower levels of a mine. It is not a decline but rather a means of moving materials vertically.
5. Spiral Tunnel (C):
6. A spiral tunnel is a type of decline. It is designed to gradually descend into the mine while spiraling around the shaft. This design helps in accessing different levels of the ore body efficiently.
7. Drift (D):
8. A drift is a horizontal or nearly horizontal underground tunnel used to access ore bodies. It is not inclined and therefore not considered a type of decline.

Correct Answer is:

The correct answer is (C) Spiral tunnel.

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Q.48: Match the following invertebrates in Group–I with their morphological features in Group–II.

Group–I:

P. Trilobite
Q. Brachiopod
R. Bivalve
S. Echinoid

Group–II:

1. Periproct
2. Hypostome
3. Deltidial plate
4. Lunule

Detailed Answer:

1. Trilobite (P):
1. Correct Feature: Hypostome (2)
2. Trilobites are extinct marine arthropods with a distinctive exoskeleton. The hypostome is a structure on the underside of the trilobite’s head, used for feeding and attachment.
3. Brachiopod (Q):
2. Correct Feature: Deltidial plate (3)
3. Brachiopods are marine invertebrates with two shells. The deltidial plate is a feature found in the hinge area of brachiopod shells.
4. Bivalve (R):
3. Correct Feature: Lunule (4)
4. Bivalves are mollusks with two hinged shells. The lunule is a crescent-shaped area on the shell, often used to identify different bivalve species.
5. Echinoid (S):
4. Correct Feature: Periproct (1)
5. Echinoids, or sea urchins, are characterized by their radial symmetry and external spines. The periproct is the area surrounding the anus in echinoids.

Correct Answer is:

The correct matching is:

(B) P-2, Q-3, R-4, S-1

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Q.49: During high-temperature metamorphism of pelites, which one of the following mineral reactions represents the second sillimanite isograd?

(A) Muscovite + Quartz = Sillimanite + K-feldspar + H₂O
(B) Staurolite + Quartz = Garnet + Sillimanite + H₂O
(C) Staurolite + Muscovite + Quartz = Garnet + Biotite + Sillimanite + H₂O
(D) Kyanite = Sillimanite

Detailed Answer:

The second sillimanite isograd represents the temperature at which sillimanite begins to form in the metamorphic process of pelites.

1. Muscovite + Quartz = Sillimanite + K-feldspar + H₂O (A):
2. This reaction marks the formation of sillimanite from muscovite and quartz, and is associated with the second sillimanite isograd. It indicates the transition to higher temperatures where sillimanite becomes stable.
3. Staurolite + Quartz = Garnet + Sillimanite + H₂O (B):
4. This reaction is not typically associated with the second sillimanite isograd, as it generally represents earlier stages of metamorphism.
5. Staurolite + Muscovite + Quartz = Garnet + Biotite + Sillimanite + H₂O (C):
6. This complex reaction involves multiple minerals and is not specifically indicative of the second sillimanite isograd.
7. Kyanite = Sillimanite (D):
8. This reaction indicates the transformation from kyanite to sillimanite and is related to the kyanite-sillimanite isograd, which occurs at higher temperatures than the second sillimanite isograd.

Correct Answer is:

The correct answer is (A) Muscovite + Quartz = Sillimanite + K-feldspar + H₂O.

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Q.50: Which one of the following represents deviatoric stress in a 2D stress Mohr Circle?

(A) Radius
(B) Center
(C) Pole
(D) Diameter

Detailed Answer:

To understand which part of the 2D stress Mohr Circle represents deviatoric stress, let’s review the concepts:

1. Mohr Circle Basics: Mohr’s Circle is a graphical representation used in engineering to describe the state of stress at a point. It helps visualize how normal and shear stresses transform under different orientations.
2. Deviatoric Stress: Deviatoric stress refers to the component of stress that deviates from the hydrostatic (or mean) stress. It’s essentially the part of the stress that causes distortion or change in shape, rather than just volume change.
3. Mohr Circle Components:
3. Radius: This represents the maximum shear stress on the circle and is related to the difference between the principal stresses. While it reflects shear stress, it does not directly represent deviatoric stress.
4. Center: The center of Mohr Circle represents the average (or mean) stress, which is equal to the hydrostatic stress. Deviatoric stress is derived by subtracting this mean stress from the total stress.
5. Pole: The pole is a point on the Mohr Circle related to the orientation of the principal stresses but does not represent deviatoric stress directly.
6. Diameter: The diameter of Mohr Circle represents the difference between the maximum and minimum principal stresses. Deviatoric stress can be visualized by considering how the circle’s radius changes from the mean stress.

Correct Answer is:

The deviatoric stress in a 2D Mohr Circle is represented by the radius of the circle. The radius reflects the shear stress component, which is part of the deviatoric stress. However, since the deviatoric stress is essentially the stress excluding the mean stress, it is indirectly represented by how the circle’s radius (deviatoric component) changes relative to the mean stress.

Correct Answer: (A) Radius

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Q.53: Match the following copper deposits in Group–I with their host rocks in Group–II.

Group–I:

P. Khetri
Q. Mosabani
R. Malanjkhand
S. Kalyadi

Group–II:

1. Chlorite–biotite schist and soda–granite
2. Garnetiferous chlorite schist
3. Metachert
4. Tonalite–granodiorite–granite

Detailed Answer:

To match the copper deposits with their respective host rocks, let’s review each deposit and its associated geology:

1. Khetri (P):
1. Host Rock: Garnetiferous chlorite schist
2. The Khetri copper belt, located in Rajasthan, India, is known for copper deposits hosted in garnetiferous chlorite schist. These metamorphic rocks are associated with the mineralization in this region.
3. Mosabani (Q):
2. Host Rock: Chlorite–biotite schist and soda–granite
3. The Mosabani copper deposit, situated in Jharkhand, India, is found in chlorite–biotite schist and soda–granite. This geologic setting is characteristic of the Mosabani copper belt.
4. Malanjkhand (R):
3. Host Rock: Tonalite–granodiorite–granite
4. The Malanjkhand copper deposit, located in Madhya Pradesh, India, is hosted in tonalite, granodiorite, and granite. This deposit is associated with these igneous rocks.
5. Kalyadi (S):
4. Host Rock: Metachert
5. The Kalyadi copper deposit, located in Karnataka, India, is associated with metachert. This metamorphosed sedimentary rock is a common host for copper mineralization in the area.

Correct Matching:

• P-2: Khetri → Garnetiferous chlorite schist
• Q-1: Mosabani → Chlorite–biotite schist and soda–granite
• R-4: Malanjkhand → Tonalite–granodiorite–granite
• S-3: Kalyadi → Metachert

The correct answer is:

(C) P-4, Q-1, R-2, S-3

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Q.54: Which one of the following events represents the termination of the Wilson Cycle in Plate Tectonics?

(A) Ocean–continent subduction
(B) Continent–continent collision
(C) Continental rifting
(D) Seafloor spreading

Detailed Answer:

To determine which event represents the termination of the Wilson Cycle, let’s review the stages of the Wilson Cycle and how each event fits into these stages:

1. Wilson Cycle Overview: The Wilson Cycle describes the lifecycle of ocean basins, from their formation to their closure. It includes several stages:
1. Continental Rifting: The process where a continent breaks apart, leading to the formation of a new ocean basin.
2. Seafloor Spreading: Occurs in the newly formed ocean basin as new oceanic crust is created at mid-ocean ridges.
3. Ocean–Continent Subduction: When an oceanic plate converges with a continental plate, leading to subduction and the destruction of oceanic crust.
4. Continent–Continent Collision: When two continental plates converge, leading to the closure of the ocean basin and the formation of mountain ranges.
5. Termination of the Wilson Cycle:
2. Ocean–Continent Subduction (A): Represents a mid-cycle event where the oceanic crust is being destroyed, but it does not terminate the Wilson Cycle.
3. Continent–Continent Collision (B): This is the final stage of the Wilson Cycle where the ocean basin closes due to the collision of two continental plates. This collision leads to the formation of mountain ranges and marks the end of the Wilson Cycle.
4. Continental Rifting (C): This marks the beginning of the Wilson Cycle, not its termination.
5. Seafloor Spreading (D): This is part of the ocean basin formation stage and does not represent the end of the Wilson Cycle.

Correct Answer is:

The event that represents the termination of the Wilson Cycle is:

(B) Continent–continent collision

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Q.55: The fraction of the incident electromagnetic energy reflected from a material is known as:

(A) acuity
(B) albedo
(C) spectral hue
(D) artifact

Detailed Answer:

To determine which term describes the fraction of incident electromagnetic energy reflected from a material, let’s define each option:

1. Acuity (A):
2. Acuity refers to the sharpness or clarity of vision or hearing, not the reflection of electromagnetic energy.
3. Albedo (B):
4. Albedo is the term used to describe the fraction of incident electromagnetic energy, such as sunlight, that is reflected by a surface. It is a measure of reflectivity and is commonly used in the context of planetary science and meteorology.
5. Spectral Hue (C):
6. Spectral hue refers to the color or wavelength of light that is perceived, not the fraction of reflected energy.
7. Artifact (D):
8. Artifact refers to any unintended alteration or feature introduced during the collection or processing of data, not the reflection of energy.

Correct Answer is:

The correct term for the fraction of incident electromagnetic energy reflected from a material is:

(B) albedo

Albedo quantifies how much light or other electromagnetic radiation is reflected by a surface compared to the amount incident upon it.

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Q.56: Which of the following statements regarding ore deposits is/are CORRECT?

(A) Both replacement and exhalative ores are possible in SEDEX type deposits
(B) Rampura–Agucha Pb–Zn deposit is a Mississippi Valley Type deposit
(C) Orogenic gold deposit is an epigenetic type deposit
(D) Fluid boiling in the early stage of magmatic crystallization is responsible for Cu–(Mo) deposits

Detailed Answer:

1. Statement (A): Both replacement and exhalative ores are possible in SEDEX type deposits
2. SEDEX (Sedimentary Exhalative) deposits are typically formed by the exhalation of metal-rich fluids into sedimentary environments. These deposits are primarily known for exhalative ores. However, SEDEX deposits can also exhibit characteristics of replacement ores where metals replace existing minerals or rock. Thus, replacement ores can occur within the context of SEDEX deposits, making this statement correct.
3. Statement (B): Rampura–Agucha Pb–Zn deposit is a Mississippi Valley Type deposit
4. The Rampura–Agucha deposit in India is classified as a SEDEX-type deposit, not a Mississippi Valley Type (MVT) deposit. MVT deposits are typically associated with carbonate rocks and are characterized by lead-zinc mineralization. Therefore, this statement is incorrect.
5. Statement (C): Orogenic gold deposit is an epigenetic type deposit
6. Orogenic gold deposits are indeed epigenetic, meaning that gold mineralization occurs after the formation of the host rocks, often associated with tectonic activity during mountain-building events (orogeny). Thus, this statement is correct.
7. Statement (D): Fluid boiling in the early stage of magmatic crystallization is responsible for Cu–(Mo) deposits
8. Copper-Molybdenum (Cu–Mo) deposits can indeed be associated with magmatic processes where fluid boiling, fluid interaction, and differentiation occur. While these processes can start in the early stages of magmatic crystallization, it is also important to note that they can continue through various stages. Thus, this statement is correct.

Correct Answer is:

The correct statements regarding ore deposits are:

(A) Both replacement and exhalative ores are possible in SEDEX type deposits
(C) Orogenic gold deposit is an epigenetic type deposit
(D) Fluid boiling in the early stage of magmatic crystallization is responsible for Cu–(Mo) deposits

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Q.57: Which of the following sedimentary structures is/are found in intertidal deposits?

(A) Ladder-back ripple
(B) Rain print
(C) Double mud drape
(D) Mud-crack

Detailed Answer:

1. Ladder-back Ripple:
2. Ladder-back ripples are distinctive for their ladder-like appearance and are often formed in environments with moderate to high energy, such as intertidal zones. They occur due to the interplay of wave and current actions. Thus, ladder-back ripples can indeed be found in intertidal deposits.
3. Rain Print:
4. Rain prints are impressions made on the sediment surface by raindrops. They are characteristic of wet environments and can be preserved in intertidal deposits where the sediment surface is exposed to rain.
5. Double Mud Drape:
6. Double mud drapes are formed when alternating layers of sediment accumulate due to varying energy conditions, typically in tidal environments. They are indicative of periodic changes in sediment deposition. Although they are associated with tidal environments, they are less commonly recognized in typical intertidal deposits compared to other structures.
7. Mud-Crack:
8. Mud-cracks form when wet mud or clay dries out and contracts, leading to a network of cracks. They are commonly found in environments where sediment undergoes regular wetting and drying cycles, including intertidal zones.

Correct Answer is:

The sedimentary structures found in intertidal deposits include:

(A) Ladder-back ripple
(B) Rain print
(D) Mud-crack

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Q.58: Which of the following materials is/are used for estimation of hydrocarbon source rock maturation based on color?

(A) Conodont
(B) Illite
(C) Spore
(D) Zircon

Detailed Answer:

1. Conodont:
2. Conodonts are extinct, microscopic, tooth-like fossils used in geology for biostratigraphy and as indicators of thermal maturity. The color of conodont elements, specifically their color alteration index (CAI), is used to estimate the degree of metamorphism and hydrocarbon source rock maturation. Therefore, conodonts are relevant for estimating hydrocarbon source rock maturation based on color.
3. Illite:
4. Illite is a common clay mineral whose color and structural changes can indicate various stages of diagenesis and metamorphism. In the context of hydrocarbon source rock maturation, illite is used to determine the illite crystallinity index, which reflects thermal maturity. While it involves color change, it is more about crystallinity rather than color alone.
5. Spore:
6. Spores, particularly their color and morphology, are used in palynology to assess the thermal maturity of source rocks. The color and preservation of spores can indicate various stages of maturation and can be used to estimate hydrocarbon source rock maturation. Thus, spores are relevant for this purpose.
7. Zircon:
8. Zircon is a robust mineral commonly used for dating and tracing geological events but is not typically used for estimating hydrocarbon source rock maturation based on color. It is more relevant for radiometric dating and does not provide information on maturation through color changes.

Correct Answer is:

The materials used for the estimation of hydrocarbon source rock maturation based on color are:

(A) Conodont
(C) Spore

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Q.59: Which of the following schist belts occur(s) to the east of the Closepet Granite in southern India?

(A) Shimoga
(B) Kolar
(C) Bababudan
(D) Hutti

Detailed Answer:

The Closepet Granite is a prominent geological feature in southern India. When considering the schist belts located to the east of this granite, we need to identify the schist belts that lie in that region relative to the Closepet Granite.

1. Shimoga Schist Belt:
2. The Shimoga Schist Belt is situated to the west of the Closepet Granite. Therefore, it does not meet the criteria of occurring to the east of the Closepet Granite.
3. Kolar Schist Belt:
4. The Kolar Schist Belt is located to the east of the Closepet Granite. It is one of the major schist belts in southern India, and its location fits the criteria specified in the question.
5. Bababudan Schist Belt:
6. The Bababudan Schist Belt is located to the west of the Closepet Granite. Hence, it does not occur to the east of the Closepet Granite.
7. Hutti Schist Belt:
8. The Hutti Schist Belt is situated to the east of the Closepet Granite. This belt is known for its gold deposits and is correctly positioned according to the criteria given.

Correct Answer is:

The schist belts that occur to the east of the Closepet Granite in southern India are:

(B) Kolar
(D) Hutti

Q.61: Which of the following statements is/are CORRECT for the M-plane of any fault?

(A) M-plane pole of a fault is located on the fault plane
(B) M-plane pole of a fault is perpendicular to the slickenline on the fault plane
(C) M-plane pole of a fault is parallel to the slickenline on the fault plane
(D) M-plane pole of a fault is perpendicular to the pole to the fault plane

Detailed Answer:

1. Statement (A): M-plane pole of a fault is located on the fault plane.
2. This statement is incorrect. The M-plane pole is not located on the fault plane but is perpendicular to it.
3. Statement (B): M-plane pole of a fault is perpendicular to the slickenline on the fault plane.
4. This statement is correct. The M-plane is the plane of maximum shear, and its pole is perpendicular to the slickenline (the line indicating slip direction) on the fault plane.
5. Statement (C): M-plane pole of a fault is parallel to the slickenline on the fault plane.
6. This statement is incorrect. The M-plane pole is perpendicular to, not parallel to, the slickenline.
7. Statement (D): M-plane pole of a fault is perpendicular to the pole to the fault plane.
8. This statement is correct. The M-plane pole is perpendicular to the pole to the fault plane, as it represents the plane of maximum shear stress.

Correct Answer is:

The correct statements regarding the M-plane of any fault are:

(A) M-plane pole of a fault is located on the fault plane
(B) M-plane pole of a fault is perpendicular to the slickenline on the fault plane
(D) M-plane pole of a fault is perpendicular to the pole to the fault plane

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Q.62: Which of the following microfossils is/are foraminifera?

(A) Miliammina
(B) Triceratium
(C) Cibicides
(D) Guembelitria

Detailed Answer:

1. Miliammina:
2. Miliammina is a genus of foraminifera. Foraminifera are single-celled protists with a shell, and Miliammina falls under this category.
3. Triceratium:
4. Triceratium is a genus of diatoms, not foraminifera. Diatoms are different from foraminifera in terms of their cell structure and shell composition.
5. Cibicides:
6. Cibicides is a genus of foraminifera. It is a well-known group of microfossils used in biostratigraphy and paleoceanography.
7. Guembelitria:
8. Guembelitria is a genus of foraminifera. Like Cibicides, it is used in geological studies to understand past environmental conditions.

Correct Answer is:

The microfossils that are foraminifera are:

(A) Miliammina
(C) Cibicides
(D) Guembelitria

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Q.63: The in situ stress at a point in a dry sandstone terrain is as follows: \( \sigma_1 = 12 \text{ MPa} \) and \( \sigma_3 = 4 \text{ MPa} \). The pore water pressure (\( p_w \)) increases due to the construction of a reservoir. The failure criterion of the sandstone is given by \( \sigma_1′ = 3.48 \text{ MPa} + 3 \sigma_3′ \), where \( \sigma_1′ \) and \( \sigma_3′ \) are the effective maximum and minimum principal stresses, respectively. Assuming failure occurs at peak stress, the minimum value of \( p_w \) (in MPa) that will cause the sandstone to fail in situ is __________ (rounded off to two decimal places). **Solution:** 1. **Determine the Effective Stresses:** The effective stress is given by: \[ \sigma’ = \sigma – p_w \] For the given stress conditions: – Maximum principal stress \( \sigma_1 = 12 \text{ MPa} \) – Minimum principal stress \( \sigma_3 = 4 \text{ MPa} \) The effective stresses are: \[ \sigma_1′ = \sigma_1 – p_w \] \[ \sigma_3′ = \sigma_3 – p_w \] 2. **Apply the Failure Criterion:** The failure criterion is: \[ \sigma_1′ = 3.48 \text{ MPa} + 3 \sigma_3′ \] Substitute the effective stresses \( \sigma_1′ \) and \( \sigma_3′ \) into the equation: \[ \sigma_1 – p_w = 3.48 \text{ MPa} + 3 (\sigma_3 – p_w) \] Rearranging the terms: \[ \sigma_1 – p_w = 3.48 \text{ MPa} + 3 \sigma_3 – 3 p_w \] Simplify: \[ \sigma_1 – 3 \sigma_3 = 3.48 \text{ MPa} – 2 p_w \] Solving for \( p_w \): \[ p_w = \frac{\sigma_1 – 3 \sigma_3 – 3.48 \text{ MPa}}{-2} \] 3. **Substitute the Given Values:** – \( \sigma_1 = 12 \text{ MPa} \) – \( \sigma_3 = 4 \text{ MPa} \) Substitute these into the equation: \[ p_w = \frac{12 \text{ MPa} – 3 \times 4 \text{ MPa} – 3.48 \text{ MPa}}{-2} \] Simplifying: \[ p_w = \frac{12 \text{ MPa} – 12 \text{ MPa} – 3.48 \text{ MPa}}{-2} \] \[ p_w = \frac{-3.48 \text{ MPa}}{-2} \] \[ p_w = 1.74 \text{ MPa} \] **Conclusion:** The minimum value of pore water pressure (\( p_w \)) required to cause the sandstone to fail in situ is \( 1.74 \text{ MPa} \).

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Q.64: If the Rb-Sr isochron formed by a suite of gabbro samples has a slope of 0.0265, then the calculated age of the gabbro in million years is _________ (in integer). [Use \( \lambda_{\text{Rb}87} = 1.42 \times 10^{-11} \text{ year}^{-1} \)] **Solution:** 1. **Understanding the Rb-Sr Isochron Equation:** The slope of the Rb-Sr isochron is related to the age of the rock sample through the equation: \[ \text{Slope} = \lambda_{\text{Rb}87} \times \text{Age} \] where: – \( \text{Slope} \) is the slope of the isochron (0.0265) – \( \lambda_{\text{Rb}87} \) is the decay constant for \( ^{87}\text{Rb} \) (1.42 × \( 10^{-11} \) year\( ^{-1} \)) – \( \text{Age} \) is the age of the sample in years 2. **Rearranging the Isochron Equation to Solve for Age:** Rearrange the equation to solve for \( \text{Age} \): \[ \text{Age} = \frac{\text{Slope}}{\lambda_{\text{Rb}87}} \] 3. **Substitute the Given Values:** Given: – \( \text{Slope} = 0.0265 \) – \( \lambda_{\text{Rb}87} = 1.42 \times 10^{-11} \text{ year}^{-1} \) Substitute these values into the equation: \[ \text{Age} = \frac{0.0265}{1.42 \times 10^{-11}} \] 4. **Perform the Calculation:** \[ \text{Age} = \frac{0.0265}{1.42 \times 10^{-11}} = 1.86 \times 10^{9} \text{ years} \] \[ \text{Age} = 1860 \text{ million years} \] **Conclusion:** The calculated age of the gabbro is **1860 million years**.

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Q.65: A soil mass comprises two horizontal layers (of equal thickness and equal width) stacked one above the other. The hydraulic conductivities of the two layers are \( 5 \times 10^{-2} \) cm/s and \( 3 \times 10^{-2} \) cm/s. Considering Darcian flow of water and the same hydraulic gradient for both layers, the effective hydraulic conductivity of the soil mass in cm/s is ___________ (rounded off to two decimal places). **Solution:** 1. **Understanding the Effective Hydraulic Conductivity for Layers in Series:** When two layers are stacked one above the other, the effective hydraulic conductivity \( K_{\text{eff}} \) for vertical flow can be found using the arithmetic mean of the hydraulic conductivities since the flow is through equal thickness layers. The formula is: \[ K_{\text{eff}} = \frac{K_1 + K_2}{2} \] where: – \( K_1 \) is the hydraulic conductivity of the first layer (\( 5 \times 10^{-2} \) cm/s) – \( K_2 \) is the hydraulic conductivity of the second layer (\( 3 \times 10^{-2} \) cm/s) 2. **Substitute the Given Values:** \[ K_{\text{eff}} = \frac{5 \times 10^{-2} + 3 \times 10^{-2}}{2} \] \[ K_{\text{eff}} = \frac{0.05 + 0.03}{2} \] 3. **Calculate the Effective Hydraulic Conductivity:** \[ K_{\text{eff}} = \frac{0.08}{2} = 0.04 \text{ cm/s} \] **Conclusion:** The effective hydraulic conductivity of the soil mass is **0.04 cm/s**.
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