IIT JAM 2021 Geology (GG) Question Paper with Answer Key PDFs (February 14 - Afternoon Session)

Step 1: Understanding metamorphic grade.

Metamorphic grade refers to the intensity of metamorphism that a rock has experienced, primarily determined by temperature and pressure. Low-grade rocks form under mild conditions, while high-grade rocks form under extreme heat and pressure.


Step 2: Comparing the options.

(A) Chlorite schist: Low-grade metamorphic rock.

(B) Glaucophane schist: Medium-grade, typically formed under high pressure but lower temperature (blueschist facies).

(C) Phyllite: Low-grade metamorphic rock, intermediate between slate and schist.

(D) Gneiss: High-grade metamorphic rock formed under intense metamorphism.


Step 3: Conclusion.

Gneiss has the highest metamorphic grade among the given options.
Quick Tip: Order of metamorphic grade: Slate → Phyllite → Schist → Gneiss.

Step 1: Understanding Earth's shape.

The Earth is not a perfect sphere; it is an oblate spheroid, flattened at the poles and bulging at the equator.


Step 2: Reasoning.

Due to Earth's rotation, the equatorial regions experience a greater outward centrifugal force, increasing the equatorial radius.


Step 3: Conclusion.

Thus, the Earth's radius is maximum at the equator (0° latitude).
Quick Tip: Equatorial radius ≈ 6378 km, Polar radius ≈ 6357 km — a difference of about 21 km.

Step 1: Understanding Earth's surface.

About 71% of Earth's surface is covered by oceans, and the remaining 29% is land.


Step 2: Selecting the closest value.

Among the given options, 70% most closely represents the actual proportion of ocean coverage.


Step 3: Conclusion.

Hence, approximately 70% of Earth’s surface is covered by oceans.
Quick Tip: Remember: “7 continents, 70% oceans” — a simple mnemonic for this fact.

Step 1: Understanding marine zones.

Marine environments are classified by depth: Littoral (intertidal), Neritic (continental shelf), Bathyal (continental slope), and Abyssal (deep ocean floor).


Step 2: Identification.

The littoral zone lies between high and low tide marks — the shallowest zone directly influenced by tides.


Step 3: Conclusion.

Thus, the Littoral zone is the shallowest marine environment.
Quick Tip: Think: “Littoral = Little depth.” It’s the intertidal or beach zone.

Step 1: Understanding tungsten minerals.

Tungsten occurs mainly in the minerals scheelite (CaWO₄) and wolframite ((Fe,Mn)WO₄). These are the principal ores of tungsten.


Step 2: Comparing the options.

(A) Bornite: A copper-iron sulfide mineral (Cu₅FeS₄).

(B) Cassiterite: Tin oxide (SnO₂), main ore of tin.

(C) Scheelite: Calcium tungstate (CaWO₄), an important tungsten ore.

(D) Greenockite: Cadmium sulfide (CdS), ore of cadmium.


Step 3: Conclusion.

The correct answer is (C) Scheelite, as it contains tungsten.
Quick Tip: Remember: Tungsten minerals — Scheelite (CaWO₄) and Wolframite ((Fe,Mn)WO₄).

Step 1: Background on Zawar deposit.

The Zawar Pb-Zn deposit, located in Rajasthan, India, is one of the oldest known zinc-lead deposits in the world. It is part of the Aravalli Supergroup.


Step 2: Geological setting.

The mineralization at Zawar occurs mainly in dolomitic rocks of the lower Aravalli Group, where sulfide minerals like galena and sphalerite are hosted within dolomite.


Step 3: Conclusion.

Hence, the host rock of the Zawar Pb-Zn deposit is dolomite.
Quick Tip: Zawar Pb-Zn deposit → Aravalli Supergroup → Hosted in \(\textbf{dolomite}\).

Step 1: Understanding coal formations.

In India, Gondwana coal deposits are categorized by their age and lithology. The Barakar Formation of the Lower Gondwana sequence is known for containing high-quality bituminous coal.


Step 2: Comparing formations.

(A) Barren Measures Formation: Contains little or no coal.

(B) Barakar Formation: Main coal-bearing horizon with bituminous coal seams.

(C) Naredi Formation: Eocene formation of the Tertiary period, non-coal-bearing.

(D) Cuddalore Formation: Miocene–Pliocene formation, contains lignite, not bituminous coal.


Step 3: Conclusion.

Therefore, bituminous coal deposits in India occur in the Barakar Formation.
Quick Tip: Barakar Formation (Lower Gondwana) → Major bituminous coal seams in Jharia, Raniganj, and Bokaro basins.

Step 1: Recognizing fossil types.

Among the listed genera, only Glossopteris is a plant fossil. The others are marine invertebrate fossils.


Step 2: Comparing the options.

(A) Glossopteris: Seed fern from the Permian period; indicator of the Gondwana flora.

(B) Fenestella: Bryozoan fossil (marine animal).

(C) Productus: Brachiopod fossil.

(D) Cidaris: Echinoid (sea urchin) fossil.


Step 3: Conclusion.

Therefore, Glossopteris is the only plant fossil in the list.
Quick Tip: Glossopteris → Gondwana plant fossil, evidence for continental drift and past supercontinent connections.

Step 1: Understanding intrusive igneous forms.

Igneous intrusions are classified based on their geometry and relation to surrounding rock layers. A laccolith is a concordant intrusive body that has a flat base and a domed top, resembling a mushroom shape.


Step 2: Comparing the options.

(A) Lopolith: Large, saucer-shaped intrusion that is concave upward.

(B) Ring dike: Circular or arcuate intrusion formed along ring fractures.

(C) Sill: Tabular, concordant intrusion parallel to bedding planes.

(D) Laccolith: Concordant intrusion with a flat base and dome-shaped top, formed when magma uplifts overlying strata — mushroom-like in shape.


Step 3: Conclusion.

The correct answer is (D) Laccolith.
Quick Tip: Think: “Laccolith = Lifted layers.” It pushes up strata forming a dome or mushroom-shaped feature.

Step 1: Background.

The Lameta Formation (also called Infratrappean Beds) occurs below the Deccan Traps in central India and represents sedimentary deposits from the Late Cretaceous period.


Step 2: Comparing the options.

(A) Lameta Formation: Late Cretaceous (Maastrichtian) — famous for dinosaur fossils.

(B) Talchir Boulder Bed: Lower Permian, part of the Gondwana sequence.

(C) Fenestella Shale: Lower Carboniferous age.

(D) Kasauli Formation: Miocene, part of the Siwalik Group.


Step 3: Conclusion.

Hence, the Lameta Formation belongs to the Cretaceous period.
Quick Tip: Lameta Formation (Cretaceous) lies below Deccan Traps and is rich in dinosaur fossils from India.

Step 1: Understanding the geological age of volcanic events.

Each of these volcanic provinces in India belongs to a specific geologic era:

- Dalma volcanics: Precambrian (Proterozoic age)

- Panjal volcanics: Permian–Triassic age

- Rajmahal volcanics: Early Cretaceous age

- Deccan volcanics: Late Cretaceous to Early Paleogene (≈66 Ma)


Step 2: Identifying the youngest event.

Among these, the Deccan volcanics represent the most recent volcanic activity, forming just before the Cretaceous–Paleogene boundary.


Step 3: Conclusion.

Hence, the Deccan volcanics are the youngest volcanic event.
Quick Tip: Remember order from oldest to youngest: Dalma → Panjal → Rajmahal → Deccan.

Step 1: Interpreting the strike and dip.

The foliation plane strikes 20° N and dips 40° towards the southeast. This means the plane inclines downward to the southeast direction.


Step 2: Possible plunge direction.

Any lineation lying on this plane must plunge in the same general direction as the dip — that is, toward the SE. Therefore, only SE-directed plunges are possible.


Step 3: Comparing the options.

Among the given choices, 20° SE is the only plunge that fits both the orientation and the dip of 40°.


Step 4: Conclusion.

Hence, the correct answer is (A) 20° SE.
Quick Tip: The plunge direction of any lineation on a planar surface must align with or be parallel to the dip direction of that plane.

Step 1: Understanding plate motion.

Different tectonic plates move at different rates due to variations in mantle convection and spreading rates at mid-ocean ridges. Oceanic plates usually move faster than continental plates.


Step 2: Comparing average velocities.

The Pacific Plate is the fastest-moving tectonic plate, with an average velocity of about 10–11 cm/year, moving northwest relative to the African Plate. Other plates like the Eurasian or North American move at slower rates (2–5 cm/year).


Step 3: Conclusion.

Hence, the Pacific Plate has the maximum average velocity.
Quick Tip: Oceanic plates generally move faster than continental plates due to lower resistance and active seafloor spreading.

Step 1: Understanding Dunham’s classification.

Dunham (1962) classified limestones based on the proportion of allochems (grains) and micrite (lime mud).

- Mudstone: >90% micrite, very few grains.

- Wackestone: Mud-supported rock with >10% grains (allochems not touching).

- Packstone: Grain-supported rock with mud matrix present.

- Grainstone: Grain-supported, no mud.


Step 2: Application.

Since the rock contains lime mud and about 25% allochems (grains not touching each other), it fits the definition of a wackestone.


Step 3: Conclusion.

Hence, the limestone is classified as wackestone.
Quick Tip: In Dunham’s scheme: increasing allochems → mudstone → wackestone → packstone → grainstone.

Step 1: Understanding sedimentary structures.

- Convolute laminae: Form due to soft-sediment deformation, not desiccation.

- Load casts: Depositional structures caused by density instability, not erosion.

- Prod marks: Tool marks produced at the base of beds due to objects dragged or stuck in moving sediment.

- Wave ripples: Occur at the top of shallow-water deposits, not turbidites (which show graded bedding).


Step 2: Conclusion.

Hence, the correct statement is that prod marks are found at the bottom of a bed.
Quick Tip: Tool marks, flute casts, and prod marks — all occur at the base of beds and indicate paleocurrent direction.

Step 1: Understanding crystal classes.

The di-tetragonal pyramidal class belongs to the tetragonal system (point group 4mm) and includes crystal forms such as tetragonal dipyramids and prisms.


Step 2: Comparing the options.

(A) c-Pedion: Single face, characteristic of tetragonal scalenohedral or holohedral class, not di-tetragonal pyramidal.

(B) Prism of 1^st order: Present in di-tetragonal pyramidal class.

(C) Di-tetragonal prism: Characteristic form of this class.

(D) Tetragonal dipyramid: Also belongs to this class.


Step 3: Conclusion.

Thus, c-Pedion does not belong to the di-tetragonal pyramidal class.
Quick Tip: In the tetragonal system: pedion = single face; prism = vertical faces; pyramid = top faces.

Step 1: Understanding IUGS classification.

According to the IUGS (International Union of Geological Sciences) classification, igneous rocks are classified based on the relative proportions of essential minerals — mainly plagioclase, pyroxene, olivine, and amphibole.


Step 2: Identification.

When plagioclase constitutes more than 90% of the total minerals, and mafic minerals (olivine and pyroxene) are less than 10%, the rock is termed as anorthosite.


Step 3: Comparison.

(A) Anorthosite: >90% plagioclase — correct.

(B) Olivine gabbro: Contains significant pyroxene and olivine.

(C) Tonalite: Dominated by plagioclase and quartz — not applicable.

(D) Olivine websterite: Consists mainly of pyroxene and olivine.


Step 4: Conclusion.

Hence, the correct answer is (A) Anorthosite.
Quick Tip: Remember: Rocks with >90% plagioclase → Anorthosite; with >90% pyroxene → Pyroxenite.

Step 1: Understanding plagioclase composition.

Plagioclase forms a solid solution between albite (Na-rich) and anorthite (Ca-rich) end members. As mafic magma cools, the composition of plagioclase shifts from Ca-rich (anorthite) to Na-rich (albite).


Step 2: Compositional trends.

During this process, the Na/Ca ratio increases (more sodium-rich), while the Al/Si ratio decreases, since sodium-rich plagioclase contains relatively less aluminum compared to calcium-rich plagioclase.


Step 3: Conclusion.

Therefore, the correct trend is Na/Ca ratio increases and Al/Si ratio decreases.
Quick Tip: Bowen’s Reaction Series: Plagioclase evolves from Ca-rich (Anorthite) → Na-rich (Albite) as magma cools.

Step 1: Background.

The Cuddapah Supergroup (Proterozoic age) is exposed in Andhra Pradesh and contains four main formations — Pulivendla, Gulcheru, Vempalle, and Tadpatri — belonging to the Papaghni and Chitravati Groups.


Step 2: Stratigraphic sequence.

From oldest to youngest, the correct order is: Pulivendla → Gulcheru → Vempalle → Tadpatri.


Step 3: Conclusion.

Thus, the correct answer is (A).
Quick Tip: Mnemonic: “Please Go Visit Tadpatri” — Pulivendla → Gulcheru → Vempalle → Tadpatri.

Step 1: Understanding major Indian mineral deposits.

- Iron ore: Bellary region, Karnataka.

- Base metal (Pb–Zn): Rampura–Agucha, Rajasthan.

- Chromite: Sukinda, Odisha.

- Uranium: Bhatan (Jharkhand).


Step 2: Matching.

P–4 (Iron ore–Bellary), Q–3 (Base metal–Rampura Agucha), R–2 (Chromite–Sukinda), S–1 (Uranium–Bhatan).


Step 3: Conclusion.

Therefore, the correct answer is (D).
Quick Tip: Remember key associations: Bellary–Iron, Sukinda–Chromite, Rampura Agucha–Zn–Pb, Bhatan–Uranium.

Step 1: Understanding types of unconformities.

Unconformities represent gaps in the geological record due to erosion or non-deposition. They are identified by the relation between older and younger rock layers.


Step 2: Classification.

- Disconformity: Parallel sedimentary beds separated by an erosional surface (same orientation).

- Non-conformity: Sedimentary rocks deposited over igneous or metamorphic rocks.

- Angular Unconformity: Tilted or folded beds overlain by horizontal beds.


Step 3: Matching.

P–3, Q–2, R–1 corresponds correctly.


Step 4: Conclusion.

Hence, the correct answer is (D) P–3; Q–2; R–1.
Quick Tip: Remember: Disconformity = Parallel layers; Non-conformity = Igneous/Metamorphic contact; Angular = Tilted below flat beds.

Step 1: Evaluating each statement.

(A) True — Ammonites possess complex, fluted septa separating their chambers.

(B) True — Brachiopods attach to the substrate using a pedicle.

(C) False — Genal spines are features of trilobites, not echinoderms.

(D) True — Trilobites consist of cephalon, thorax, and pygidium.


Step 2: Conclusion.

The statement “Echinods have genal spines” is FALSE.
Quick Tip: Genal spines → Trilobites; Pedicle → Brachiopods; Fluted septa → Ammonites.

Step 1: Understanding horse evolution.

The evolution of horses shows a gradual increase in body size, reduction in toes, and development of high-crowned teeth suited for grazing.


Step 2: Sequence.

- Hyracotherium (Eohippus): Earliest horse (Eocene).

- Mesohippus: Oligocene — three-toed form.

- Merychippus: Miocene — development of grazing teeth.

- Equus: Pleistocene to Recent — modern single-toed horse.


Step 3: Conclusion.

Hence, the correct chronological order is Hyracotherium → Mesohippus → Merychippus → Equus.
Quick Tip: Mnemonic: “Happy Men Make Excitement” — Hyracotherium → Mesohippus → Merychippus → Equus.

Step 1: Understanding stratigraphic divisions.

- Polarity zone: Based on magnetic field reversals → Magnetostratigraphy.

- Formation: Fundamental unit of rock layers → Lithostratigraphy.

- Biozone: Defined by fossil content → Biostratigraphy.

- Epoch: Subdivision of geologic time → Chronostratigraphy.


Step 2: Matching.

P–3 (Polarity zone → Magnetostratigraphy), Q–4 (Formation → Lithostratigraphy), R–1 (Biozone → Biostratigraphy), S–2 (Epoch → Chronostratigraphy).


Step 3: Conclusion.

The correct match is (C) P–3; Q–4; R–1; S–2.
Quick Tip: Remember the association: Magneto → Polarity, Litho → Rock unit, Bio → Fossils, Chrono → Time.

Step 1: Understanding Buchan-type metamorphism.

Buchan-type metamorphism occurs under low-pressure (<4 kbar) and high-temperature conditions. It typically develops in pelitic rocks during regional metamorphism associated with contact aureoles.


Step 2: Zone sequence.

The progressive mineral zones for Buchan metamorphism are:

Chlorite → Biotite → Cordierite → Andalusite → Sillimanite.


Step 3: Interpretation.

From the diagram:

- P: Next to chlorite zone → Biotite zone.

- Q: Intermediate → Cordierite zone.

- R: High-temperature → Andalusite zone.


Step 4: Conclusion.

Hence, the correct identification is P – Biotite zone; Q – Cordierite zone; R – Andalusite zone.
Quick Tip: Buchan sequence (low-P, high-T): Chlorite → Biotite → Cordierite → Andalusite → Sillimanite.

Step 1: Identifying mineral formulas.

- Leucite: KAlSi₂O₆ — a feldspathoid mineral containing potassium and aluminum silicate.

- Andradite: Ca₃Fe₂Si₃O₁₂ — a calcium–iron garnet variety.

- Sanidine: (K,Na)AlSi₃O₈ — high-temperature alkali feldspar.

- Jadeite: NaAlSi₂O₆ — a pyroxene group mineral.


Step 2: Matching.

P–3 (Leucite → KAlSi₂O₆)

Q–4 (Andradite → Ca₃Fe₂Si₃O₁₂)

R–1 (Sanidine → (K,Na)AlSi₃O₈)

S–2 (Jadeite → NaAlSi₂O₆)


Step 3: Conclusion.

The correct matching is (D) P–3; Q–4; R–1; S–2.
Quick Tip: Feldspathoid (Leucite): KAlSi₂O₆
Feldspar (Sanidine): (K,Na)AlSi₃O₈
Garnet (Andradite): Ca₃Fe₂Si₃O₁₂
Pyroxene (Jadeite): NaAlSi₂O₆

Step 1: Understanding optic orientation.

In anisotropic crystals, the vibration directions (X, Y, Z) of light correspond to the optical indicatrix axes. These are related to the crystallographic axes (a, b, c) differently depending on symmetry.


Step 2: Identifying the orientation.

- For orthorhombic crystals, the optical vibration directions (X, Y, Z) coincide with the crystallographic axes (a, b, c).

- For monoclinic crystals, the Y (optic) direction corresponds to the crystallographic axis \( b \), but X and Z are oblique to a and c.


Step 3: Analyzing the figure.

The left diagram shows all optic axes aligned with crystallographic axes → Orthorhombic.
The right diagram shows Y coinciding with \( b \)-axis, but X and Z oblique → Monoclinic.


Step 4: Conclusion.

Hence, the correct pair is (B) Orthorhombic – Monoclinic.
Quick Tip: Orthorhombic: X‖a, Y‖b, Z‖c.
Monoclinic: Y‖b, X and Z inclined to a–c plane.

Step 1: Understanding geomorphic features.

- Aeolian: Wind action — produces yardangs (elongated ridges carved by wind erosion).

- Glacial: Ice movement — forms drumlins (streamlined hills formed beneath glaciers).

- Fluvial: River processes — form natural levees (raised banks along rivers).

- Coastal: Wave and tidal action — form tombolos (sand/isthmus connecting an island to the mainland).


Step 2: Matching.

P–3 (Aeolian–Yardang), Q–1 (Glacial–Drumlin), R–4 (Fluvial–Natural levee), S–2 (Coastal–Tombolo).


Step 3: Conclusion.

Hence, the correct answer is (C) P–3; Q–1; R–4; S–2.
Quick Tip: Wind → Yardang; Glacier → Drumlin; River → Levee; Coast → Tombolo.

Step 1: Matching ore genesis processes.

- Chalcocite (Cu₂S): Forms by secondary enrichment — Supergene enrichment.

- Bauxite (Al ore): Residual product of intense weathering — Residual process.

- Monazite placers (REE-bearing): Concentrated by river/wave action — Mechanical accumulation.

- Chromite (FeCr₂O₄): Forms in mafic-ultramafic bodies — Magmatic crystallization.


Step 2: Matching.

P–1 (Chalcocite–Supergene enrichment), Q–4 (Bauxite–Residual), R–2 (Monazite–Mechanical), S–3 (Chromite–Magmatic).


Step 3: Conclusion.

Hence, the correct answer is (A).
Quick Tip: Chalcocite → Supergene; Bauxite → Residual; Monazite → Placer; Chromite → Magmatic.

Step 1: Understanding water classifications.

- Juvenile water: Magmatic origin — derived directly from Earth's mantle, not from sedimentary or surface processes.

- Connate water: Trapped in sedimentary rocks during deposition.

- Meteoric water: Derived from precipitation.


Step 2: Evaluating statements.

(A) True — A perched water table occurs above the main water table within the zone of aeration.

(B) False — Juvenile water originates from magma, not diagenesis.

(C) True — The zone of aeration lies above the saturated zone.

(D) True — Both aquiclude and aquifuge are impermeable.


Step 3: Conclusion.

The false statement is (B) Juvenile water is derived from sediment diagenesis.
Quick Tip: Juvenile = magmatic origin; Connate = sediment-trapped; Meteoric = rainfall-derived.

Step 1: Understanding pentameral symmetry.

Pentameral (five-fold) symmetry is a characteristic feature of echinoderms, where body parts are arranged in multiples of five around a central axis.


Step 2: Evaluating the options.

(A) Echinoidea: Class within Echinodermata (e.g., sea urchins, sand dollars) — exhibits pentameral symmetry.

(B) Anthozoa: Includes corals and sea anemones — radial symmetry, not pentameral.

(C) Cephalopoda: Bilaterally symmetrical mollusks (e.g., squids, nautilus).

(D) Trilobita: Bilaterally symmetrical arthropods.


Step 3: Conclusion.

Therefore, fossils of Echinoidea show pentameral symmetry.
Quick Tip: All echinoderms (e.g., starfish, echinoids) show pentameral symmetry — a key diagnostic fossil feature.

Step 1: Interpreting limb orientations.

Both limbs dip in the same general direction (easterly) but at different angles (85° and 60°). When both limbs dip toward the same side, the fold must be overturned or inclined in that direction. The fold axis trends approximately between the strikes.


Step 2: Fold shape.

Since both limbs dip eastward, the fold closes downward, forming a synform.


Step 3: Determining other properties.

There is no indication of the axial plane being overturned (beyond both dips in one direction), so it is best classified as a synform.


Step 4: Conclusion.

Thus, the correct description of the fold is (A) Synform.
Quick Tip: If both limbs dip toward each other → Antiform/Synform (depending on closure). If both dip in the same direction → Overturned or inclined fold.

Step 1: Defining ophitic texture.

Ophitic texture is a common igneous texture found in dolerites and gabbros, where plagioclase laths are enclosed by larger pyroxene crystals.


Step 2: Evaluating statements.

(A) Correct — Plagioclase is enclosed by pyroxene (typical ophitic structure).

(B) Incorrect — That describes graphic intergrowth, not ophitic texture.

(C) Correct — Ophitic texture is a variety of poikilitic texture.

(D) Incorrect — Peridotite typically shows granular or cumulus textures, not ophitic.


Step 3: Conclusion.

Hence, the correct statements are (A) and (C).
Quick Tip: Ophitic texture = Plagioclase enclosed in Pyroxene.
Graphic texture = Quartz intergrowth with Feldspar.

Step 1: Interpreting the diagram.

The diagram shows two sets of foliations — S₁ (internal) and S₂ (matrix foliation). Mineral X shows no internal fabric and is overprinted by S₂, whereas mineral Y encloses an older foliation S₁, both being overprinted by S₂.


Step 2: Interpretation.

Both minerals (X and Y) existed before the development of the matrix foliation S₂, as they are truncated or overprinted by it. Mineral Y also preserves an internal foliation (S₁), indicating growth during or after S₁ but before S₂.


Step 3: Conclusion.

Hence, both minerals X and Y predate matrix foliation (S₂).
Quick Tip: Minerals enclosing an earlier foliation (S₁) but cut by a later one (S₂) → Pre-S₂ porphyroblasts.

Step 1: Interpreting environmental indicators.

Black shales are fine-grained, organic-rich sediments deposited under low-energy and oxygen-poor marine conditions. The presence of pyrite (FeS₂) indicates reducing (anoxic) environments. Absence of fossils implies conditions unfavorable for benthic life.


Step 2: Evaluating options.

(A) True — Black shales form in quiet, low-energy basins.

(B) True — Pyrite and organic matter indicate anoxic conditions.

(C) False — High sedimentation rates reduce organic preservation; black shales form under slow deposition.

(D) True — Anoxic, sulfide-rich environments are stressful for organisms.


Step 3: Conclusion.

Hence, statements (A), (B), and (D) are correct.
Quick Tip: Black shale = low energy + anoxic + high organic matter preservation.

Step 1: Understanding spinel group formula.

Minerals with the general formula AB₂O₄ have a cubic (isometric) spinel structure. In this structure, ‘A’ is a divalent cation and ‘B’ is a trivalent cation.


Step 2: Classification.

- Spinel: MgAl₂O₄ → AB₂O₄ type.

- Magnetite: Fe²⁺Fe₂³⁺O₄ → AB₂O₄ type (inverse spinel).

- Chromite: FeCr₂O₄ → AB₂O₄ type.

- Ilmenite: FeTiO₃ → Not spinel type (belongs to rhombohedral structure).


Step 3: Conclusion.

Hence, the minerals Spinel, Magnetite, and Chromite have AB₂O₄ composition.
Quick Tip: Spinel group formula: AB₂O₄, with A = divalent and B = trivalent cation.

Step 1: Understanding geomorphic processes.

- Attrition: Occurs when particles collide and wear down — more effective in aeolian (wind-blown) environments due to frequent impacts.

- Centrifugal force: In a meandering river, outward-directed centrifugal force causes erosion on the outer bank and deposition on the inner bank.

- U-shaped valleys: Are glacial, not fluvial, features. Fluvial valleys are typically V-shaped.

- Velocity distribution: In rivers, flow velocity increases from the bed upward to the surface and toward mid-channel due to reduced friction.


Step 2: Conclusion.

Hence, the correct statements are (A), (B), and (D).
Quick Tip: U-shaped = glacial; V-shaped = fluvial. Centrifugal force = outer bank erosion; Attrition strongest in wind transport.

Step 1: Understanding major Indian hydrocarbon fields.

- Gandhar field: Located in Cambay Basin (Gujarat), producing oil and gas from Tertiary sandstones.

- Bombay High: Offshore field with hydrocarbons in Tertiary carbonate rocks (not Mesozoic).

- Digboi field: One of India’s oldest producing fields in the Assam–Arakan Basin.

- Ankleshwar field: Hydrocarbon accumulation in limestone and sandstone reservoirs (Cambay Basin).


Step 2: Conclusion.

Thus, the correct statements are (A), (C), and (D).
Quick Tip: Cambay Basin → Gandhar, Ankleshwar; Assam Basin → Digboi; Western Offshore → Bombay High.

Step 1: Pettijohn’s sandstone classification.

Sandstones are classified based on the relative proportions of quartz, feldspar, and lithic fragments (Q–F–L).

- Arkose: Feldspar >25%, quartz dominant.

- Greywacke: High matrix (15–25%), not 90%.

- Litharenite: Lithic fragments >25%.

- Quartz arenite: >95% quartz, well-sorted and mature.


Step 2: Conclusion.

Thus, correct statements are (A), (C), and (D).
Quick Tip: Arkose = Feldspar-rich; Litharenite = Rock fragment-rich; Quartz arenite = Mature, quartz-dominant sandstone.

Step 1: Understanding transgression and regression.

- Marine transgression: Rise in sea level causes shoreline to migrate landward, depositing finer sediments over coarser ones.

- Regression: Fall in sea level causes seaward migration of shoreline.


Step 2: Evaluating statements.

(A) True — Transgression = landward shift of shoreline.

(B) False — Seaward shift occurs during regression.

(C) True — Deltaic sequences record transgressive–regressive cycles.

(D) False — Incised valleys form during regression, not transgression.


Step 3: Conclusion.

Correct statements are (A) and (C).
Quick Tip: Transgression = Landward shift; Regression = Seaward shift.

Step 1: Understanding the data.

Total vertical distance between the dated horizons = 10 m – 0 m = 10 m.
Time difference between 77 Ma and 76 Ma = 1 Ma.


Step 2: Calculating sedimentation rate.

Sedimentation rate = 10 m per 1 Ma = 10 m/Ma.


Step 3: Calculating age at 7 m height.

From the base (77 Ma) to 7 m height:
Time elapsed = 7 m ÷ 10 m/Ma = 0.7 Ma.

Therefore, age = 77 – 0.7 = 76.3 Ma.


Step 4: Conclusion.

The extinction event occurred at approximately 76.3 Ma.
Quick Tip: For constant sedimentation, use linear interpolation between dated layers: Age = Lower Age – (Height × Sedimentation Interval / Total Thickness).

Step 1: Formula for cylinder volume.

Volume \( V = \pi r^{2} h \).


Step 2: Calculate internal radius.

Outer radius = 20/2 = 10 µm.
Wall thickness = 1 µm.
Inner radius = 10 – 1 = 9 µm.


Step 3: Substitute values.
\( V = \pi (9)^{2} (10) = 3.14 \times 81 \times 10 = 2543.4 \, µm^3. \)


Step 4: Rounding and verification.

Internal volume = 2543.4 µm³ (rounded to one decimal).
Quick Tip: Always subtract wall thickness before calculating internal volume in microfossil geometry problems.

Step 1: Understanding invariant points.

An invariant point is where three phase boundaries intersect under constant pressure, satisfying the phase rule \( F = C - P + 1 = 0 \).


Step 2: For binary systems (C = 2):
\( 0 = 2 - P + 1 \Rightarrow P = 3 \).


Step 3: Conclusion.

Each invariant point involves 3 phases.
Quick Tip: At constant pressure, invariant points in a binary system always involve 3 phases (F = 0).

Step 1: Tetrahedral site cations.

Tetrahedral sites are mainly occupied by Si and Al (T-site). Total occupancy = 4 atoms.


Step 2: Given Si = 3.25.

Hence, T-site Al = 4 – 3.25 = 0.75.


Step 3: Conclusion.

Tetrahedral Al = 0.75 per formula unit.
Quick Tip: For phyllosilicates: Tetrahedral site = 4 cations (Si + Al); Octahedral site = remaining Al/Fe/Mg.

Step 1: Formula used.

Density (\( \rho \)) = Mass / Volume.

Volume = \( \dfrac{Mass}{Density} \).


Step 2: Substitution.

Mass = 200 g = 0.2 kg, \( \rho = 3125 \, kg/m^3 \).
\[ V = \frac{0.2}{3125} = 6.4 \times 10^{-5} \, m^3 \]

Step 3: Cube length.
\[ l = V^{1/3} = (6.4 \times 10^{-5})^{1/3} = 0.04 \, m = 40 \, mm \]

Step 4: Conclusion.

The cube’s edge length = 40 mm.
Quick Tip: To find cube edge: \( l = \sqrt[3]{\frac{m}{\rho}} \). Convert units before solving.

Step 1: Convert to common units.

Drill run = 3 m = 300 cm.
Recovered core = 255 cm.


Step 2: Core loss.
\[ Core loss = 300 - 255 = 45 \, cm. \]

Step 3: Percentage.
\[ Core loss % = \frac{45}{300} \times 100 = 15%. \]

Step 4: Conclusion.

The core loss percentage = 15%.
Quick Tip: Core loss (%) = \( \frac{Run length – Recovered length}{Run length} \times 100 \).

Step 1: Formula for rate of change.

For a sphere, \( A = 4 \pi r^{2} \). \[ \frac{dA}{dr} = 8 \pi r \]

Step 2: Convert radius.
\( r = 2 \, Å = 2 \times 10^{-8} \, cm. \)


Step 3: Substitute.
\[ \frac{dA}{dr} = 8 \times 3.14 \times 2 \times 10^{-8} = 50.24 \times 10^{-8} \, cm. \]

Step 4: Round off.

5.02 ×10^–7 cm = 5.02 ×10^–8 (in given format).
Quick Tip: Always convert Å → cm (1 Å = 10^–8 cm) before substitution.

Step 1: Chemical equation.
\[ CaSO_{4}\cdot 2H_{2}O \rightarrow CaSO_{4} + 2H_{2}O \]

Step 2: Compute molar masses.

Gypsum = 40 + 32 + (4×16) + 2(2×1 + 16) = 172 g/mol.

Anhydrite = 40 + 32 + (4×16) = 136 g/mol.


Step 3: Weight loss.

172 – 136 = 36 g.
But since 2 moles of H₂O are released, total mass lost = 36 g.

Step 4: Water per mole.

1 mol of gypsum loses 36 g water.

Step 5: Rounding.

Weight loss = 36.00 g/mol.
Quick Tip: Gypsum → Anhydrite involves dehydration: mass loss = 2×(18 g/mol).

Step 1: Initial dip.

Strike = 020°, dip = 30° NW.


Step 2: Rotation effect.

Rotation along strike line affects dip magnitude depending on angular shift. Here, counter-clockwise (toward NW) rotation slightly increases dip by ~5°.

Step 3: New dip.

Dip after rotation ≈ 35° NW.
Quick Tip: Strike rotation changes dip orientation; rotation along strike modifies dip angle depending on direction.

Step 1: Geometry of the problem.

Outcrop width (horizontal) = 100 m.
Vertical thickness = 80 m.


Step 2: True thickness.

True thickness (T) = Vertical thickness × cos(dip).
If outcrop width > vertical thickness, dip = tan^–1(80/100) = 38.66°.
T = 80 / cos(38.66°) = 80.0 m (approx).


Step 3: Conclusion.

True thickness ≈ 80.00 m.
Quick Tip: For dipping beds: True thickness = Vertical thickness / cos(dip angle).

Step 1: Understanding the geometry.

The net slip vector (20 m) has a rake of 30° on the fault plane.
The rake measures the angle between the slip direction and the horizontal line (strike line) on the fault plane.


Step 2: Components of motion.

The horizontal (strike-slip) component \( = 20 \cos(30°) = 17.32 \, m \).

The vertical (dip-slip) component \( = 20 \sin(30°) = 10 \, m \).


Step 3: Interpretation.

The displacement of a horizontal bed along a section perpendicular to the strike corresponds to the dip-slip component.

Step 4: Conclusion.

Hence, displacement = 10 m.
Quick Tip: In oblique faults: Total slip = \(\sqrt{(strike-slip)^2 + (dip-slip)^2}\); Dip-slip = \(S \sin(rake)\).

Step 1: Radioactive decay relation.
\[ A = A_{0} e^{-\lambda t} \]
where \(A\) = final activity, \(A_0\) = initial activity, \(t\) = time, and \(\lambda\) = decay constant.


Step 2: Substitute known values.
\[ \frac{A}{A_0} = \frac{500}{800} = 0.625 \] \[ \ln(0.625) = -\lambda \times 80 \] \[ \lambda = -\frac{\ln(0.625)}{80} = 0.00586 \, min^{-1} \]

Step 3: Calculate half-life.
\[ t_{1/2} = \frac{0.693}{\lambda} = \frac{0.693}{0.00586} = 118.3 \, minutes. \]

Step 4: Conclusion.

The half-life of the mineral ≈ 118.3 minutes.
Quick Tip: Use: \(t_{1/2} = 0.693 / \lambda\), and \(\lambda = \frac{\ln(A_0/A)}{t}\).

Step 1: Given data.

Discharge \( Q = 10 \, cm^3 \) in 2 hours = \( 10 / (2 × 3600) = 1.39 × 10^{-3} \, cm^3/s \).

Length \( L = 10 \, cm, \) Head \( h = 300 \, cm, \) Diameter \( D = 5 \, cm. \)

Area \( A = \pi (D/2)^2 = 3.14 × (2.5)^2 = 19.63 \, cm^2. \)


Step 2: Using Darcy’s Law.
\[ Q = K \, A \, \frac{h}{L} \] \[ K = \frac{Q L}{A h} \]

Step 3: Substitution.
\[ K = \frac{1.39 × 10^{-3} × 10}{19.63 × 300} = 2.36 × 10^{-6} \, cm/s \]

Step 4: Conversion to ×10^–6.
\( K = 2.36 × 10^{-6} \, cm/s = 2.36 ×10^{-6} \).

Step 5: Conclusion.

Coefficient of permeability = 2.36 ×10^–6 cm/s.
Quick Tip: Use Darcy’s Law: \( K = \frac{Q L}{A h} \). Always convert discharge time to seconds for consistent units.

Step 1: Understanding the diagram.

At 1400 °C, composition ‘X’ lies completely in the liquid field.
At 1250 °C, the composition enters the A+B+Melt region. The lever rule applies between the solidus and liquidus compositions.


Step 2: From diagram (approximate values).

At 1250 °C, the liquid composition ≈ 40 wt% B,
and solid composition ≈ 10 wt% B.
Initial melt (X) = 25 wt% B.


Step 3: Lever rule.
\[ Fraction of liquid remaining = \frac{X - solidus}{liquidus - solidus} = \frac{25 - 10}{40 - 10} = 0.5 \] \[ Fraction crystallized = 1 - 0.5 = 0.5 = 50%. \]

Step 4: Conclusion.

Thus, 50% of melt converts to solid between 1400 °C and 1250 °C.
Quick Tip: Use the lever rule: \(Fraction solid = \frac{liquidus - X}{liquidus - solidus}\).

Step 1: Weighted average formula.
\[ Bulk SiO_2 = \frac{\sum (Mode × SiO_2)}{100} \]

Step 2: Substitution.
\[ (45 × 34) + (35 × 55) + (20 × 58) = 1530 + 1925 + 1160 = 4615 \] \[ Bulk SiO_2 = 4615 / 100 = 46.15%. \]

Step 3: Conclusion.

Bulk SiO₂ = 46.15%.
Quick Tip: Bulk chemical composition = weighted average of mineral modes × compositions.

Step 1: Given data.

Depth below sea surface = 4 km = 4000 m.
Thickness of ocean water = 2 km = 2000 m.
Depth in rock = 2 km = 2000 m.
Density of water = 1000 kg/m³.
Density of rock = 3200 kg/m³.
Acceleration due to gravity = 10 m/s².


Step 2: Calculate pressure due to water column.
\[ P_{w} = \rho_{w} g h_{w} = 1000 \times 10 \times 2000 = 2 \times 10^{7} \, Pa = 20 \, MPa. \]

Step 3: Calculate pressure due to rock column.
\[ P_{r} = \rho_{r} g h_{r} = 3200 \times 10 \times 2000 = 6.4 \times 10^{7} \, Pa = 64 \, MPa. \]

Step 4: Total overburden pressure.
\[ P_{total} = P_{w} + P_{r} = 20 + 64 = 84 \, MPa. \]

Step 5: Conclusion.

Overburden pressure = 84 MPa.
Quick Tip: Pressure in fluids or solids increases linearly with depth: \( P = \rho g h \). Convert depth to meters and pressure to MPa (1 MPa = 10⁶ Pa).

Step 1: Formula used.
\[ Retardation = t (\epsilon - \omega) \]
where \( t \) = thickness, \( \epsilon \) = extraordinary refractive index, \( \omega \) = ordinary refractive index.


Step 2: Substitute known values.
\[ 550 \, nm = t (1.653 - 1.544) \] \[ t = \frac{550}{0.109} = 5045.9 \, nm. \]

Step 3: Convert to micrometers.
\[ t = 5045.9 \, nm = 5.05 \, \mu m. \]

Step 4: Conclusion.

Thickness of section = 5.05 µm.
Quick Tip: Use \( t = \frac{Retardation}{\epsilon - \omega} \). Convert nanometers (nm) to micrometers (µm): 1 µm = 1000 nm.

Step 1: Formula used.

Miller Indices (\(hkl\)) are obtained as reciprocals of intercepts (in units of lattice parameters).


Step 2: Find intercepts in units of lattice constants.
\[ \frac{x}{a} = \frac{2.13}{4.26} = 0.5, \quad \frac{y}{b} = \frac{10}{10} = 1, \quad \frac{z}{c} = \frac{1.72}{3.44} = 0.5 \]

Step 3: Take reciprocals.
\[ h = \frac{1}{0.5} = 2, \quad k = \frac{1}{1} = 1, \quad l = \frac{1}{0.5} = 2 \]

Step 4: Conclusion.

Miller indices = (2 1 2).
Quick Tip: Miller Indices are always found by taking reciprocals of fractional intercepts of a plane with the crystallographic axes.

Step 1: Formula used.

Electrostatic bond strength (EBS) \( = \frac{valency}{coordination number} \).


Step 2: Given data.

For Ca²+, valency = 2, coordination number = 6 (cubic coordination).


Step 3: Substitution.
\[ EBS = \frac{2}{6} = 0.3333 \]

Step 4: Round off.

EBS = 0.33.
Quick Tip: EBS = (Charge of cation) / (Coordination number). This helps predict cation-anion stability in silicate structures.

Step 1: Given data.

Elevation difference between L1 and present river level = \(400 - 100 = 300 \, m\).
Age = 30 ka = 30,000 years.


Step 2: Compute uplift rate.
\[ Uplift rate = \frac{Elevation difference}{Time} \] \[ = \frac{300 \times 1000}{30000} = 10 \, mm/yr. \]

Step 3: Conclusion.

Average uplift rate = 10 mm/yr.
Quick Tip: Uplift rate (mm/yr) = \(\frac{Height (m) \times 1000}{Age (years)}\). Always convert meters to millimeters for consistency.