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Section A: Long Answer Questions

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5 questions
1long8 marks

Define a mineral and explain the principal physical properties used to identify rock-forming minerals in hand specimen. Using these properties, distinguish between quartz and calcite, and explain why the Mohs scale is only an ordinal (relative) scale of hardness.

Definition of a mineral

A mineral is a naturally occurring, inorganic, homogeneous solid with a definite (but generally variable within limits) chemical composition and an ordered internal atomic (crystalline) structure.

Principal physical properties for identification

  1. Colour – overall hue; unreliable alone because of impurities (e.g., quartz can be clear, milky, rose, smoky).
  2. Streak – colour of the fine powder on an unglazed porcelain plate; more diagnostic than colour (e.g., hematite is grey-black but streaks reddish-brown).
  3. Lustre – appearance of reflected light: metallic vs non-metallic (vitreous/glassy, pearly, resinous, earthy, silky, adamantine).
  4. Hardness – resistance to scratching, graded on Mohs scale (1 Talc – 10 Diamond).
  5. Cleavage and fracture – cleavage is the tendency to break along planes of weak bonding (described by number of directions and the angles between them); fracture (conchoidal, uneven, hackly) is breakage where no cleavage exists.
  6. Crystal form/habit – external geometric shape (e.g., cubic, prismatic, tabular).
  7. Specific gravity (relative density) – heft compared with water.
  8. Special properties – reaction with dilute HCl (effervescence), magnetism, double refraction, taste, feel, fluorescence.

Distinguishing quartz from calcite

PropertyQuartz (SiO2_2)Calcite (CaCO3_3)
Hardness7 (scratches glass)3 (scratched by a knife)
CleavageNone – conchoidal fracturePerfect rhombohedral, 3 directions
Reaction with dilute HClNoneVigorous effervescence (CO2_2)
Double refractionWeak/absentStrong (clear calcite shows it clearly)
LustreVitreousVitreous to pearly

The decisive field tests: calcite is soft (H = 3), has three perfect cleavages, and fizzes with dilute HCl; quartz is hard (H = 7), has no cleavage (conchoidal fracture), and does not react with acid.

Why Mohs scale is only ordinal

The Mohs scale ranks minerals by which scratches which: a higher number scratches every lower one. It tells us order but not magnitude. The absolute (e.g., indentation/Vickers) hardness differences between successive steps are unequal — for instance, the jump from corundum (9) to diamond (10) is far larger than from talc (1) to gypsum (2). Hence Mohs hardness can be used to compare but not to quantify how much harder one mineral is than another; it is a relative, non-linear ranking.

mineralsphysical-propertiesrock-forming-minerals
2long8 marks

Describe the geometry of a fold with a neat sketch, defining axial plane, fold axis, limbs, hinge and interlimb angle. Classify folds on the basis of the dip of the axial plane and interlimb angle.

A bed dips 3535^{\circ} towards S60ES60^{\circ}E. Compute the apparent dip measured along a vertical cliff face that trends due East–West (090090^{\circ}).

Fold geometry and elements

A fold is a wave-like bend produced in originally planar rock layers by compressive (and sometimes shear) stress.

        Axial plane
           |
   limb \  |  / limb
         \ | /
          \|/  <- hinge line / fold axis
   ---------+---------  (crest of antiform)
  • Limbs: the two sides (flanks) of a fold sharing a common hinge.
  • Hinge line / fold axis: the line of maximum curvature where the limbs meet; the axis is the line that, moved parallel to itself, generates the folded surface.
  • Axial plane (axial surface): the plane (or surface) that contains the hinge lines of successive folded layers and bisects the interlimb angle.
  • Interlimb angle: the angle measured between the two limbs.

Classification by dip of axial plane

TypeAxial planeLimbs
Symmetrical (upright)VerticalDip equally in opposite directions
AsymmetricalInclinedLimbs dip at unequal angles
OverturnedSteeply inclinedBoth limbs dip same direction
RecumbentHorizontalLimbs sub-horizontal
IsoclinalParallel to limbsLimbs parallel

Classification by interlimb angle

Interlimb angleClass
180180^{\circ}120120^{\circ}Gentle
120120^{\circ}7070^{\circ}Open
7070^{\circ}3030^{\circ}Close
3030^{\circ}00^{\circ}Tight
00^{\circ}Isoclinal

Apparent dip calculation

The relationship between true dip δ\delta, apparent dip α\alpha, and the angle β\beta between the strike direction and the line of section is:

tanα=tanδsinβ\tan\alpha = \tan\delta \cdot \sin\beta

Step 1 — Find the strike. The dip direction is S60ES60^{\circ}E, i.e., azimuth 18060=120180^{\circ}-60^{\circ}=120^{\circ}. Strike is perpendicular: 12090=030120^{\circ}-90^{\circ}=030^{\circ} (NNE–SSW).

Step 2 — Find β\beta, the angle between the strike (030030^{\circ}) and the section line (090090^{\circ}, E–W):

β=090030=60.\beta = 090^{\circ}-030^{\circ}=60^{\circ}.

Step 3 — Apply the formula with δ=35\delta = 35^{\circ}:

tanα=tan35sin60=0.7002×0.8660=0.6064.\tan\alpha = \tan 35^{\circ}\cdot\sin 60^{\circ} = 0.7002 \times 0.8660 = 0.6064. α=arctan(0.6064)=31.2.\alpha = \arctan(0.6064)= 31.2^{\circ}.

Apparent dip 31.2\approx 31.2^{\circ} along the E–W cliff face. (As expected, the apparent dip is less than the true dip of 3535^{\circ}.)

foldsstructural-geologydip-strike
3long8 marks

Define a fault and explain its principal elements (fault plane, hanging wall, footwall, slip, throw, heave, hade). Classify faults according to the relative movement of the blocks with sketches.

A normal fault has a fault plane dipping at 6060^{\circ}. A marker bed displaced across the fault shows a net dip-slip of 12 m along the fault plane. Calculate the vertical throw and the horizontal heave.

Definition

A fault is a fracture or zone of fractures in rock along which there has been appreciable relative displacement of the two sides parallel to the fracture surface.

Elements of a fault

  • Fault plane: the surface along which displacement occurs.
  • Hanging wall: the block resting above the inclined fault plane.
  • Footwall: the block lying below the fault plane.
  • Net slip: total relative displacement measured on the fault plane between two formerly adjacent points.
  • Dip-slip: component of slip along the dip of the fault plane.
  • Throw: the vertical component of the dip separation.
  • Heave: the horizontal component of the dip separation.
  • Hade: the angle between the fault plane and the vertical (hade=90dip\text{hade} = 90^{\circ}-\text{dip}).

Classification by relative movement

NORMAL              REVERSE/THRUST        STRIKE-SLIP
HW moves down       HW moves up           horizontal, along strike
  \                   /                    --> <--
   \  HW            HW /                    (no vertical sep.)
FW  \               / FW
(tensional)        (compressional)        (shear couple)
  • Normal fault: hanging wall moves down relative to footwall (extension/tension).
  • Reverse fault: hanging wall moves up (compression); a low-angle reverse fault (dip < 45°) is a thrust.
  • Strike-slip (transcurrent) fault: predominantly horizontal movement along strike; dextral or sinistral.
  • Oblique-slip fault: combination of dip-slip and strike-slip.

Engineering significance (brief): faults create zones of crushed/weak rock (gouge), control groundwater flow, may be seismically active, and are a major hazard for dams, tunnels and foundations.

Calculation

Given: dip of fault plane θ=60\theta = 60^{\circ}, dip-slip s=12 ms = 12\ \text{m} (movement is along the fault plane in the dip direction).

The dip-slip resolves into a vertical (throw) and horizontal (heave) component:

Throw=ssinθ=12×sin60=12×0.8660=10.39 m.\text{Throw} = s\sin\theta = 12\times\sin 60^{\circ} = 12\times 0.8660 = 10.39\ \text{m}. Heave=scosθ=12×cos60=12×0.5000=6.00 m.\text{Heave} = s\cos\theta = 12\times\cos 60^{\circ} = 12\times 0.5000 = 6.00\ \text{m}.

Check: 10.392+6.002=107.9+36.0=143.9=12.0 m\sqrt{10.39^2+6.00^2}=\sqrt{107.9+36.0}=\sqrt{143.9}=12.0\ \text{m}

Vertical throw 10.39 m\approx 10.39\ \text{m}; horizontal heave =6.00 m= 6.00\ \text{m}.

faultsfault-throwengineering-significance
4long8 marks

Explain the theory of plate tectonics, the types of plate boundaries, and the driving mechanisms. Relate this framework to the collision of the Indian and Eurasian plates and the formation of the Himalaya, and state two engineering implications for civil works in Nepal.

Plate tectonics – the theory

The lithosphere (rigid crust + uppermost mantle, ~100 km thick) is broken into a number of rigid plates that move over the weak, ductile asthenosphere. Plate tectonics unifies continental drift (Wegener) and sea-floor spreading (Hess) and explains the global distribution of earthquakes, volcanoes, mountain belts and ocean basins.

Types of plate boundary

  1. Divergent (constructive) – plates move apart; new lithosphere forms by upwelling magma. Features: mid-ocean ridges, rift valleys (e.g., Mid-Atlantic Ridge, East African Rift).
  2. Convergent (destructive) – plates move together. Three sub-types:
    • Oceanic–continental: subduction, volcanic arc (e.g., Andes).
    • Oceanic–oceanic: subduction, island arc (e.g., Japan).
    • Continental–continental: collision, mountain building (e.g., Himalaya).
  3. Transform (conservative) – plates slide past each other; lithosphere neither created nor destroyed (e.g., San Andreas Fault).

Driving mechanisms

  • Mantle convection currents – thermal convection in the asthenosphere drags plates.
  • Ridge push – gravitational sliding of plates away from elevated ridges.
  • Slab pull – the dominant force; dense subducting slab sinks and pulls the trailing plate.

Indian–Eurasian collision and the Himalaya

The Indian plate broke from Gondwana (~140 Ma) and drifted north, consuming the Tethys Ocean by subduction beneath Eurasia. Around 50–55 Ma the two continental plates collided. Because continental crust is buoyant, neither could fully subduct; instead the crust crumpled, thickened and was thrust southward along major faults, raising the Himalaya. Continued northward convergence (~4–5 cm/yr) keeps the range rising and seismically active.

The collision produced Nepal's principal E–W tectonic zones and bounding thrusts:

  • Main Central Thrust (MCT) – separates Higher Himalaya from Lesser Himalaya.
  • Main Boundary Thrust (MBT) – separates Lesser Himalaya from the Siwaliks (Sub-Himalaya).
  • Main Frontal Thrust (MFT) – the active southern front against the Gangetic plain.

Engineering implications in Nepal

  1. High seismic hazard: ongoing convergence stores strain released as large earthquakes (e.g., 1934, 2015 Gorkha) — structures, dams and tunnels must be designed for strong ground motion.
  2. Weak, fractured, fault-bounded rock and steep slopes: thrusting and rapid uplift produce sheared rock masses, landslides and slope-instability hazards that govern road, hydropower and tunnel design.
plate-tectonicshimalayageology-of-nepal
5long8 marks

Differentiate between physical (mechanical) and chemical weathering, listing the main processes of each. Explain how climate controls the dominant style of weathering, and discuss the engineering significance of weathering for foundations and slopes.

A fresh granite has a unit weight of 26.5 kN/m326.5\ \text{kN/m}^3. After chemical weathering the rock loses 18%18\% of its solid mineral mass and the void volume increases so that porosity rises from 1%1\% to 22%22\%. Estimate the dry unit weight of the weathered granite, assuming the specific gravity of the remaining solids is unchanged.

Physical vs chemical weathering

Physical (mechanical)Chemical
NatureDisintegration into smaller fragments; composition unchangedDecomposition; minerals altered to new chemical compounds
Main processesFrost wedging (freeze–thaw), thermal expansion/insolation, exfoliation (unloading), salt crystallisation, abrasion, biological prising by rootsHydrolysis (feldspar → clay), oxidation, hydration, carbonation (dissolving CaCO3_3), solution, chelation by organic acids
ProductAngular rock fragments, screeClay minerals, soluble salts, residual soils, iron oxides

Climatic control

  • Cold/high-altitude and arid climates → physical weathering dominates (frost shattering in mountains; thermal/salt action in deserts) because little liquid water and low temperatures slow chemical reactions.
  • Hot, humid (tropical) climates → chemical weathering dominates and is intense; abundant warm water and vegetation accelerate hydrolysis and oxidation, producing thick residual (laterite) soils.

Engineering significance

  • Foundations: weathering reduces rock strength and bearing capacity; a deeply weathered, variable rock profile (corestones in a soft matrix) makes settlement uneven; chemical alteration to expansive clays causes heave; carbonation/solution creates cavities (karst) and differential support.
  • Slopes: weathering deepens the regolith, weakens discontinuities and reduces shear strength, promoting landslides; weathered seams act as planes of failure. Weathering grade (I fresh – VI residual soil) must be logged for excavation, cut-slope and tunnel design.

Calculation

Take a representative volume of fresh rock V0=1 m3V_0 = 1\ \text{m}^3.

Fresh rock: porosity n0=1%=0.01n_0 = 1\% = 0.01, so solids occupy 0.99 m30.99\ \text{m}^3. Dry unit weight =26.5 kN/m3= 26.5\ \text{kN/m}^3 (assume fresh rock essentially dry/solid-dominated), so weight of solids in 1 m31\ \text{m}^3:

Ws,0=26.5 kN.W_{s,0}=26.5\ \text{kN}.

Unit weight of the solid mineral substance:

γs=Ws,0Vs,0=26.50.99=26.77 kN/m3.\gamma_s = \frac{W_{s,0}}{V_{s,0}} = \frac{26.5}{0.99}=26.77\ \text{kN/m}^3.

Weathered rock: 18% of solid mass is removed (dissolved/leached), so remaining solid weight:

Ws,1=26.5×(10.18)=26.5×0.82=21.73 kN.W_{s,1}=26.5\times(1-0.18)=26.5\times0.82=21.73\ \text{kN}.

The specific gravity of solids is unchanged, so the remaining solids occupy:

Vs,1=Ws,1γs=21.7326.77=0.8117 m3.V_{s,1}=\frac{W_{s,1}}{\gamma_s}=\frac{21.73}{26.77}=0.8117\ \text{m}^3.

New porosity n1=22%=0.22n_1 = 22\% = 0.22, meaning solids form (10.22)=0.78(1-0.22)=0.78 of the total volume. Total weathered volume:

V1=Vs,10.78=0.81170.78=1.0407 m3.V_1=\frac{V_{s,1}}{0.78}=\frac{0.8117}{0.78}=1.0407\ \text{m}^3.

Dry unit weight of weathered granite:

γd,1=Ws,1V1=21.731.0407=20.88 kN/m3.\gamma_{d,1}=\frac{W_{s,1}}{V_1}=\frac{21.73}{1.0407}=20.88\ \text{kN/m}^3.

Dry unit weight of the weathered granite 20.9 kN/m3\approx 20.9\ \text{kN/m}^3 (a drop of about 21% from the fresh value of 26.5 kN/m326.5\ \text{kN/m}^3, reflecting both mass loss and increased voids).

weatheringsoil-formationengineering-geology
B

Section B: Short Answer Questions

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6 questions
6short8 marks

Classify igneous rocks on the basis of mode of occurrence and silica content. Explain how cooling rate controls texture, and give one named example for each of intrusive vs extrusive and acidic vs basic rocks.

Igneous rocks form by the solidification of magma/lava.

Classification by mode of occurrence

  • Intrusive (plutonic) – solidify within the crust; slow cooling → coarse, fully crystalline (phaneritic) texture. e.g., granite, gabbro. Bodies: batholith, stock, dyke, sill, laccolith.
  • Hypabyssal – solidify at shallow depth; medium-grained, sometimes porphyritic. e.g., dolerite (in dykes/sills).
  • Extrusive (volcanic) – solidify on the surface as lava; rapid cooling → fine-grained (aphanitic) or glassy texture. e.g., basalt, rhyolite, obsidian (glassy), pumice (vesicular).

Classification by silica (SiO2_2) content

ClassSiO2_2Colour/densityExample
Acidic (felsic)> 66%Light, low densityGranite / Rhyolite
Intermediate52–66%Diorite / Andesite
Basic (mafic)45–52%Dark, denseGabbro / Basalt
Ultrabasic< 45%Very darkPeridotite

Cooling rate and texture

  • Slow cooling (deep) → atoms have time to organise → large, well-formed crystals (coarse-grained, phaneritic), e.g., granite.
  • Fast cooling (surface) → little time for growth → very small crystals (fine-grained, aphanitic), e.g., basalt.
  • Very rapid (quenching)no crystals, natural glass, e.g., obsidian.
  • Two-stage cooling → large crystals (phenocrysts) in a fine groundmass = porphyritic texture.

Examples: Intrusive = granite; Extrusive = basalt; Acidic = granite/rhyolite; Basic = gabbro/basalt.

igneous-rocksrock-classificationtexture
7short8 marks

Define geomorphology and describe the fluvial (river) cycle of erosion through its youth, mature and old stages, naming the characteristic landforms of each stage. Mention how the river gradient and load change downstream.

Geomorphology is the scientific study of landforms — their form, origin, and the processes (weathering, erosion, transport, deposition, tectonics) that shape the Earth's surface.

Fluvial (Davisian) cycle of erosion

W. M. Davis's "geographical cycle" describes how a landscape evolves through stages after uplift, as rivers wear it down toward base level (a peneplain).

1. Youth stage (upper course)

  • Steep gradient, fast turbulent flow, dominant vertical (downward) erosion.
  • Landforms: V-shaped valleys, gorges/canyons, rapids, waterfalls, plunge pools, potholes, interlocking spurs.
  • Load: coarse boulders/gravel; valley deeper than wide.

2. Mature stage (middle course)

  • Reduced gradient; lateral erosion becomes important alongside transport.
  • Landforms: meanders, wider valley floor, river cliffs and slip-off slopes, beginnings of a flood plain, river terraces, alluvial fans where tributaries enter.
  • Load: mixed sand and gravel; valley becomes wider, graded profile developing.

3. Old stage (lower course)

  • Very gentle gradient; deposition dominates.
  • Landforms: broad flood plains, large meanders, ox-bow lakes, natural levees, braided channels, deltas at the mouth.
  • Load: fine silt/clay carried in suspension; deposited as alluvium.

Downstream changes

  • Gradient: steep in youth → progressively gentler downstream (concave long profile, the "graded profile").
  • Load: particle size decreases downstream (coarse → fine by attrition and sorting), while total quantity/discharge of load and water generally increases.

The end product of the complete cycle is a low, almost-flat peneplain close to base level, with isolated residual hills (monadnocks).

geomorphologyfluvial-landformsriver-erosion
8short8 marks

Describe the major physiographic/tectonic zones of Nepal from south to north, giving for each its bounding thrust, dominant rock types and approximate age. Why is the Siwalik (Sub-Himalaya) zone particularly problematic for highway and infrastructure construction?

Tectonic / physiographic zones of Nepal (south → north)

ZoneBounding fault (north side)Dominant rocksApprox. age
Terai (Gangetic) PlainMFT (north margin)Recent alluvium (gravel, sand, silt)Quaternary (Recent)
Siwalik (Sub-Himalaya)MBTMudstone, siltstone, sandstone, conglomerate (molasse)Mio-Pliocene
Lesser HimalayaMCTSlate, phyllite, schist, quartzite, limestone/dolomite, some granite/gneissPrecambrian – Eocene
Higher (Greater) HimalayaSTDS (South Tibetan Detachment)High-grade gneiss, schist, migmatite, leucogranitePrecambrian (crystalline core)
Tibetan-Tethys (Tethyan) HimalayaFossiliferous marine sediments: limestone, shale, sandstoneCambrian – Eocene

Bounding thrusts from south to north: MFT → MBT → MCT → STDS.

Why the Siwalik zone is problematic for construction

  1. Weak, young, poorly consolidated rocks – soft mudstones/siltstones and loosely cemented sandstone–conglomerate that weather and erode rapidly.
  2. High erodibility and gullying – heavy monsoon rain causes severe sheet/gully erosion and debris flows.
  3. Frequent landslides and slope failures – steeply dipping beds, fault-disturbed and fractured strata, and clay-rich seams give low shear strength.
  4. Active tectonics – it lies between the active MFT and MBT, so it is uplifting, fractured and seismically active.
  5. Flash floods and unstable river channels along the dun valleys and outwash fans.

These conditions make Siwalik road cuts, bridges and foundations prone to instability, demanding careful slope protection, drainage and bioengineering.

geology-of-nepaltectonic-zonesstratigraphy
9short8 marks

Define a joint and distinguish it from a fault. Classify joints by origin and by geometry (relationship to bedding). Explain why joints and other discontinuities are the most important factor controlling the engineering behaviour of a rock mass, and list the discontinuity parameters that should be recorded in the field.

Joint – a fracture or crack in rock along which no appreciable displacement has taken place (or only opening normal to the surface). It contrasts with a fault, where there is measurable shear displacement parallel to the fracture.

FeatureJointFault
DisplacementNegligible / noneAppreciable, parallel to plane
OccurrenceIn sets, very commonLess common, often single major plane/zone
Associated rockUsually intact wallsCrushed gouge, breccia, slickensides

Classification by origin

  • Tectonic joints – from tectonic stress (shear joints and tension/extension joints).
  • Non-tectonic / contraction jointscooling joints (columnar joints in basalt), drying/mud cracks in sediments.
  • Unloading (sheeting/exfoliation) joints – release of overburden pressure, parallel to surface.

Classification by geometry (vs bedding/structure)

  • Strike joints – parallel to the strike of beds.
  • Dip joints – parallel to the dip direction.
  • Oblique (diagonal) joints – at an angle to strike and dip.
  • Bedding joints – parallel to bedding planes.
  • Master joints (large, persistent) vs minor joints; a parallel group is a joint set, intersecting sets form a joint system.

Why discontinuities control rock-mass behaviour

Intact rock is usually strong; it is the discontinuities (joints, bedding, faults, foliation) that govern the mass because they:

  • provide planes of weakness along which sliding, toppling and wedge failures occur in slopes and tunnels;
  • drastically lower the shear strength and deformability of the mass compared with intact rock;
  • act as the main pathways for groundwater, raising water pressure and reducing effective stress;
  • divide the mass into blocks, controlling rock-fall, blasting and excavatability.

Discontinuity parameters to record (after ISRM)

  1. Orientation (dip and dip direction)
  2. Spacing
  3. Persistence (continuity/length)
  4. Roughness
  5. Aperture (opening width)
  6. Infilling (gouge/material)
  7. Wall strength
  8. Seepage / water condition
  9. Number of sets
  10. Block size / shape
jointsdiscontinuitiesrock-mass
10short4 marks

Explain the rock cycle with a labelled flow diagram, showing how the three major rock groups (igneous, sedimentary, metamorphic) are interrelated and the processes that convert one to another.

Rock cycle

The rock cycle is the continuous set of geological processes by which rock material is created, destroyed and recycled among the three major rock groups over geological time, driven by the Earth's internal heat (tectonics, melting) and external agents (weathering, erosion).

              melting
   MAGMA  <---------------------  (any rock)
     |                                ^
     | crystallisation                | melting (heat)
     v                                |
  IGNEOUS ROCK                   METAMORPHIC ROCK
     |                                ^
     | weathering & erosion           | heat & pressure
     v                                | (metamorphism)
  SEDIMENT  --(transport, deposition,
     |          compaction & cementation = lithification)-->  SEDIMENTARY ROCK
     |                                ^        |
     +--------------------------------+        |
              (further weathering / uplift)----+

Key processes

  1. Crystallisation / solidification – cooling magma or lava → igneous rock.
  2. Weathering & erosion – any rock breaks down into sediment.
  3. Transport, deposition, compaction & cementation (lithification) – sediment → sedimentary rock.
  4. Heat & pressure (metamorphism) – any pre-existing rock is recrystallised in the solid state → metamorphic rock.
  5. Melting – any rock taken deep enough re-melts into magma, closing the cycle.

Uplift and exposure can return buried igneous or metamorphic rock to the surface to be weathered again, so the cycle has many possible paths, not a single fixed loop.

sedimentary-rocksrock-cyclemetamorphism
11short4 marks

State and explain any four fundamental principles of stratigraphy / relative dating (e.g., superposition, original horizontality, lateral continuity, cross-cutting relationships, faunal succession). Use a short example to show how cross-cutting relationships establish the relative age of an igneous dyke.

Principles of relative dating (stratigraphy)

  1. Law of Superposition – In an undisturbed sequence of sedimentary (or volcanic) layers, each bed is younger than the one below it and older than the one above. The oldest is at the bottom.

  2. Principle of Original Horizontality – Sediments are deposited in nearly horizontal layers under gravity. Therefore, tilted or folded strata indicate later tectonic deformation after deposition.

  3. Principle of Lateral Continuity – A layer of sediment originally extends laterally in all directions until it thins out or reaches a barrier. Matching beds on opposite sides of a valley were once continuous.

  4. Principle of Cross-Cutting Relationships – A geological feature (fault, dyke, intrusion) that cuts across another rock body is younger than the rock it cuts.

  5. Principle of Faunal Succession – Fossil organisms succeed one another in a definite, recognisable order; strata can be correlated and relatively dated by their fossil content (index fossils).

  6. (Also) Principle of Inclusions – Fragments (inclusions/xenoliths) contained within a rock are older than the rock that encloses them.

Example – cross-cutting dyke

  ===== Bed C (sandstone, top) =====
  ----- Bed B (shale) --------||----
  ::::: Bed A (limestone) ::::||::::
                              ||  <- vertical igneous DYKE

The dyke slices through beds A and B but is truncated by (does not enter) bed C. By cross-cutting, the dyke is younger than beds A and B (which it cuts). Because bed C lies across the top of the dyke undisturbed, bed C was deposited after the dyke was emplaced and eroded.

Relative order, oldest → youngest: A → B → dyke → C.

geological-timestratigraphyrelative-dating

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