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Section A: Long Answer Questions

Attempt all questions.

5 questions
1long10 marks

Define a mineral and explain why a substance must satisfy all five criteria of the standard mineral definition. Describe the principal physical properties used to identify minerals in hand specimen (lustre, hardness, cleavage, fracture, colour, streak, specific gravity). Using the Mohs scale, explain how relative hardness is determined, and identify the mineral that would scratch fluorite (H = 4) but be scratched by orthoclase (H = 6). Finally, distinguish the silicate structures of quartz and mica and relate their structures to their cleavage behaviour.

Definition of a mineral (2 marks). A mineral is a naturally occurring, inorganic, homogeneous solid with a definite (but not fixed) chemical composition and an ordered internal atomic (crystalline) structure.

The five criteria, and why each is required:

CriterionWhy required
Naturally occurringExcludes lab-synthesised crystals (e.g. synthetic diamond).
InorganicExcludes organic products (e.g. sugar, coal).
SolidExcludes liquids/gases (native mercury is the conventional exception).
Definite chemical compositionComposition expressible by a formula, e.g. quartz SiO2SiO_2; allows limited substitution (solid solution).
Ordered atomic structureA repeating crystal lattice — distinguishes a mineral from a glass (amorphous solid).

Physical properties for identification (4 marks).

  • Lustre: appearance of reflected light — metallic (galena, pyrite) vs non-metallic (vitreous, pearly, resinous, earthy).
  • Hardness: resistance to scratching, measured on Mohs scale 1–10.
  • Cleavage: tendency to break along planes of weak bonding, giving smooth surfaces; described by number and angle of planes (e.g. mica = 1 perfect; feldspar = 2 at ~90°; calcite = 3 rhombohedral).
  • Fracture: breakage NOT along cleavage planes — conchoidal (quartz), uneven, hackly.
  • Colour: often diagnostic but unreliable for many silicates (e.g. quartz varies widely).
  • Streak: colour of the powdered mineral on an unglazed porcelain plate — more reliable than colour (haematite = red-brown streak even when grey).
  • Specific gravity: ratio of mineral mass to mass of equal volume of water; e.g. galena ≈ 7.5, quartz ≈ 2.65.

Mohs scale and the unknown (2 marks). Mohs scale is an ordinal (relative) scale of ten reference minerals:

1 Talc   2 Gypsum  3 Calcite  4 Fluorite  5 Apatite
6 Orthoclase  7 Quartz  8 Topaz  9 Corundum  10 Diamond

A mineral of higher number scratches all lower ones. Relative hardness is found by testing which reference minerals the unknown scratches and which scratch it.

The unknown scratches fluorite (4) → unknown H > 4, and is scratched by orthoclase (6) → unknown H < 6. Therefore 4<H<64 < H < 6, i.e. H = 5 → apatite.

Quartz vs mica silicate structure (2 marks).

  • Quartz is a framework (tectosilicate): every SiO44SiO_4^{4-} tetrahedron shares all four oxygen corners, giving a strong three-dimensional network with bonds equally strong in all directions. Consequently quartz has no cleavage and breaks by conchoidal fracture.
  • Mica (e.g. muscovite, biotite) is a sheet (phyllosilicate): tetrahedra share three of four oxygens to form continuous 2-D sheets, weakly bonded to adjacent sheets by interlayer cations (K⁺). The weak inter-sheet bonds give mica one perfect basal cleavage, so it splits into thin elastic flakes.

Thus crystal structure directly controls the mechanical breakage behaviour observed in hand specimen.

mineralsphysical-propertiesrock-forming-minerals
2long10 marks

Differentiate between physical (mechanical) and chemical weathering, giving at least three processes under each. Explain the chemical weathering reaction of feldspar to clay and write a balanced reaction. Discuss how climate controls the dominant weathering type, and explain the engineering significance of a deep weathering profile (residual soil) for a foundation engineer in the hills of Nepal. A rock cube of side 0.50 m0.50\ \text{m} is progressively broken by frost action into cubes of side 0.05 m0.05\ \text{m}. Calculate the increase in total exposed surface area and comment on its weathering significance.

Physical vs chemical weathering (3 marks). Physical (mechanical) weathering breaks rock into smaller fragments without changing its chemical composition. Processes:

  1. Frost wedging (freeze–thaw of water in joints).
  2. Thermal expansion/exfoliation (insolation, sheeting due to unloading).
  3. Salt crystallisation and biological/root wedging.

Chemical weathering decomposes minerals by altering their composition. Processes:

  1. Hydrolysis (feldspar → clay).
  2. Oxidation (Fe-minerals → iron oxides/hydroxides).
  3. Carbonation/solution (calcite dissolved by carbonic acid) and hydration.

Feldspar → clay reaction (2 marks). Hydrolysis of potassium feldspar (orthoclase) by carbonic acid:

2KAlSi3O8+2H2CO3+9H2OAl2Si2O5(OH)4+4H4SiO4+2K++2HCO32KAlSi_3O_8 + 2H_2CO_3 + 9H_2O \rightarrow Al_2Si_2O_5(OH)_4 + 4H_4SiO_4 + 2K^+ + 2HCO_3^-

Products: kaolinite (clay), dissolved silica, and potassium + bicarbonate ions carried away in solution.

Climatic control (2 marks).

  • Hot, humid climates → chemical weathering dominates (abundant water + high temperature accelerate reactions → deep lateritic/residual soils).
  • Cold or arid climates → physical weathering dominates (frost action in alpine zones; thermal/insolation in deserts) with little chemical alteration.
  • In Nepal: high Himalayan zone → frost-dominated mechanical weathering; mid-hills with monsoon → intense chemical weathering and deep residual soils.

Engineering significance of deep weathering profile (1 mark). Deep residual soils in the mid-hills mean foundations may rest on weak, variable, partly-weathered material; reduced bearing capacity, high susceptibility to landslides on monsoon-saturated slopes, difficulty locating sound bedrock, and the need for deeper foundations / slope stabilisation.

Surface-area calculation (2 marks). Original cube, side L1=0.50 mL_1 = 0.50\ \text{m}:

A1=6L12=6×(0.50)2=6×0.25=1.5 m2A_1 = 6 L_1^2 = 6 \times (0.50)^2 = 6 \times 0.25 = 1.5\ \text{m}^2

Number of small cubes (side 0.05 m0.05\ \text{m}) from the big cube:

N=(0.500.05)3=103=1000N = \left(\frac{0.50}{0.05}\right)^3 = 10^3 = 1000

Surface area of one small cube:

a=6×(0.05)2=6×0.0025=0.015 m2a = 6 \times (0.05)^2 = 6 \times 0.0025 = 0.015\ \text{m}^2

Total new surface area:

A2=N×a=1000×0.015=15 m2A_2 = N \times a = 1000 \times 0.015 = 15\ \text{m}^2

Increase:

ΔA=A2A1=151.5=13.5 m2 (a 10-fold increase).\Delta A = A_2 - A_1 = 15 - 1.5 = \mathbf{13.5\ m^2}\ \text{(a 10-fold increase).}

Significance: mechanical fragmentation hugely increases the surface area exposed to air and water, dramatically accelerating subsequent chemical weathering. Physical and chemical weathering therefore act synergistically.

weatheringengineering-geologysoil-formation
3long10 marks

Define strike and dip and explain the rule of VV's for outcrop patterns. Classify folds on the basis of (i) the attitude of the axial plane and (ii) the dip of the limbs, with neat sketches described in text. Classify faults according to the relative movement of the hanging wall and footwall. A bed has a true dip of 3535^{\circ}; calculate the apparent dip measured along a section oriented at 4040^{\circ} to the strike direction. State two engineering consequences of unfavourable dip orientation for a road cut.

Strike and dip (2 marks).

  • Strike: the compass direction (bearing) of the horizontal line formed by the intersection of an inclined bed with a horizontal plane.
  • Dip: the angle of maximum inclination of the bed measured from horizontal, in a vertical plane perpendicular to strike (true dip). Strike and the true-dip direction are always at 90°.
  • Rule of V's: where an inclined bed crosses a valley, its outcrop trace forms a V; the V points (closes) in the direction of dip for beds dipping upstream, and the relationship of the V to topographic contours indicates dip direction and approximate amount (a vertical bed shows no V; a bed dipping downstream more steeply than the valley floor V's downstream).

Classification of folds (3 marks). (i) By attitude of the axial plane:

  • Symmetrical (upright): axial plane vertical, limbs dip equally in opposite directions.
  • Asymmetrical: axial plane inclined, limbs dip unequally.
  • Overturned: axial plane strongly inclined, both limbs dip in the same direction (one limb inverted).
  • Recumbent: axial plane horizontal.

(ii) By limb dip / form:

  • Anticline: up-arched, oldest beds in core.
  • Syncline: down-warped, youngest beds in core.
  • Isoclinal: limbs parallel (equal dip same direction).
  • Monocline: local step-like flexure in otherwise flat beds.

Sketch (text): Anticline ∩ (limbs diverge downward), Syncline ∪ (limbs converge downward); axial plane bisects the fold.

Classification of faults (2 marks).

  • Normal (gravity) fault: hanging wall moves DOWN relative to footwall; due to tension/extension.
  • Reverse fault: hanging wall moves UP relative to footwall; due to compression (a low-angle reverse fault, <45°, is a thrust).
  • Strike-slip (transcurrent) fault: horizontal movement parallel to strike (dextral/sinistral).
  • Oblique-slip fault: combination of dip-slip and strike-slip.

Apparent dip calculation (2 marks). Formula relating apparent dip δ\delta, true dip θ\theta, and angle β\beta between the section line and the strike:

tanδ=tanθsinβ\tan\delta = \tan\theta \cdot \sin\beta

Given θ=35\theta = 35^{\circ}, β=40\beta = 40^{\circ}:

tanδ=tan35×sin40=0.7002×0.6428=0.4501\tan\delta = \tan 35^{\circ} \times \sin 40^{\circ} = 0.7002 \times 0.6428 = 0.4501 δ=tan1(0.4501)=24.2\delta = \tan^{-1}(0.4501) = \mathbf{24.2^{\circ}}

The apparent dip (24.2°) is less than the true dip (35°), as expected for any section not perpendicular to strike.

Engineering consequences of unfavourable dip (1 mark).

  1. Daylighting beds dipping out of (towards) the cut face promote planar sliding/rock falls along bedding.
  2. Adverse dip increases excavation support cost and requires flatter cut slopes, rock bolting, or benching for stability.
geological-structuresfoldsfaults
4long8 marks

Explain the theory of plate tectonics and describe the three types of plate boundaries with one example each. Using plate tectonics, explain the origin of the Himalaya and the present-day seismicity of Nepal. If the Indian Plate is converging towards Eurasia at 45 mm/year45\ \text{mm/year} and about 20 mm/year20\ \text{mm/year} is absorbed as crustal shortening along the Main Himalayan Thrust, estimate the slip deficit accumulated over a 250250-year seismic gap and comment on its hazard implication.

Plate tectonics theory (2 marks). The Earth's rigid outer shell (the lithosphere) is broken into about a dozen major plates that move over the weaker, ductile asthenosphere. Plate motion is driven by mantle convection, ridge-push (at spreading centres) and slab-pull (at subduction zones). New lithosphere is created at mid-ocean ridges and consumed at subduction zones, conserving surface area. It unifies sea-floor spreading, continental drift and palaeomagnetic evidence.

Three plate boundaries (3 marks).

BoundaryMotionFeaturesExample
Divergent (constructive)Plates move apartMid-ocean ridges, rift valleys, new crustMid-Atlantic Ridge; East African Rift
Convergent (destructive)Plates collideSubduction zones, trenches, fold mountains, arcsIndia–Eurasia (Himalaya); Andes
Transform (conservative)Plates slide pastCrust neither created nor destroyed; shallow quakesSan Andreas Fault

Origin of the Himalaya & Nepal seismicity (2 marks). The Himalaya is a continent–continent collision orogen. The Indian Plate, after closure of the Tethys Ocean, collided with the Eurasian Plate (~50 Ma). Continental crust is buoyant and resists subduction, so the collision produced enormous crustal shortening, thickening and uplift, forming the Himalaya. The major thrusts — MCT (Main Central Thrust), MBT (Main Boundary Thrust) and MFT (Main Frontal Thrust) — sole into the Main Himalayan Thrust (MHT) at depth. Ongoing convergence locks the MHT, accumulating strain that is periodically released as great earthquakes (e.g. 1934 Bihar–Nepal, 2015 Gorkha), making Nepal one of the most seismically active regions on Earth.

Slip-deficit estimate (1 mark). Shortening rate accommodated on the MHT (the slip that must eventually be released) =20 mm/year= 20\ \text{mm/year}. Over a seismic gap of t=250 yearst = 250\ \text{years}:

D=20 mm/yr×250 yr=5000 mm=5.0 mD = 20\ \text{mm/yr} \times 250\ \text{yr} = 5000\ \text{mm} = \mathbf{5.0\ m}

Hazard implication: a locked segment storing ~5 m of slip deficit could rupture in a single great (M ≈ 8+) earthquake, since coseismic slip of several metres corresponds to such magnitudes. This quantifies the well-documented Himalayan seismic-gap hazard and underlines the need for earthquake-resistant design in Nepal.

plate-tectonicsgeology-of-nepalhimalaya
5long8 marks

Define geomorphology and explain the concept of the geomorphic cycle (cycle of erosion) with its stages of youth, maturity and old age for a fluvial landscape. Describe the major physiographic divisions of Nepal from south to north. A river drops 1500 m1500\ \text{m} over a channel length of 60 km60\ \text{km}; compute its average gradient as a ratio and in m/km, and state what this gradient implies about the stage and erosive power of the river.

Geomorphology (1 mark). Geomorphology is the scientific study of landforms, their origin, evolution, form and the surface processes (weathering, erosion, transport, deposition) and tectonics that shape the Earth's surface.

Geomorphic cycle of erosion (3 marks). Proposed by W. M. Davis: an uplifted landmass is progressively worn down through stages toward a low, flat peneplain.

  • Youth: steep gradients, rapid downcutting (vertical erosion dominant), V-shaped valleys, waterfalls and rapids, few tributaries, narrow valley floors.
  • Maturity: gradient reduced, lateral erosion increases, widest variety of landforms; well-integrated drainage, meanders begin, valley widens, relief at maximum then declines.
  • Old age: very low gradient, broad flat floodplains, large meanders, ox-bow lakes, levees; deposition dominant, landscape reduced toward a peneplain with residual monadnocks.

Physiographic divisions of Nepal (south → north) (3 marks).

  1. Terai (Gangetic Plain): flat alluvial plain, <300 m; recent sediments, fertile.
  2. Siwaliks (Churia Hills): youngest molasse sediments (sandstone, mudstone, conglomerate); fragile, landslide-prone; bounded by MFT (south) and MBT (north).
  3. Lesser Himalaya (Mahabharat Range): low-grade metamorphic and sedimentary rocks; between MBT and MCT.
  4. Higher (Greater) Himalaya: high-grade gneiss, schist, migmatite above the MCT; highest peaks incl. Everest.
  5. Tibetan-Tethys (Inner) Himalaya: fossiliferous Tethyan sedimentary sequence north of the Greater Himalaya (e.g. Thakkhola, Manang).

Gradient calculation (1 mark). Vertical drop h=1500 mh = 1500\ \text{m}; channel length L=60 km=60000 mL = 60\ \text{km} = 60000\ \text{m}. Gradient as a ratio:

S=hL=150060000=140=1:40S = \frac{h}{L} = \frac{1500}{60000} = \frac{1}{40} = \mathbf{1:40}

Gradient in m/km:

S=1500 m60 km=25 m/kmS = \frac{1500\ \text{m}}{60\ \text{km}} = \mathbf{25\ m/km}

Interpretation: 25 m/km (1:40) is a steep gradient, characteristic of a youthful, high-energy mountain river with strong vertical erosion (downcutting), high transport capacity and V-shaped valleys — typical of Himalayan headwater streams in Nepal.

geomorphologyriversgeology-of-nepal
B

Section B: Short Answer Questions

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6 questions
6short6 marks

Explain the rock cycle with a labelled flow described in text, and classify igneous rocks on the basis of mode of occurrence and silica content. Give one example of an extrusive and one of a plutonic igneous rock.

Rock cycle (3 marks). The rock cycle describes the continuous transformation among the three rock families driven by internal (tectonic, magmatic) and external (weathering, erosion) processes.

Text flow:

   Magma --(crystallisation/cooling)--> IGNEOUS ROCK
      ^                                      |
      | (melting)                            | (weathering, erosion, transport)
      |                                      v
  METAMORPHIC ROCK <--(heat & pressure)-- SEDIMENT
      ^                                      |
      |                                      | (deposition, compaction, cementation)
      +--(heat & pressure)-- SEDIMENTARY ROCK
  • Magma cools → igneous rock.
  • Any rock weathers/erodes → sediment → compaction & cementation → sedimentary rock.
  • Any rock subjected to heat & pressure → metamorphic rock.
  • Deep burial/heat → melting → back to magma. Any rock can be converted to any other.

Classification of igneous rocks (2 marks). By mode of occurrence:

  • Extrusive (volcanic): cooled at surface, fine-grained/glassy (rapid cooling) — e.g. basalt, rhyolite.
  • Intrusive (hypabyssal/shallow): medium-grained, in dykes/sills — e.g. dolerite.
  • Plutonic (deep intrusive): slow cooling, coarse-grained — e.g. granite.

By silica (SiO₂) content:

  • Acidic (>66% SiO₂, light) — granite, rhyolite.
  • Intermediate (52–66%) — diorite, andesite.
  • Basic (45–52%, dark) — gabbro, basalt.
  • Ultrabasic (<45%) — peridotite, dunite.

Examples (1 mark): Extrusive = basalt; Plutonic = granite.

rocksrock-cycleigneous-rocks
7short6 marks

Define a joint and distinguish joints from faults. Describe the engineering importance of joints in rock masses. A rock mass contains three joint sets with average spacings of 0.40 m0.40\ \text{m}, 0.50 m0.50\ \text{m} and 1.0 m1.0\ \text{m}. Compute the volumetric joint count JvJ_v and comment qualitatively on the resulting block size.

Definition and joint vs fault (2 marks). A joint is a fracture (crack) in rock along which there has been no appreciable displacement parallel to the fracture surface; joints usually occur in parallel sets.

FeatureJointFault
DisplacementNegligible / noneAppreciable relative movement
OccurrenceIn sets, very commonSingle surfaces or zones
Scalecm–mm–km

Engineering importance of joints (2 marks).

  • Control strength and deformability of the rock mass (intact rock is far stronger than jointed mass).
  • Provide pathways for groundwater seepage (permeability) — important for dams, tunnels.
  • Govern slope and tunnel stability: control wedge/planar failures and overbreak.
  • Determine block size and ease of excavation/rippability and rock support requirements.
  • Influence weathering depth (water ingress along joints).

Volumetric joint count (2 marks). JvJ_v is the total number of joints per cubic metre, summed over all sets:

Jv=1S1+1S2+1S3J_v = \frac{1}{S_1} + \frac{1}{S_2} + \frac{1}{S_3}

With S1=0.40 mS_1 = 0.40\ \text{m}, S2=0.50 mS_2 = 0.50\ \text{m}, S3=1.0 mS_3 = 1.0\ \text{m}:

Jv=10.40+10.50+11.0=2.5+2.0+1.0=5.5 joints/m3J_v = \frac{1}{0.40} + \frac{1}{0.50} + \frac{1}{1.0} = 2.5 + 2.0 + 1.0 = \mathbf{5.5\ joints/m^3}

Comment: On the standard ISRM block-size scale, JvJ_v between 3 and 10 indicates large blocks (low joint density). The rock mass is moderately blocky, generally favourable for stability, requiring relatively light support.

jointsrock-massengineering-geology
8short5 marks

Define metamorphism and list its principal agents. Distinguish between contact and regional metamorphism. Explain the terms foliation and lineation, and give the metamorphic equivalents of shale, limestone and granite.

Metamorphism and agents (1.5 marks). Metamorphism is the solid-state transformation of a pre-existing rock (protolith) into a new rock with different texture and/or mineralogy in response to changed physical/chemical conditions, without complete melting. Principal agents: heat (temperature), pressure (lithostatic and directed/stress), and chemically active fluids.

Contact vs regional metamorphism (1.5 marks).

FeatureContact (thermal)Regional (dynamothermal)
CauseHeat from an igneous intrusionHigh T and directed P over large areas (orogeny)
ExtentLocal aureole around plutonHundreds of km (mountain belts)
TextureNon-foliated (hornfels)Strongly foliated (slate→schist→gneiss)
PressureLow directed stressHigh directed stress

Foliation and lineation (1 mark).

  • Foliation: planar, parallel arrangement of platy minerals (e.g. mica) or compositional banding produced by directed pressure — slaty cleavage, schistosity, gneissic banding.
  • Lineation: a linear (1-D) fabric — parallel alignment of elongate/needle minerals or fold-axis-related lines within the rock.

Metamorphic equivalents (1 mark).

ProtolithMetamorphic equivalent
Shale (mudstone)Slate → phyllite → schist → gneiss
LimestoneMarble
GraniteGranitic gneiss
metamorphismmetamorphic-rockstexture
9short5 marks

Explain karst topography and the geological conditions required for its development. Describe three characteristic karst landforms. Briefly state why a dam site on highly karstic limestone poses an engineering problem.

Karst topography (2 marks). Karst is a distinctive terrain produced by the dissolution (carbonation/solution) of soluble rocks — mainly limestone (and dolomite, gypsum) — by slightly acidic groundwater:

CaCO3+H2O+CO2Ca(HCO3)2 (soluble)CaCO_3 + H_2O + CO_2 \rightarrow Ca(HCO_3)_2\ \text{(soluble)}

Conditions required:

  1. Soluble rock (thick, pure limestone) at or near the surface.
  2. Well-developed joints/bedding for water to penetrate.
  3. Humid climate with adequate rainfall and vegetation (source of CO2CO_2).
  4. Water table below the surface allowing vertical and lateral flow / a hydraulic gradient.

Three karst landforms (2 marks).

  • Sinkholes (dolines): closed surface depressions where rock has dissolved or collapsed; surface drainage disappears underground.
  • Caves and caverns: underground solution channels and chambers, often with stalactites/stalagmites (dripstone).
  • Disappearing (sinking) streams / springs: surface streams sink underground and re-emerge as large springs; also limestone pavements and underground drainage networks.

Dam-site problem on karstic limestone (1 mark). Karstic limestone is riddled with solution cavities and conduits, giving very high secondary permeability. A reservoir would suffer severe water leakage through the foundation/abutments, potential piping, sinkhole collapse and loss of storage — requiring extensive (and costly) grouting/cut-off treatment or site rejection.

geomorphologygroundwaterweathering
10short6 marks

Describe how sedimentary rocks form and classify them on the basis of mode of origin (clastic, chemical, organic) with examples. Define the principle of superposition and the unconformity, naming the three main types of unconformity. Relate the Siwalik Group of Nepal to sedimentary processes.

Formation of sedimentary rocks (1.5 marks). Sediments are produced by weathering and erosion of pre-existing rocks, transported by water/wind/ice, deposited in basins, and converted to rock by lithification = compaction (removal of pore water under burial) + cementation (precipitation of silica, calcite, iron oxide between grains). Chemical/biological precipitation also produces sediment.

Classification by origin (2 marks).

ClassOriginExamples
Clastic (detrital)Mechanical accumulation of rock/mineral fragmentsConglomerate, sandstone, siltstone, shale
ChemicalPrecipitation from solutionRock salt, gypsum, some limestone, chert
Organic (biogenic)Accumulation of organic remainsCoal, fossiliferous limestone, coquina

Superposition and unconformity (1.5 marks).

  • Law of superposition: in an undisturbed sequence of sedimentary strata, each bed is younger than the one beneath it and older than the one above — gives relative age.
  • Unconformity: a buried erosion/non-deposition surface representing a gap (hiatus) in the geological record. Three main types:
  1. Angular unconformity — tilted/folded older beds overlain by horizontal younger beds.
  2. Disconformity — erosion surface between parallel strata.
  3. Nonconformity — sedimentary rocks overlying eroded igneous/metamorphic basement.

Siwalik Group of Nepal (1 mark). The Siwaliks (Churia) are Mio-Pliocene to Pleistocene molasse sediments — alternating sandstone, mudstone and conglomerate — deposited by braided/meandering rivers in the foreland basin as the rising Himalaya shed its erosional debris southward. They demonstrate the clastic sedimentary cycle (uplift → weathering → fluvial transport → deposition → lithification) and are notoriously weak and landslide-prone.

sedimentary-rocksstratigraphygeology-of-nepal
11short6 marks

Outline the main tectonic / geological zones of Nepal and the major boundary thrusts separating them. Briefly explain the engineering-geological significance of the Main Boundary Thrust (MBT) zone. Why is geological investigation essential before siting major infrastructure (tunnels, dams, highways) in Nepal?

Tectonic/geological zones of Nepal and boundary thrusts (3 marks). From south to north, separated by major north-dipping thrusts:

  Indo-Gangetic Plain (Terai)
  ---- MFT (Main Frontal Thrust) ----
  Sub-Himalaya (Siwalik / Churia)
  ---- MBT (Main Boundary Thrust) ----
  Lesser Himalaya (Mahabharat)
  ---- MCT (Main Central Thrust) ----
  Higher (Greater) Himalaya
  ---- STDS (South Tibetan Detachment System) ----
  Tibetan-Tethys Himalaya
  • Terai/Indo-Gangetic plain: Quaternary alluvium.
  • Sub-Himalaya (Siwaliks): young, weak molasse; bounded by MFT and MBT.
  • Lesser Himalaya: low-grade metasediments; MBT to MCT.
  • Higher Himalaya: high-grade crystallines (gneiss, schist) above the MCT.
  • Tethys Himalaya: fossiliferous sediments above the STDS.

Significance of the MBT zone (1.5 marks). The MBT is a major active thrust separating weak Siwaliks from Lesser Himalayan rocks. The zone is characterised by intense fracturing, shearing, crushed/sheared rock, fault gouge, high seismicity and abundant landslides. Engineering significance: poor and highly variable rock quality, instability of cut slopes and tunnel faces, high water inflow along the fracture zone, and elevated seismic hazard — all demanding special design and support.

Why geological investigation is essential (1.5 marks). Nepal's terrain is young, tectonically active, seismically hazardous, with fragile lithologies and steep, landslide-prone slopes. Prior geological investigation is essential to:

  • identify active faults, weak/sheared zones and groundwater conditions;
  • assess slope stability, bearing capacity and excavability;
  • choose safe alignments/sites and appropriate support, reducing the risk of failure, cost overruns and loss of life. Skipping investigation in such terrain greatly increases the chance of dam leakage, tunnel collapse and highway landslides.
geology-of-nepaltectonic-zonesengineering-geology

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