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Section A: Long Answer Questions

Attempt all questions.

5 questions
1long10 marks

Define a mineral and explain why minerals are the fundamental building blocks of rocks. Describe, with engineering relevance, the principal physical properties used to identify minerals in hand specimen (crystal form, cleavage, fracture, hardness, lustre, colour, streak, and specific gravity). Using Mohs scale, explain the concept of relative hardness and determine which of two minerals will scratch the other for the pair: quartz (H = 7) and orthoclase feldspar (H = 6).

Definition of a mineral

A mineral is a naturally occurring, inorganic, homogeneous solid with a definite (but not always fixed) chemical composition and an ordered internal atomic arrangement (crystalline structure). Examples: quartz (SiO2SiO_2), calcite (CaCO3CaCO_3), orthoclase (KAlSi3O8KAlSi_3O_8).

Why minerals are the building blocks of rocks

A rock is an aggregate of one or more minerals (plus sometimes mineraloids/glass). The identity, proportion, grain size and bonding of the constituent minerals control the rock's strength, durability, permeability and weathering response. Hence rock engineering behaviour is ultimately rooted in mineralogy.

Principal physical properties (with engineering relevance)

PropertyDescriptionEngineering relevance
Crystal formExternal geometry from internal lattice (e.g. cubic galite, hexagonal quartz)Indicates mineral identity and anisotropy
CleavageTendency to break along planes of weak bonding (1, 2, 3 directions)Planes of weakness reduce intact rock strength; mica cleavage weakens schist
FractureIrregular breakage (conchoidal in quartz) where no cleavageControls aggregate angularity and crushing behaviour
HardnessResistance to scratching (Mohs 1-10)Governs abrasion resistance of aggregate, drilling/cutting cost
LustreAppearance of reflected light (metallic, vitreous, pearly)Aids identification
ColourBulk colour (often unreliable)Quick but ambiguous indicator
StreakColour of powder on porcelainMore reliable than colour for ore minerals
Specific gravityRatio of mineral density to water densityAffects rock unit weight, dam/foundation loading

Mohs scale and relative hardness

Mohs scale (1 talc -> 10 diamond) is an ordinal, relative scale: a mineral can scratch any mineral of equal or lower Mohs number and is scratched by any mineral of higher number. It does NOT give absolute hardness (the absolute hardness gap from 9 corundum to 10 diamond is very large).

Quartz (H = 7) vs Orthoclase (H = 6):

Since 7>67 > 6, quartz will scratch orthoclase, and orthoclase cannot scratch quartz. This is consistent with the durability of quartz-rich rocks (quartzite, quartz sandstone) as engineering aggregate.

mineralsphysical-propertiessilicates
2long10 marks

Explain folds as geological structures. Describe the principal parts of a fold (limbs, axial plane, hinge, fold axis, crest, trough) and classify folds on the basis of (i) the attitude of the axial plane and (ii) the inter-limb angle. A planar bed has a true dip of 3030^{\circ} measured perpendicular to the strike. Determine the apparent dip seen in a vertical cliff face that trends at 4040^{\circ} to the strike of the bed. (Use tan(δapp)=tan(δtrue)sinβ\tan(\delta_{app}) = \tan(\delta_{true})\cdot\sin\beta, where β\beta is the angle between the cliff face and the strike.)

Folds

Folds are wave-like bends produced in originally planar rock layers (bedding, foliation) by compressive tectonic stress (ductile deformation). They are primary indicators of crustal shortening.

Principal parts of a fold

        axial plane
            |
  limb \    | hinge   / limb
        \   v        /
   ------ \----.----/ ------  (crest of anticline)
           \__/
            fold axis (line of max curvature)
  • Limbs: the two sloping sides of the fold.
  • Axial plane: the plane (or surface) that bisects the inter-limb angle, passing through hinge lines of successive beds.
  • Hinge: the zone/line of maximum curvature on a folded surface.
  • Fold axis (hinge line): line of intersection of the axial plane with a bedding surface.
  • Crest: highest point of an upfold (anticline).
  • Trough: lowest point of a downfold (syncline).

Classification (i): by attitude of 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, one limb rotated past vertical.
  • Recumbent: axial plane nearly horizontal.
  • Isoclinal: both limbs parallel (axial plane parallel to limbs).

Classification (ii): by inter-limb angle

ClassInter-limb angle
Gentle120180120^{\circ}-180^{\circ}
Open7012070^{\circ}-120^{\circ}
Close307030^{\circ}-70^{\circ}
Tight0300^{\circ}-30^{\circ}
Isoclinal0\approx 0^{\circ}

Apparent dip calculation

Given: true dip δtrue=30\delta_{true} = 30^{\circ}, angle between cliff face and strike β=40\beta = 40^{\circ}.

tan(δapp)=tan(δtrue)sinβ\tan(\delta_{app}) = \tan(\delta_{true})\cdot\sin\beta tan(δapp)=tan(30)sin(40)=0.57735×0.64279=0.37113\tan(\delta_{app}) = \tan(30^{\circ})\cdot\sin(40^{\circ}) = 0.57735 \times 0.64279 = 0.37113 δapp=tan1(0.37113)=20.36\delta_{app} = \tan^{-1}(0.37113) = 20.36^{\circ}

The apparent dip in the cliff face is approximately 20.420.4^{\circ}. (Note: apparent dip < true dip, as expected since the cliff is oblique to the strike.)

geological-structuresfoldsdip-strike
3long10 marks

Explain the theory of plate tectonics and the three types of plate boundaries. Apply the theory to explain the origin of the Himalaya and the high seismicity of Nepal. Identify the principal tectonic feature along which the Indian Plate is underthrusting beneath the Tibetan (Eurasian) Plate.

Plate tectonics theory

The Earth's rigid outer shell (lithosphere) is broken into about a dozen major plates that move (a few cm/yr) over the weaker, partly ductile asthenosphere. The driving mechanisms are mantle convection, ridge-push and slab-pull. Plates interact at their boundaries, producing earthquakes, volcanism and mountain building.

Three types of plate boundaries

  1. Divergent (constructive): plates move apart; new oceanic lithosphere forms (mid-ocean ridges, e.g. Mid-Atlantic Ridge). Shallow earthquakes, basaltic volcanism.
  2. Convergent (destructive): plates move together. Sub-types:
    • Ocean-continent / ocean-ocean: subduction, deep trenches, volcanic arcs (e.g. Andes).
    • Continent-continent: collision, no subduction of buoyant crust, fold mountains (e.g. Himalaya).
  3. Transform (conservative): plates slide laterally past each other; lithosphere is neither created nor destroyed (e.g. San Andreas Fault). Shallow strike-slip earthquakes.

Origin of the Himalaya

About 50-55 million years ago the northward-drifting Indian Plate collided with the Eurasian Plate after the intervening Tethys Ocean was consumed by subduction. Because both are buoyant continental crust, the collision caused intense crustal shortening, stacking and uplift, raising the Himalayan range. Convergence continues today (~18-20 mm/yr of India-Tibet shortening), so the mountains are still rising.

High seismicity of Nepal

Nepal lies on the active collision front. The continuing underthrusting of India beneath Tibet builds elastic strain that is released as large earthquakes (e.g. 1934 Bihar-Nepal, 2015 Gorkha). Stress concentrates along major thrusts (MFT, MBT, MCT) and along the locked Main Himalayan Thrust ramp, making Nepal one of the most seismically hazardous regions in the world.

Principal underthrusting feature

The Indian Plate underthrusts beneath the Tibetan Plate along the Main Himalayan Thrust (MHT) - the basal decollement to which the surface thrusts MFT, MBT and MCT merge at depth.

plate-tectonicsgeology-of-nepalhimalaya
4long10 marks

Differentiate between physical (mechanical) and chemical weathering and describe at least three processes of each. Explain how weathering degrades the engineering properties of rock. A fresh granite core has a dry density of 2650kg/m32650\,\text{kg/m}^3. After prolonged chemical weathering, a sample of the same granite shows a dry density of 2120kg/m32120\,\text{kg/m}^3. Assuming the mineral grain (solid) density is unchanged at 2700kg/m32700\,\text{kg/m}^3, compute the porosity of the fresh rock and of the weathered rock, and the percentage increase in porosity.

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

  1. Frost wedging (freeze-thaw): water in cracks freezes, expands ~9%, prising rock apart.
  2. Thermal expansion/insolation: repeated heating-cooling causes differential expansion and exfoliation.
  3. Unloading (pressure release): removal of overburden allows expansion -> sheet jointing.
  4. (Also: salt crystallisation, root/biological wedging.)

Chemical weathering alters the mineral composition, producing new (usually clay) minerals. Processes:

  1. Hydrolysis: feldspar + water/acid -> clay minerals (kaolinite). The dominant decay of silicates.
  2. Oxidation: Fe2+Fe^{2+} minerals react with oxygen -> iron oxides (rust staining).
  3. Carbonation/dissolution: CO2+H2O>H2CO3CO_2 + H_2O -> H_2CO_3, dissolving carbonates (karst).
  4. (Also: hydration.)

Effect on engineering properties

Weathering reduces intact strength and stiffness, increases porosity and water absorption, lowers durability, transforms hard minerals into weak clays (reducing shear strength, increasing plasticity and swelling), and opens discontinuities - all of which degrade foundation, slope and aggregate quality.

Numerical solution

Porosity n=1ρdryρsolidn = 1 - \dfrac{\rho_{dry}}{\rho_{solid}} (the dry bulk density reflects solids in the total volume; assuming negligible bound water in the dry state).

Fresh granite:

nfresh=126502700=10.98148=0.01852=1.85%n_{fresh} = 1 - \frac{2650}{2700} = 1 - 0.98148 = 0.01852 = \mathbf{1.85\%}

Weathered granite:

nweath=121202700=10.78519=0.21481=21.48%n_{weath} = 1 - \frac{2120}{2700} = 1 - 0.78519 = 0.21481 = \mathbf{21.48\%}

Percentage increase in porosity:

nweathnfreshnfresh×100=0.214810.018520.01852×100=0.196300.01852×100=1060% (10.6×)\frac{n_{weath} - n_{fresh}}{n_{fresh}}\times 100 = \frac{0.21481 - 0.01852}{0.01852}\times 100 = \frac{0.19630}{0.01852}\times 100 = \mathbf{1060\%\ (\approx 10.6\times)}

The porosity rises from about 1.85% to about 21.5%, an increase of roughly 1060% (about an 11-fold rise), reflecting severe weakening of the rock.

weatheringengineering-propertiesporosity
5long8 marks

Describe the major fluvial (river) landforms and the cycle of river erosion (youthful, mature, old stages). Discuss the engineering significance of fluvial geomorphology in the siting of dams and bridges in Nepal. A river drops 1500m1500\,\text{m} over a channel length of 50km50\,\text{km}. Calculate the average channel gradient (i) as a ratio, (ii) in m/km, and (iii) in percent.

Major fluvial landforms

  • Erosional: V-shaped valleys, gorges/canyons, waterfalls and rapids, potholes, river terraces, meander cut-banks.
  • Depositional: alluvial fans, flood plains, natural levees, point bars, braided bars, deltas, ox-bow lakes.

Cycle (stages) of river erosion

StageProfileDominant processFeatures
YouthfulSteep gradientVertical (downcutting)V-valleys, gorges, rapids, waterfalls
MatureModerate gradientLateral erosion beginsWider valley, meanders, flood plain forms
OldGentle/near base levelDeposition dominatesBroad flood plain, ox-bows, levees, deltas

(In Nepal the Himalayan rivers are largely youthful/mature due to active uplift, giving deep gorges ideal for high dams.)

Engineering significance for dams and bridges in Nepal

  • Narrow youthful gorges (e.g. on the Trishuli, Karnali) give short, strong dam axes but pose problems of high sediment load, debris flows and bank instability.
  • River terraces provide founding levels for bridge abutments but may be undercut.
  • Meandering/braided reaches in valleys require careful scour-depth design for bridge piers.
  • Aggradation/degradation trends, sediment yield and active faulting near river courses must be assessed for reservoir life and structure safety.

Gradient calculation

Vertical drop =1500m= 1500\,\text{m}, horizontal length =50km=50000m= 50\,\text{km} = 50000\,\text{m}.

(i) As a ratio:

gradient=150050000=133.331:33.3\text{gradient} = \frac{1500}{50000} = \frac{1}{33.33} \approx \mathbf{1:33.3}

(ii) In m/km:

1500m50km=30 m/km\frac{1500\,\text{m}}{50\,\text{km}} = \mathbf{30\ m/km}

(iii) In percent:

150050000×100=3%\frac{1500}{50000}\times 100 = \mathbf{3\%}

The average channel gradient is 1:33.3, i.e. 30 m/km, i.e. 3% - a steep, youthful-stage gradient typical of Nepal's Himalayan rivers.

geomorphologyriver-erosiongeology-of-nepal
B

Section B: Short Answer Questions

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6 questions
6short6 marks

Define the three major rock groups (igneous, sedimentary, metamorphic) and explain the rock cycle with a labelled sketch (described in text). Give one common engineering example of each rock group used in construction in Nepal.

Three major rock groups

  1. Igneous rocks: form by cooling and solidification of molten magma/lava. Intrusive (slow, coarse, e.g. granite) or extrusive (fast, fine, e.g. basalt).
  2. Sedimentary rocks: form by weathering, transport, deposition and lithification (compaction + cementation) of sediments, or by chemical/organic precipitation. E.g. sandstone, limestone, shale.
  3. Metamorphic rocks: form by solid-state alteration of pre-existing rocks under heat and/or pressure. E.g. gneiss, schist, slate, marble, quartzite.

Rock cycle (described sketch)

   Magma --cooling--> IGNEOUS
     ^                    |
     |              weathering/erosion
   melting               v
     |              SEDIMENT --lithification--> SEDIMENTARY
     |                                              |
  METAMORPHIC <----heat & pressure-----------------+
     |                                              |
     +----------heat & pressure---------------------+

Any rock can be transformed into another: igneous, sedimentary and metamorphic rocks are continuously recycled through melting, uplift, weathering, deposition and metamorphism, driven by internal (tectonic) and external (surface) energy.

Engineering examples in Nepal

  • Igneous: granite - dimension stone, aggregate.
  • Sedimentary: limestone/sandstone - limestone for cement (e.g. Udayapur, Chobhar), sandstone for masonry.
  • Metamorphic: gneiss/slate - gneiss as aggregate and building stone; slate for traditional roofing.
rocksrock-cycleclassification
7short6 marks

Define a fault and explain its main elements (fault plane, hanging wall, foot wall, throw, heave, slip). Classify faults on the basis of relative movement. A normal fault has a fault plane dipping at 6060^{\circ} and a net dip-slip displacement (slip along the plane) of 40m40\,\text{m}. Calculate the throw (vertical component) and the heave (horizontal component) of the fault.

Fault: a fracture (or fracture zone) in rock along which there has been appreciable relative displacement of the two sides.

Main elements

  • Fault plane: the surface along which movement occurs (described by dip and strike).
  • Hanging wall: the block resting above the inclined fault plane.
  • Foot wall: the block lying below the inclined fault plane.
  • Slip (net slip): total relative displacement measured along the fault plane.
  • Throw: vertical component of the dip-slip displacement.
  • Heave: horizontal component of the dip-slip displacement.

Classification by relative movement

  1. Normal fault: hanging wall moves down relative to foot wall (tensional/extensional).
  2. Reverse fault: hanging wall moves up relative to foot wall (compressional); a low-angle reverse fault is a thrust.
  3. Strike-slip (transcurrent) fault: horizontal movement parallel to strike (dextral/sinistral).
  4. Oblique-slip fault: combination of dip-slip and strike-slip.

Numerical solution

For dip-slip on a plane dipping at angle θ\theta with net slip SS:

Throw=Ssinθ,Heave=Scosθ\text{Throw} = S\sin\theta, \qquad \text{Heave} = S\cos\theta

Given S=40mS = 40\,\text{m}, θ=60\theta = 60^{\circ}:

Throw=40×sin60=40×0.86603=34.64 m\text{Throw} = 40\times\sin 60^{\circ} = 40\times 0.86603 = \mathbf{34.64\ m} Heave=40×cos60=40×0.5=20.0 m\text{Heave} = 40\times\cos 60^{\circ} = 40\times 0.5 = \mathbf{20.0\ m}

Check: 34.642+20.02=1199.9+400.0=1599.940m\sqrt{34.64^2 + 20.0^2} = \sqrt{1199.9 + 400.0} = \sqrt{1599.9} \approx 40\,\text{m} = net slip. Correct.

Throw = 34.64 m and heave = 20.0 m.

faultsgeological-structuresthrow-heave
8short5 marks

What are joints? Distinguish joints from faults. Explain how joint sets, spacing and orientation affect the engineering behaviour of a rock mass (slope stability, permeability, excavation). A rock mass has three joint sets with mean spacings of 0.50m0.50\,\text{m}, 0.40m0.40\,\text{m} and 0.25m0.25\,\text{m}. Estimate the volumetric joint count JvJ_v (joints per cubic metre) and comment on the block size.

Joints: fractures in rock along which there has been no appreciable displacement parallel to the fracture surface (essentially opening/tensional or shrinkage cracks).

Joints vs faults

FeatureJointFault
DisplacementNegligible/none along planeAppreciable relative movement
ScaleUsually small, very abundantLarger, fewer
OriginCooling, unloading, tectonic stressShear failure under tectonic stress

Effect on rock-mass behaviour

  • Slope stability: joints oriented adversely (daylighting out of a slope, dip < slope angle) create planar/wedge/toppling failures.
  • Permeability: joints are the main flow paths in hard rock; closely spaced, open joints raise mass permeability and seepage/leakage (important for dams, tunnels).
  • Excavation: spacing and orientation control blockiness, overbreak, support needs and ease of ripping/blasting.

Volumetric joint count JvJ_v

Jv=1S1+1S2+1S3J_v = \frac{1}{S_1}+\frac{1}{S_2}+\frac{1}{S_3} Jv=10.50+10.40+10.25=2.0+2.5+4.0=8.5 joints/m3J_v = \frac{1}{0.50}+\frac{1}{0.40}+\frac{1}{0.25} = 2.0 + 2.5 + 4.0 = \mathbf{8.5\ joints/m^3}

Comment: By the standard JvJ_v description, JvJ_v of 3103-10 indicates a blocky / medium-sized block rock mass (roughly Jv=8.5J_v=8.5 is at the small-to-medium end). Approximate block volume 0.50×0.40×0.25=0.05m3\approx 0.50\times0.40\times0.25 = 0.05\,\text{m}^3, i.e. blocks about half a metre across - a moderately fractured mass requiring attention to block stability in slopes and tunnels.

jointsdiscontinuitiesrock-mass
9short5 marks

Describe the major physiographic-tectonic zones of Nepal from south to north and the principal boundary faults separating them. Name one representative rock type or formation in each zone.

From south to north, Nepal is divided into five major physiographic-tectonic zones, separated by major thrust faults:

Zone (S -> N)Brief geologyRepresentative rock/formation
Terai (Gangetic Plain)Recent alluvium of Himalayan riversQuaternary gravel, sand, silt
Siwalik (Sub-Himalaya)Mio-Pliocene molasse, foreland-basin sedimentsSandstone, mudstone, conglomerate
Lesser HimalayaLow-to-medium grade metasediments, some unfossiliferous sequencesSlate, phyllite, quartzite, dolomite (e.g. Nawakot/Kunchha)
Higher HimalayaHigh-grade crystalline rocks (Tibetan slab)Gneiss, schist, migmatite, marble
Tethys (Tibetan-Tethyan) HimalayaFossiliferous marine sedimentary sequenceLimestone, shale, sandstone

Principal boundary faults (thrusts), from south to north:

  • MFT (Main Frontal Thrust / Himalayan Frontal Thrust): separates Terai from Siwalik.
  • MBT (Main Boundary Thrust): separates Siwalik from Lesser Himalaya.
  • MCT (Main Central Thrust): separates Lesser Himalaya from Higher Himalaya.
  • STDS (South Tibetan Detachment System): separates Higher Himalaya from the Tethys Himalaya (a normal-sense detachment).

These zones and bounding thrusts reflect the southward propagation of the India-Asia collision and control Nepal's seismicity, slope stability and rock-engineering conditions.

geology-of-nepaltectonic-zonesstratigraphy
10short5 marks

Define mass wasting (slope movements) and classify the main types (fall, topple, slide, flow, creep) by mechanism and rate. Discuss the principal causes of landslides in the hills of Nepal and one engineering mitigation measure for each cause.

Mass wasting: the downslope movement of rock, soil and debris under the direct action of gravity, without a primary transporting medium such as a stream or glacier.

Classification of main types

TypeMechanismTypical rate
FallFree fall of detached blocks from steep faces/cliffsVery rapid
ToppleForward rotation of columns/blocks about a baseRapid
SlideMovement on a defined failure surface (planar/rotational)Slow to rapid
FlowInternal deformation, mass behaves like a fluid (debris/mud flow)Rapid to very rapid
CreepVery slow, continuous, imperceptible soil movementExtremely slow

Principal causes in Nepal's hills + one mitigation each

CauseMitigation
Steep slopes & active upliftSlope flattening / benching / retaining walls
Intense monsoon rainfall & high pore pressureSurface and subsurface drainage (catch drains, weep holes)
Weak, fractured/weathered rock & adverse jointsRock bolting, shotcrete, slope reinforcement
Seismic shakingSeismic-resistant design, avoid undercutting, gabion check structures
Toe erosion by riversRiver training, gabion/revetment toe protection
Human activity (road cuts, deforestation, irrigation)Proper cut-slope design, bio-engineering (vegetation), controlled land use

Mass wasting is the dominant geohazard along Nepal's hill roads and settlements, and integrated drainage + bio-engineering + structural support is the standard control strategy.

geomorphologymass-wastinglandslides
11short5 marks

Define specific gravity of a mineral and explain its engineering significance. A mineral specimen weighs 0.085N0.085\,\text{N} in air and 0.053N0.053\,\text{N} when fully submerged in water. Determine its specific gravity and identify the most likely mineral from this list: quartz (G \approx 2.65), calcite (G \approx 2.71), galena (G \approx 7.5).

Specific gravity (G): the ratio of the weight (or density) of a mineral to the weight of an equal volume of water at 4C4^{\circ}C. It is dimensionless.

Engineering significance: G controls the unit weight of rock, hence dead loads on foundations and dams, the design of mineral-processing and the identification of heavy (ore) versus light (rock-forming) minerals; it also feeds into porosity/void calculations.

Numerical solution (Archimedes' principle)

Using the submerged-weight method:

G=WairWairWwaterG = \frac{W_{air}}{W_{air} - W_{water}}

The denominator WairWwaterW_{air}-W_{water} equals the weight of water displaced (the buoyant force).

Given Wair=0.085NW_{air} = 0.085\,\text{N}, Wwater=0.053NW_{water} = 0.053\,\text{N}:

WairWwater=0.0850.053=0.032NW_{air} - W_{water} = 0.085 - 0.053 = 0.032\,\text{N} G=0.0850.032=2.656G = \frac{0.085}{0.032} = 2.656

Specific gravity 2.66\approx 2.66.

Identification: G=2.66G = 2.66 matches quartz (G \approx 2.65) most closely (calcite is a bit higher at 2.71, galena is far heavier at 7.5). The specimen is most likely quartz.

mineralsspecific-gravityengineering-properties

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