Gate 2018 GG Solution (Geology Option)

© 2018 by Lawrence Kanyan. All rights reserved.
GATE 2018 (GEOLOGY) SOLUTION KEY
BY: LAWRENCE KANYAN
(AIR – 1, GATE 2017)
PART A: COMPULSORY SECTION FOR ALL CANDIDATES
Q.1) Which one of the following periods has the longest time duration?
(A) Ordovician (B) Cretaceous (C) Jurassic (D) Silurian  
Solution: (B) Cretaceous
Ordovician covers from 485.4 mya to 443.8 mya (span of 41.2 my)
Cretaceous covers from 145 mya to 66 mya (span of 79 my)
Jurassic covers from 201.3 mya to 145 mya (span of 56.3 my)
Silurian covers from 443.8 mya to 419.2 mya (span of 24.6 my)
Cretaceous has the longest time duration.
Q.2) A siliciclastic sedimentary rock consisting predominantly of the same type of gravel-sized
clasts is called  
(A) Polymict conglomerate (B) Arkose (C) Oligomict conglomerate. (D) Petromict
conglomerate
Solution: (C) Oligomict conglomerate
There are two key words to focus on: “same type” and “gravel-sized.” The classification
Polymict and Oligomict is based on the type of clasts: are the clast of same rock/mineral or not.
Conglomerates are named pebble-conglomerate, gravel-conglomerate etc based on clast size.
Polymict: conglomerate consisting of clasts of more than one type
Arkose: Type of sandstone, not a type of conglomerate
Oligomict: conglomerate consisting of clasts made up of a single rock/mineral
Petromict: This type of conglomerated contains clasts of unstable and metastable rocks/minerals.
Q.3) Brown coal that has high moisture content and commonly retains many of the original
wood fragments is called

(A) Anthracite (B) Bituminous coal (C) Lignite (D) Peat
Solution: (C) Lignite
Anthracite: Shiny black, not brown. Highly altered and does not retain original wood fragments.
It has very low moisture content.
Bituminous: Mostly black in colour. More mature than lignite and thus has low moisture content.
Lignite: Immature coal, and thus has brown colour. It has high moisture content and retains
original fragments of wood
Peat: It is not coal, but a precursor to coal.
Q.4) The speed of revolution of the Earth around the Sun is
(A) maximum at Perihelion (B) minimum at Perihelion (C) maximum at Aphelion (D) equal at
Aphelion and Perihelion
Solution: (A) maximum at Perihelion
Planetary orbits around the sun are not perfectly circular; these are rather elliptical with sun
located at one of the foci of the ellipse. Hence the distance of the planet from the sun is not
constant but depends upon its position in the orbit. The point at which the planet (earth in this
case) is closest to the sun is called Perihelion and the point at which it is farthest from the sun is
called Aphelion.
Kepler’s second law states that: A line joining a planet and the Sun sweeps out equal areas
during equal intervals of time; in other words, aerial velocity of the planet around sun is constant.
Velocity and angular velocity vary depending upon the position of the planet. Since the radial
distance is closer during perihelion, the planet must cover more distance in the same time for the
aerial velocity to be constant. This translates into faster speed of revolution at perihelion.
If a planet moves infinitesimal distance dx in infinitesimal time dt, the area covered can be
approximated by a triangle of base dx and height r, where r is the distance of the planet form sun.
𝐴"# =
1
2
r. dx

𝐴"# =
1
2
r. v. dt
𝐴"# =
1
2
r. v. dt
Where, 𝐴"# is area covered in time dt; area of triangle of height r and base v×dt.
Since 𝐴"# is constant, v is inversely proportional to r.
Q.5) The geometrical factor for the following electrode configuration is
(A) 𝜋𝑎 (B) 2𝜋𝑎 (C) 3𝜋𝑎 (D) 4𝜋𝑎
Solution: (D) 4𝜋𝑎
Different electrode configurations in resistivity surveys are: Wenner, Schlumberger,
Dipole-Dipole, Pole-dipole, and pole-pole configurations. In the given problem, we have a
pole-dipole electrode configuration, i.e. the second current electrode (infinite electrode) is
at a great distance from the measurement electrodes.
Geometrical factor (Kg) is a numerical multiplier that depends upon the configuration of
electrodes and is used to find out the apparent resistivity (𝜌O) using:

𝜌O = 𝐾Q
𝑉
𝐼
For a general 4-electrode setting, 𝜌O is given by:
𝜌O = 2𝜋
𝑉
𝐼
1
1
𝑟UV
−
1
𝑟XV
−
1
𝑟UY
−
1
𝑟XY
Hence, general formula for the geometrical factor is: 2𝜋
[

]^_
`

]a_
`

]^b
`

]ab

Given, 𝑟UV = 𝑎, 𝑟XV = ∞, 𝑟UY = 2𝑎, 𝑟XY = ∞
Substituting respective values, we get
𝐾Q = 2𝜋
1
1
𝑎
−
1
∞
−
1
2𝑎
−
1
∞

[
d
is basically zero. Therefore,
𝐾Q = 2𝜋
1
1
𝑎
−
1
2𝑎

𝐾Q = 4𝜋𝑎
Q.6) Which one of the following geophysical methods uses the physical property ‘Dielectric
Constant’?
(A) Gravity (B) Ground Penetrating Radar (C) Seismic (D) Self-Potential

Solution: (B) Ground Penetrating Radar
Gravity method does not use electrical properties, it relies on density variations in the sb-surface.
GPR is a geophysical method that uses electromagnetic waves in the microwave band of the
radio spectrum to image the subsurface. The EM waves are reflected from surfaces where a
change in dielectric constant of the subsurface material is encountered.
Seismic survey uses acoustic waves, not electromagnetic wave. Hence, no use of Dielectric
Constant
Self-Potential is a well logging method that uses naturally occurring electric potential difference
in earth. SP arises from streaming potential and electrochemical potential.
Q.7) Pascal second is a unit of
(A) seepage force (B) dynamic viscosity (C) kinematic viscosity (D) permeability
Solution: (B) Dynamic Viscosity
Pascal is unit of pressure and second in unit of time. Pressure is Force/Area. Therefore, Pa.s is
force per unit area per unit time.
Seepage Force is force per unit volume produced due to the viscous drag of soil. No time unit is
involved in its definition, disregard this option.
Dynamic viscosity is the measurement of the fluid's internal resistance to flow while kinematic
viscosity refers to the ratio of dynamic viscosity to density.
𝐹 = 𝜇𝐴.
𝜕𝑢
𝜕𝑦
𝜇 =
𝐹
𝐴
×
𝜕𝑦
𝜕𝑢
𝜇 is dynamic viscosity.
F/A has units Pascals and the inverse of velocity gradient
mn
mo
has units second (
p
pqr
). Therefor 𝜇
(Kinematic Viscosity) has units Pascals Seconds. Kinematic viscosity is 𝜇/𝑑𝑒𝑛𝑠𝑖𝑡𝑦 and has units
m2
/s.
Permeability defines the ease with which a fluid travels through a porous medium.

Q.8) Which one of the following statements is CORRECT?
(A) Strength of a rock decreases with increase in confining pressure.
(B) Strength of a rock increases with increase in temperature.
(C) Strength of a rock increases with increase in strain rate.
(D) Strength of a rock increases with increase in pore water pressure.
Solution: (C) Strength of a rock increases with increase in strain rate.
Factors that increase rock strength: confining pressure, strain rate
Factors that decrease rock strength: temperature, pore water.
Increase in confining pressure moves the Mohr circle to the right, away from the failure
envelope. Increase in pore pressure decreases the effective stress and moves the Mohr circle to
the left, i.e. towards the failure envelope.
Q.9) The geomorphic feature ‘horns’ are formed by
(A) wind erosion. (B) river erosion. (C) wind deposition. (D) glacial erosion.
Solution: (D) Glacial Erosion
An arête is a thin, crest of rock left after two adjacent glaciers have worn a steep ridge into the
rock. A horn forms when glaciers erode three or more arêtes, usually forming a sharp-edged
peak.
Arete Matterhorn peak, Alps

Q.10) A melanocratic porphyritic rock containing phenocrysts of biotite, with feldspar restricted
to the groundmass, is called
(A) trachyte. (B) dacite. (C) andesite. (D) lamprophyre.
Solution: (D) lamprophyre
Melanocratic means dark coloured.
Trachyte, dacite, andesite and lamprophyre can all have phenocrysts, however the appearance
and mineralogy is different in each case.
Trachite: Variable coloured, but is generally light coloured with light coloured phenocrysts.
Mineral content - orthoclase phenocrysts in a groundmass of orthoclase with minor plagioclase,
biotite, hornblende, augite etc..
Dacite: Variable colour, usually bluish grey or pale grey, groundmass generally of plagioclase
with amphibole (hornblende), biotite, pyroxene (augite), quartz, and glass; phenocrysts of
plagioclase, amphibole and often quartz.
Andesite: Andesite has variable colour, but is usually grey (lighter than basalt). So, andesite is
generally mesocratic, not melanocratic. Plagioclase is more common than feldspar
Lamprophyre: Dark Coloured Ultrapotassic igneous rocks which are silica-undersaturated, mafic
or ultramafic.
Primary mineralogy- amphibole or biotite, and with feldspar in the groundmass. These rocks do
not fit into the commonly known classification systems due to their peculiar mineralogy.
Q.11) The supercontinent that existed in the late Mesoproterozoic to early Neoproterozoic time
was
(A) Kenorland. (B) Columbia. (C) Rodinia. (D) Pangaea.
Solution: (C) Rodinia
Kenorland formed during NeoArchean (~2.72 Ga)
Columbia existed in PaleoProterozoic (~2.5-1.5 Ga)
Rodinia existed in late Mesoporterozoic to early Neoporterozoic time (~1.3-0.63 Ga)
Pangea existed during the late Paleozoic and early Mesozoic eras (~335-173 million years ago)
Q.12) The figure below shows the triple junction between three plates A, B and C. The boundary
between the plates A and B is a ridge with a half-spreading rate of 4 cm/year. The A-C and B-C
boundaries are collinear and orthogonal to the A-B ridge. The A-C boundary is a dextral
transform fault with a relative velocity of 6 cm/year. The boundary between plates B and C is a

(A) dextral transform fault with a relative velocity of 10 cm/year.
(B) dextral transform fault with a relative velocity of 2 cm/year.
(C) sinistral transform fault with a relative velocity of 2 cm/year.
(D) sinistral transform fault with a relative velocity of 6 cm/year.
Solution: (C) sinistral transform fault with a relative velocity of 2 cm/year.
Assume the mid-oceanic-ridge (MOR) to be the stationary frame of reference. W.r.t the MOR,
plate C is moving south at velocity of 2cm/year. Let the velocity towards the north be positive
and towards south be negative. Since, plate B is moving south at 4cm/year w.r.t the MOR:
VA = 4, VB = -4
VA – VC = 6 => 4 – VC = 6 => VC = -2
Therefore,
VB – VC = -4 – (-2) = -2

Negative sign indicates that plate B moves south w.r.t. plate C in an anti-clockwise manner.
Therefore, the boundary between plate B and plate C should be a left-lateral (sinistral) transform
fault with a relative velocity of 2cm/year.
Q.13) A rock follows Mohr-Coulomb failure criterion. Which one of the Mohr-Coulomb failure
envelopes shown below allows failure of the rock under stress state Y, but not under stress state
X?
(A) PP’ (B) QQ’ (C) RR’ (D) SS’
Solution: (A) PP’
Rock will fail when the Mohr circle touches the failure envelope.
PP’ – No failure under X, Failure under Y
QQ’ – No failure under Y, No failure under X

RR’ – Failure under X, No failure under Y
SS’ – Failure under X, Failure under Y. However, it will not be possible to generate stress states
X and Y in this case as the rock will rupture at much lower differential stresses for the respective
confining pressures.
Q.14) The maximum and the minimum principal stresses are denoted by σ1 and σ3, respectively.
The differential stress can have an absolute value greater than σ1 when
(A) σ1 and σ3 are both compressive.
(B) σ1 is compressive and σ3 is tensile.
(C) σ1 and σ3 are equal.
(D) σ1 and σ3 are both tensile.
Solution: (B) σ1 is compressive and σ3 is tensile.
Differential stress, σd = σ1 - σ3 => σ3 = σ1 - σd
since σd > σ1, σ3 must be negative. Therefore, σ3 is tensile.
Q.15) The geoid can be best defined as
(A) an oblate spheroid that best approximates the shape of the earth.
(B) a surface over which the value of gravity is constant.
(C) the physical surface of the earth.
(D) an equipotential surface of gravity of the earth.
Solution: (D) an equipotential surface of gravity of the earth, (a completely correct version
should be: an equipotential surface of gravity of the earth approximating mean-sea-level. An
equipotential surface of gravity can be found at any depth below the surface of the earth up to the
core, but only the one that approximates mean-sea-level is called geoid).
• an oblate spheroid that best approximates the shape of the earth: ellipsoid
• a surface over which the value of gravity is constant: geoid is an equipotential surface,
not equal gravity surface.
• the physical surface of the earth: topographic surface
Q.16) For a layered isotropic medium with a flat horizontal free surface, match the wave types
listed in Group-I with their corresponding polarizations listed in Group II.

Group-I Group-II
P. P-waves 1. Particle motion is transverse to the direction of wave propagation.
Q. S-waves 2. Particle motion is transverse to the direction of wave propagation
and confined to the horizontal plane.
R. Rayleigh waves 3. Particle motion is parallel to the direction of wave propagation.
S. Love waves 4. Particle motion is elliptical.
(A) P-1; Q-3; R-4; S-2
(B) P-3; Q-1; R-4; S-2
(C) P-3; Q-1; R-2; S-4
(D) P-2; Q-3; R-1; S-4
Solution: (B)
P-waves: Longitudinal waves with particle motion parallel to the direction of wave propagation
S-waves: Transverse waves with particle motion perpendicular to the direction of wave
propagation
Rayleigh waves: an undulating wave that travels over the surface of a solid, especially of the
ground in an earthquake, with a speed independent of wavelength, the motion of the particles
being in ellipses.
Love waves: Horizontally polarized surface waves in which particle motion is perpendicular to
the direction of wave propagation
P-waves and S-waves are ‘body waves’ as these penetrate inside the earth. Rayleigh waves and
Love waves are ‘surface waves’ as these do not penetrate the surface of earth.
Q.17) A ‘gentle’ fold with an interlimb angle equal to 160° appears tight (apparent interlimb
angle equal to 20°) in horizontal section. According to the plunge of the fold axis, it can also be
classified as
(A) horizontal fold. (B) gently plunging fold. (C) steeply plunging fold. (D) vertical fold.
Solution: (B) gently plunging fold

Since the apparent interlimb angle (20 degrees) is very less compared to the true interlimb angle
of 160 degrees, the fold must be gently plunging.
Q.18) The unit of shear modulus (rigidity modulus) is
(A) kg m-1
s-2
(B) m2
s-2
(C) kg m-2
s-2
(D) m-1
Solution: (C) kg m-2
s-2
rigidity modulus = the ratio of shear stress to shear strain
𝑆 =
𝑠𝑡𝑟𝑒𝑠𝑠
𝑠𝑡𝑟𝑎𝑖𝑛
𝑆 =
𝐹𝑜𝑟𝑐𝑒/𝐴𝑟𝑒𝑎
𝐿𝑒𝑛𝑔𝑡ℎ/𝐿𝑒𝑛𝑔𝑡ℎ
𝑆 =
𝑘𝑔. 𝑚. 𝑠`‚
/𝑚‚
1
S has units: kg m-2
s-2

Q.19) With increasing activity of silica, the CORRECT order of appearance of minerals in a
weathering environment with constant ratio of activities of K+ and H+ is
(A) gibbsite → kaolinite → pyrophyllite
(B) gibbsite → pyrophyllite → kaolinite
(C) kaolinite → gibbsite → pyrophyllite
(D) pyrophyllite → gibbsite → kaolinite
Solution: (A) gibbsite → kaolinite → pyrophyllite
Gibbsite – Al(OH)3, Kaolinite – Al2Si2O5(OH)4, Pyrophyllite – Al2Si4O10(OH)2
This is a SiO2-Al2O3-K2O-H2O system. The sequence of appearance of minerals with increasing
activity of silica can be found out using the ion activity diagram given below (Meunier, 2005).
As we move along the red arrow from left to right (constant ratio of activities of K+ and H+, and
increasing activity of silica) we first get gibbsite followed by kaolinite followed by pyrophyllite.
Q.20) Match the items listed in Group-I with those listed in Group-II.
Group-I Group-II
P. Mica 1. Beldih
Q. Uranium 2. Koderma

R. Phosphate 3. Agucha
S. Zinc 4. Gogi
(A) P-2, Q-3, R-4, S-1
(B) P-2, Q-4, R-1, S-3
(C) P-4, Q-2, R-3, S-1
(D) P-3, Q-1, R-4, S-2
Solution: (B)
Factual, requires no solution.
Q.21) Which one of the following corrections is always added during reduction of the observed
gravity data?
(A) Latitude (B) Free-air (C) Bouguer (D) Terrain
Solution: (D) Terrain
Latitude – due to the oblate shape of earth and centrifugal force from rotation of earth, g
increases with Latitude (at poles - minimum radius, no centrifugal force). Hence, Latitude
correction is subtracted from the observed value to correct for this increase in g away from the
equator.
Free-air – corrects for change in g due to height. Can be added or subtracted depending upon
whether measurement is made above the datum or below the datum
Bouguer – In free air correction, no account in taken for the gravitational effect of the rock
present between the datum line and the observation point. Bouguer correction removes this effect
by approximating the rock layer beneath the observation point to an infinite horizontal slab with
a thickness equal to the elevation of the observation above datum. On land, BC must be
subtracted.
Terrain – BC assumes that the topography around the gravity station is flat, which is rarely the
case. Hence TC is make for topographic relief in the vicinity of the gravity station. Since, BC
always over-corrects (subtracted more than required), TC must always be added.
Q.22) The magnitudes of the total geomagnetic field at the equator and pole are denoted by BE
and BP, respectively. Which one of the following is TRUE?
(A) BP ≈ 4 BE (B) BP ≈ 2 BE (C) BP ≈ BE (D) BP ≈ 1/2 BE

Solution: (B) BP ≈ 2 BE
This question is fact based. Magnetic intensity tends to decrease from poles to the equators.
Magnetic intensity at poles is ~60000 nT, and it is ~30000 nT near the equator.
Q.23) Assume a flat earth with crustal thickness of 35 km and average crustal and upper mantle
P-wave velocities of 6.4 km.s–1 and 8.1 km.s–1, respectively. The minimum distance from the
epicenter of a near surface earthquake at which Pn-waves are observed is _______ km.
Solution: 90.25 km
Minimum distance from the epicentre at which the Pn-waves will be observed is equal to the
distance at which the internally reflected wave at critical angle reaches the surface.
Using snell’s law:
sin 𝑖
sin 𝑟
=
𝑣„…oq#
𝑣pO†#‡ˆ
=
6.4
8.1
At critical angle of refraction, rc = 90o
. Therefore, sin rc = 1
sin 𝑖 =
𝑥
𝐴𝐵
64
81
=
𝑥
𝑥‚ + 35‚
solving for x, we get x = 45.124 km
Minimum distance at which Pn observed = 2x = 90.25 km
Q.24) Given that the velocity of P-waves in a sandstone matrix is 5600 m/s and that in oil is

1200 m/s, the velocity of P-wave propagation in oil saturated sandstone with 30% porosity is
___________ m/s. (Use Wyllie time average equation.)
Solution: 2666.7 m/s
Total time taken to travel through the rock = time taken to travel through matrix + time taken to
travel through the pores.
𝑇 = 𝑇p + 𝑇‘
Let the distance to be travelled = a. Since the porosity is 30%, 0.3a distance will be covered
through pores and 0.7a distance will be covered through the matrix
𝑎
𝑣
=
. 7𝑎
5600
+
. 3𝑎
1200
(using time = distance/speed)
Solving for v gives v = 2666.7 m/s
Q.25) If the total porosity of a soil is 20%, its void ratio (%) is _________.
Solution: 25%
Void ratio is defined as the ratio of volume occupied by voids (𝑉”) to the volume occupied by
grains or solids (𝑉”)
𝑉𝑅 =
𝑉”
𝑉q
𝑝𝑜𝑟𝑜𝑠𝑖𝑡𝑦 =
𝑉”
𝐵𝑢𝑙𝑘 𝑉𝑜𝑙𝑢𝑚𝑒
=
𝑉”
𝑉” + 𝑉q
0.2 =
𝑉”
𝑉” + 𝑉q
1
. 2
=
𝑉” + 𝑉q
𝑉”
= 1 +
𝑉q
𝑉”

4 =
𝑉q
𝑉”
𝑉𝑜𝑖𝑑 𝑅𝑎𝑡𝑖𝑜 =
1
4
= .25 => 25%
PART B (SECTION 1): FOR GEOLOGY CANDIDATES ONLY

Q.26) Which one of the following Himalayan lithounits predates India-Eurasia collision?
(A) Kasauli sandstone (B) Rangit Pebble Slate (C) Annapurna granite (D) Lower Karewa
sandstone
Solution: (B) Rangit Pebble Slate
Kasauli sandstone – Tertiary; part of early Himalayan foreland sediments together with Subhatu
and Dagshai formations.
Rangit Pebble Slate – lower Gondwana; exposed in a window in Rangit Valley, Darjeeling.
Annapurna Granite – Tertiary (Miocene); emplaced during partial melting of mid-crustal rocks
as a result of Himalayan tectonics.
Lower Karewa sandstone – Cenozoic glacial sediments; Kashmir
Q.27) Which one of the following ore minerals shows internal reflection?
(A) Orpiment (B) Magnetite (C) Pyrite (D) Molybdenite
Solution: (A) Orpiment
Orpiment (AS2S3) is arsenic ore. It has very high Refractive Index, very high birefringence, and
very strong internal reflection.
Other minerals with internal reflections are:
Mineral Internal Reflection
Sphalerite Yellow to brown (sometimes red to green)
Cinnabar Blood red
Proustite-pyrargyrite Ruby red
Rutile Clear yellow to deep red-brown
Anatase Blue
Azurite Blue
Malachite Green
Cassiterite Yellow brown to yellow
Hematite Blood red

Wolframite Deep brown
Chromite Very deep brown
Q.28) Which one is CORRECT for the following equilibrium reaction between quartz and
magnetite:
Si16
O16
O + Fe3
16
O3
18
O = Si16
O18
O + Fe3
16
O4 ?
(A) 1000 ln a = ∆qtz-mag where a = K1/n
(K is the equilibrium constant at the specified
temperature and n is a constant quantity)
B) 1000 ln a = ∆qtz-mag where a = Kn
(K is the equilibrium constant at the specified temperature
and n is the number of exchangeable atomic sites)
(C) (ln a/1000) = ∆qtz-mag where a = K1/n
temperature and n is a constant quantity)
(D) 1000 ln a = ∆qtz-mag where a = K1/n
temperature and n is the number of exchangeable atomic sites)
Solution: (D)
The reaction:
Si16
O16
O + Fe3
16
O3
18
O = Si16
O18
O + Fe3
16
O4
Deriving the relation between ∆›#œ`pOQ and 𝛼.
𝛿›#œ =
𝑂
[
𝑂
[¡
›#œ
( 𝑂
[ 𝑂)
[¡
q#"
− 1 ×1000
n¢ˆ‡"q
𝑂
[
𝑂
[¡
›#œ =
(𝛿›#œ + 1000)
1000
×( 𝑂
[
𝑂)
[¡
q#"
𝛿pOQ =
𝑂
[
𝑂
[¡
pOQ
( 𝑂
[ 𝑂)
[¡
q#"
− 1 ×1000
n¢ˆ‡"q
𝑂
[
𝑂
[¡
pOQ =
(𝛿pOQ + 1000)
1000
×( 𝑂
[
𝑂)
[¡
q#"
𝛼 =
𝛿›#œ + 1000
𝛿pOQ + 1000
subtracting 1 from each side:
𝛼 − 1 =
£¤¥¦`£§¨©
£§¨©ª[«««
≈
£¤¥¦`£§¨©
[«««
because 𝛿pOQ ≪ 1000
Value of 𝛼 ≈1
1000𝑙𝑛𝛼 = 𝛿›#œ − 𝛿pOQ = ∆›#œ`pOQ (using approximation: for x≈ 1, 𝑙𝑛𝑥 ≈ 𝑥 − 1)

𝐾 =
®¯°±²± ×[´µ¶
°
·¸]
®¯°±°± ×[´µ¶
°
·¶
²
±]
𝛼 =
( ·
² ·
° )¤¥¦
( ·
² ·)
°
§¨©
𝛼 is related to K by the following relation:
α = K1/n
the derivation of the above relation is complex. It is more convenient to just remember the
relation.
Q.29) Match the modes of life (listed in Group I) with the corresponding bivalve genera (listed
in Group II).
Group-I Group-II
P. Burrowing 1. Mytilus
Q. Recumbent unattached 2. Pecten
R. Byssally attached 3. Mya
S. Swimming 4. Gryphaea
(A) P-4, Q-3, R-2, S-1
(B) P-3, Q-4, R-1, S-2
(C) P-2, Q-3, R-1, S-4
(D) P-3, Q-1, R-4, S-2
Solution: (B)
Go through the description of bivalves and their mode of habitats on this link:
https://www.fossilhunters.xyz/fossil-classification/bivalves.html
Q.30) Based on the hypothetical litholog given below that shows a continuous succession of
sedimentary rocks, which one of the following statements is CORRECT?

(A) The rocks range from Cambrian to Cretaceous and show change in depositional
environment from marine to continental.
(B) The rocks range from Cambrian to Triassic and show change in depositional environment
from marine to continental.
(C) The rocks range from Cambrian to Cretaceous and show change in depositional environment
from continental to marine.
(D) The rocks are Palaeozoic in age and show change in depositional environment from marine
to continental
Solution: (A)
This problem can be solved by looking at the fossils and lithology at different stratigraphic
intervals..
Depositional Environment:
At the base mud-rock alternates with sandstone and limestone – an indication that the
environment is marine, which is confirmed by the presence of fossil Agnostus (a trilobite). The
presence of bones of Tyrannosaurus-rex (a dinosaur species commonly known as T-rex), plant
roots, and paleosol confirms that the depositional environment near the top of the section is
continental. This eliminates option C.
Age:
Since all the trilobites became extinct by Permian end, the bottom part of the section must be
paleozoic. T-rex indicates that the age is Cretaceous. The only option that satisfies all the
conditions is A.

Q.31) Which one of the following cladograms shows the CORRECT interrelationships among
the major groups of vertebrates?
(A) Cladogram I (B) Cladogram II (C) Cladogram III (D) Cladogram IV
Diapsid reptiles, which includes dinosaurs is most commonly related to the birds (Aves) and they
had a common ancestor. Only Cladogram I satisfies this condition, hence, A must be the correct
answer.
Q.32 Which one of the following stratigraphic successions is in the CORRECT chronological
order (from older to younger)?
(A) Rajmahal, Dubrajpur, Barakar
(B) Fenestella Shale, Muth Quartzite, Syringothyris Limestone
(C) Bagh Bed, Lameta Formation, Deccan Traps
(D) Singhbhum Granite, Kolhan Group, Older Metamorphic Gneiss

(A) Correct chronological order in Rajmahal Basin (Cretaceous, West Bengal): Barakar,
Dubrajpur, Rajmahal
(B) Correct chronological order (Paleozoic, Kashmir) should be: Muth Quartzite, Syringothyris
Limestone, Fenestella Shale
(C) CORRECT Bagh Bed, Lameta Formation, Deccan Traps (Cretaceous, Gujarat, MP,
Maharashtra)
http://shodhganga.inflibnet.ac.in/bitstream/10603/91316/1/chapter%201%20introduction.pdf
(D) Correct chronological order (Precambrian, Singhbhum Craton) should be: Older
Metamorphic Gneiss, Singhbhum Granite, Kolhan Group.
Q.33) Match the items listed in Group I with their appropriate description listed in Group II.
Group-I Group-II
P. Peloids 1. Grains containing nucleus surrounded by
irregular, non- concentric laminae.
Q. Ooids 2. Small micritic grains without internal
structure
R. Oncoids 3. Rounded grains with a thin coating of micrite.
S. Cortoids 4. Spherical grains consisting of regular laminae
in concentric rings
(A) P-3, Q-1, R-4, S-2
(B) P-2, Q-4, R-1, S-3
(C) P-3, Q-4, R-1, S-2
(D) P-2, Q-3, R-4, S-1
Solution: (B)
Peloids: These are spherical aggregates of microcrystalline calcite of coarse silt to fine sand size.
Most appear to be fecal pellets from burrowing benthic organisms. As these organisms burrow
through the muddy carbonate-rich sediment, they ingest material in search of nutritional organic
compounds resulting in waste products containing microcrystalline calcite. The peloids are
much easier seen in thin section than in hand specimen because of their small size.
(http://www.tulane.edu/~sanelson/eens212/carbonates.htm)
Ooids: These are spherical sand sized particles that have a concentric or radial internal structure.
The central part of each particle consists of a grain of quartz or other carbonate particle

surrounded by thin concentric layers of chemically precipitated calcite. The layers or coatings
are formed in agitated waters as the grain rolls around.
(http://www.tulane.edu/~sanelson/eens212/carbonates.htm)
Oncoids: Blue-green algal (cyanobacteria) coated grains with irregularly laminated cortex of
mud of any mineralogy.
Cortoids: Cortoids are basically rounded clasts with thin micrite envelopes. The clasts could be
bioclasts, lithoclasts, peloids, or other particles.
Q.34) Which one of the following is an image rectification technique?
(A) Histogram equalization
(B) Density slicing
(C) Histogram normalization
(D) Rubbersheeting
Solution: (D) Rubbersheeting
Histogram equalization is a technique that is used to adjust image intensities to enhance contrast.
Histogram equalization is a nonlinear normalization that stretches the area of histogram with
high abundance intensities and compresses the area with low abundance intensities. This
technique therefore enhances visualization and does not rectify the image.
(https://stackoverflow.com/questions/41118808/difference-between-contrast-stretching-and-
histogram-equalization).
Source: Wikipedia
Density Slicing is also a visualization technique that converts a greyscale values to a series of
intervals of different colours (slices). It is thus an image visualization technique, not image
rectification.

Histogram normalization is also called histogram stretching. It is a is a linear normalization that
stretches an arbitrary interval of the intensities of an image and fits the interval to an another
arbitrary interval (usually the target interval is the possible minimum and maximum of the
image, like 0 and 255). (https://stackoverflow.com/questions/41118808/difference-between-
contrast-stretching-and-histogram-equalization). This technique enhances image visualization
and does not involve any image rectification.
Rubbersheeting: A procedure for adjusting the coordinates of all the data points in a dataset to
allow a more accurate match between known locations and a few data points within the dataset.
Rubber sheeting preserves the interconnectivity between points and objects through stretching,
shrinking, or reorienting their interconnecting lines. This technique thus rectifies the image to
more accurately capture the reality.
https://support.esri.com/en/other-resources/gis-dictionary/term/rubber%20sheeting
Q.35) Match the items listed in Group I with those in Group II.

Group-I Group-II
P. Determination of coefficient of
compressibility of soils
1. Brazilian test
Q. A method of slope stabilization 2. Overcoring
R. A method of in situ stress determination 3. Oedometer test
S. Determination of indirect tensile strength
of rocks
4. Shotcreting
(A) P-4, Q-2, R-3, S-1
(B) P-1, Q-4, R-2, S-3
(C) P-3, Q-2, R-1, S-4
(D) P-3, Q-4, R-2, S-1
Solution: (D)
Brazilian Tet: It is a geotechnical laboratory test for indirect measurement of tensile strength of
rocks.
Overcoring: Methods that measure in situ stress based on the stress relief around the borehole.
Oedometer test: An oedometer test is a kind of geotechnical investigation performed in
geotechnical engineering that measures a soil's consolidation properties. Oedometer tests are
performed by applying different loads to a soil sample and measuring the deformation response.
(wikepedia definition)
Shotcreting: Spraying concrete through a hose at high velocity on a surface.
Group-I Group-II
P. Crevasse 1. River
Q. Yardang 2. Groundwater
R. Mesa 3. Wind 
S. Stalactite 4. Glacier

(A) P-4, Q-3, R-1, S-2
(B) P-3, Q-1, R-4, S-2
(C) P-4, Q-2, R-3, S-1
(D) P-1, Q-2, R-3, S-4
Solution: (A)
Crevasse: Deep, narrow cracks that develop in the brittle zone of a glacier.
Yardang: a sharp irregular ridge of sand lying in the direction of the prevailing wind in exposed
desert regions, formed by the wind erosion of adjacent material which is less resistant.
Mesa: An isolated flat-topped hill with steep sides, found in landscapes with horizontal strata.
Stalactite: a tapering structure hanging like an icicle from the roof of a cave, formed of calcium
salts deposited by dripping water.
Q.37) In the hypothetical isobaric ternary liquidus projection diagram given below, solid phases
A, B, C, D and E exist in equilibrium with liquid. The reaction taking place at the isobaric
invariant point W is
(A) Liquid (at W) = B + D + E
(B) Liquid (at W) = A + B + D
(C) Liquid (at W) + E= B + D

(D) Liquid (at W) + B + D = E
Solution: (D)
This is a peritectic type reaction system. w is an invariant point. Available solids are B, D, and E.
Therefore, option (A) and option (B) can be eliminated because the reactions given in these
options involve phases A and B, which are not present at w.
The reaction at point w can be found out by dividing the diagram into triangular regions as
shown in the figure below.
Note that the liquid composition at w lies outside triangle ABE and BDE, but inside triangle
AEC. Therefore, the solid B and D cannot crystallize out as given in option (C). Only solid E can
crystallize out. Hence, option (D) must be correct. The reaction should be:
Liquid (at W) + B + D = E
Q.38) Match the optical properties listed in Group I with the corresponding mineral in Group II.
Group-I Group-II
P. Brown colour, very high refractive
index, very high birefringence, biaxial
positive
1. Apatite
Q. Colourless, very high refractive index, 2. Rutile

low birefringence, uniaxial negative
R. Deep reddish-brown colour, very high
refractive index, very high birefringence,
uniaxial positive
3. Zircon
S. Colourless, very high refractive index,
very high birefringence, uniaxial positive
4. Titanite
(A) P-4, Q-1, R-2, S-3
(B) P-1, Q-2, R-4, S-3
(C) P-3, Q-4, R-1, S-2
(D) P-4, Q-1, R-3, S-2
Solution: (A)
Fact based, requires no explanation.
Q.39) The reaction:
muscovite + quartz = K-feldspar + sillimanite + water
(A) takes place within the greenschist facies.
(B) takes place within the amphibolite facies.
(C) takes place within the eclogite facies.
(D) takes place within the granulite facies.
Solution: (B)
The given reaction takes place during the metamorphism of pellitic sediments. Sillimanite is not
a stable phase in greenschist facies, so option (A) is eliminated. The reaction takes place along
the curve passing through point 11(blue circle) in the petrogenetic grid for the KMASH system
(see below). This lies in upper amphibolite facies. Granulite facies is marked by appearance of
orthopyroxene (green circle) which forms from the reaction of biotite and quartz.

Petrogenetic grid for the system KFMASH. After Spear (1999)
Group-I Group-II
P. Diopside-tremolite-forsterite 1. Pelite (low P, high T)
Q. Talc-phengite-kyanite 2. Metabasite (low P, high T)
R. Hornblende-cummingtonite-plagioclase 3. Calc-silicate (moderate P, T)
S. Andalusite-cordierite-biotite 4. Pelite (high P, low T)
(A) P-3, Q-4, R-1, S-2
(B) P-1, Q-2, R-4, S-3
(C) P-3, Q-4, R-2, S-1
(D) P-1, Q-2, R-3, S-4
Solution: (C)

Factual
Q.41) The figure below is a schematic section showing the initial stages of development of a
thrust fault (FF’) having a typical ramp and flat geometry, with the thrust sheet being transported
from east to west. With respect to the synform and antiform created in Stage 2, which one of the
options below is CORRECT for the next increment of movement on the fault plane?
(A) The synform and the antiform will both move westward.
(B) The synform will remain in position, while the antiform will grow in amplitude.
(C) Both synform and antiform will grow in amplitude.
(D) The geometry will remain unchanged.
Solution: (B) The synform will remain in position, while the antiform will grow in amplitude.
This is a fault bend fold, the stage c will be as shown in the figure below. Anticline (red circle)
grows in amplitude and the synform (green circle) remains in position as the fold grows.

Q.42) Which one of the following is the CORRECT chronological sequence for Iron formations?
(A) Algoma type > Superior type > Rapitan type > Minette type
(B) Superior type > Algoma type > Rapitan type > Minette type
(C) Rapitan type > Minette type > Algoma type > Superior type
(D) Algoma type > Minette type > Superior type > Rapitan type
Solution: (A) Algoma type > Superior type > Rapitan type > Minette type
Algoma-type is the oldest (Archaean) and seems to be associated with volcanic arcs.
Superior-type formed on stable continental shelves during the paleo-proterozoic.
Rapitan-type formed due to environmental changes associated with glaciation during the neo-
proterozoic.
Minette type formed during the Jurassic in fluvial/marine setting.
Therefore, the correct chronological sequence should be:
Algoma type > Superior type > Rapitan type > Minette type
Option (A) is correct.

Q.43) Assertion (a): High-temperature, low-pressure metamorphism occurs on the over-riding
plate near convergent plate margins.
Reason (r): Partial melting in the mantle wedge generates magmas that rise to form the arc.
(A) (a) is true but (r) is false.
(B) (a) is false but (r) is true.
(C) Both (a) and (r) are true and (r) is the correct reason for (a).
(D) Both (a) and (r) are true and (r) is not the correct reason for (a).
Solution: (C)
The magma generated by partial melting of the subducting plate rises to the over-riding plate and
produces high T, low P metamorphism. Therefore, the assertion that high T, low P
metamorphism occurs on the over-riding plate near convergent plate margins is correct. r is the
correct explanation for the assertion a. Option (C) is correct.
Q.44) Two coeval primary aqueous biphase fluid inclusions, X (liquid-rich) and Y (vapour-rich),
occur in the same grain of the host mineral. Which one of the following situations most likely
indicates boiling of the fluid?
(A) X homogenizes to liquid and Y homogenizes to vapour at different temperatures.
(B) Both homogenize to liquid at the same temperature.
(C) Both homogenize to vapour at the same temperature.
(D) X homogenizes to liquid and Y homogenizes to vapour at the same temperature.
Solution: (D)
During boiling, liquid phase and gaseous phase co-exist at the same temperature. Either a single
phase liquid inclusion (X) or a single phase gaseous inclusion (Y) can be trapped during boiling.
This inclusion will subsequently separate into two phases during cooling (the liquid inclusion
will be liquid rich and the gaseous inclusion will be vapour rich). Hence, during homogenization,
X will homogenize to liquid and Y will homogenize to vapour at the same temperature, the
boiling point.
Q.45) During bench blasting in a quarry, 50 kg of an explosive with a yield of 5 MegaJoule/kg is
required to break 100 m3
of marble. In this case, the energy expended in breaking a unit volume
of marble in MegaNewton/m3
would be _________________.

Solution: 2.5 MegaNewton/m3
Total energy expended = mass X yield = 50 X 5 MegaJoules
Total Volume = 100m3
Energy required for breaking a unit volume = (5 X 50)/100 = 2.5 MegaNewton/m3
Q.46) The stretching lineation on the axial plane (S2) of a reclined fold on the S1 foliation makes
an angle of 30o
with the S1/S2 intersection lineation. The rake of the stretching lineation on the
axial plane in degrees is ______________.
Solution: 60 degrees
Since the fold is a reclined fold, the hinge line dips along the axial plane with the same plunge as
the dip of the axial plane. Intersection of the foliation (S1) and the axial plane (S2) is basically
the hinge line. The stretching lineation forms an angle of 30o
with the hinge line. It is clear from
the figure given below that the rake of the stretching lineation on the axial plane is 60 degrees.
Remember, rake of a line on a given plane is the angle made by the line with the strike of the
plane, measured on the plane itself.
Q.47) A basaltic magma has an initial nickel concentration of 300 ppm. Olivine crystallizes from
this magma by equilibrium crystallization (Case I) or fractional crystallization (Case II). Then,
the absolute value of the difference between the nickel concentrations of the liquids remaining
after 25% crystallization in these two cases is _______________.
(Use KD,Ni olivine/melt = 10).
Solution: 69.78
During equilibrium crystallization the newly formed crystals are in equilibrium with the melt and
Strike of plane
Axial Plane
(S2)
Stretching lineation on
the axial plane
Axial Plane (S2)
Foliation (S1)
Intersection of
S1/S2
Stretching
Lineation
Rake of
stretching
lineation on
S2 = 60 deg
30 deg

the crystal composition keeps on varying until the crystallization has reached completion. In
contrast, during fractional crystallization the crystals are removed from the system as soon as
they are formed.
CASE I:
Equilibrium crystallization model is described by the same equation as that of batch melting:
Vº
V»
=
[
Y¼ [`½ ª½
where F is weight fraction of solid, and is equal to 1 – weight fraction of liquid.
For 0.25 liquid fraction, F = 1 - 0.25 = 0.75
Vº
¾««
=
[
[« [`.¿À ª.¿À
n¢ˆ‡"q
𝐶Â =
¾««
¾.‚À
= 92.30
CASE II:
Fractional Crystallization model equation:
Vº
V»
= 𝐹(Y¼`[)
𝐶Â
300
= 0.75([«`[)
n¢ˆ‡"q
𝐶Â = 300×. 75Ä
= 22.52
Difference in CL between CASE I and CASE II = 92.30 – 22.52 = 69.78
Q.48) The difference in the number of faces in forms {hkl} and {111} in the holosymmetric
class of the isometric system is _______________.
Solution: 40
The Isometric crystal system has three equivalent crystallographic axes perpendicular to each
other. The objects of this crystal system appear as if they could be fit into a sphere. This system
has 6 crystal classes. The holosymmetric class is characterised by following symmetry
operations: 4/m 3* 2/m.
The {111} class is an octahedron possessing 8 faces. The {hkl} class is a Hexakisoctahedron
possessing 48 faces. In {111} class we generate a plane cutting all three axes at equal intercepts
(a : a : a) and apply the symmetric operations to form a closed figure 8 faces. In {hkl} class we
generate a plane cutting the three axes are unequal intercepts (a : na : ma) and apply the
symmetric operations to form a closed figure with 48 faces.

Octahedron (8 faces) Hexakisooctahedron (48 faces)
Q.49) An inclined cylindrical confined aquifer has coefficient of permeability of 40 m/day. The
horizontal distance between two vertical wells penetrating the aquifer is 800 m. The water
surface elevations in the wells are 50 m and 45 m above a common horizontal datum. The
absolute value of Darcy flux through the aquifer is ______________ m/day.
Solution: 0.25
𝑄
𝐴
=
𝐾(ℎ‚ − ℎ[)
∆𝐿
Where Q/A is the darcy flux. Given,
K = 40m/day
∆𝐿 = 800m
h2 = 50m and h1 = 45m
Substituting in the Darcy’s equation
𝑄
𝐴
=
40(50 − 45)
800
= 0.25𝑚/𝑑𝑎𝑦
Q.50) The mass and volume of a natural soil sample are 2.1 kg and 1×10-3
m3
, respectively.
When fully dried, the mass of the soil sample becomes 2 kg without any change in volume.
Assuming the specific gravity of soil particles to be 2.5, and water density of 1000 kg/m3
, the
degree of saturation of the natural soil sample is _______________%.

Solution: 50%
mbulk = mmatrix + mvoid
voids are filled with water and air, hence
mbulk = mmatrix + mwater + mair
mbulk = 2.1 kg, mmatrix = 2kg, mair = 0
mwater = 𝜌ÆO#ˆ…×𝑉ÆO#ˆ… = 𝜌ÆO#ˆ…×𝑉”Ç¢"×𝑆Æ = 1000(𝑉Èo‡É − 𝑉pO#…¢Ê)𝑆Æ
2.1 = 2 + 1000(10`¾
−
2
2500
)𝑆Æ
0.1 = (1 −
‚
‚.À
)𝑆Æ. This gives
Sw = 0.5, or Sw = 50%
Q.51) For a granitic rock mass, joint set number (Jn) = 9, joint water reduction factor (Jw) = 1,
joint alteration number (Ja) = 1, stress reduction factor (SRF) = 1, rock quality designation (%) =
84 and joint roughness number (Jr) = 3. The Q-value as per Barton’s Q-system of rock mass
classification (year 1974) is _______________.
Solution: 28
𝑄 =
𝑅𝑄𝐷
𝐽†
×
𝐽…
𝐽O
×
𝐽Æ
𝑆𝑅𝐹
substituting the respective values provided in the problem statement:
𝑄 =
84
9
×
3
1
×
1
1
= 28
Q.52) A sun synchronous satellite is at an altitude of 300 km and the spectrometer makes an
angular coverage angle of 12o
. The Swath (GFOV) of the satellite is ______________ km.
Solution: 63.77 km

𝐺𝐹𝑂𝑉 = 𝐻×𝑡𝑎𝑛𝜃 = 300×𝑡𝑎𝑛12° = 63.77 𝑘𝑚
Alternatively:
since 𝜃 is small, tan𝜃 = 𝜃 in radians
Therefore,
𝐺𝐹𝑂𝑉 = 300×
12
360
×2𝜋 = 100×2×3.14 = 62.8 𝑘𝑚
Q.53) The stability field boundary between two minerals A and B is linear with a positive slope
in P-T space. The molar entropy of A and B are 85.5 and 92.5 Joules K-1
, respectively and their
respective molar volumes are 35.5 and 38.2 cc. The slope of the phase boundary in P-T space is
_____________ bar K-1
.
Solution: 25.9 bar K-1
According to Clapeyron equation:
𝑑𝑃
𝑑𝑇
=
∆𝑆
∆𝑉
∆𝑆 = 92.5 – 85.5 = 7 Joules K-1
since 𝑉 ∝ 𝑛
∆𝑉 = 38.2 – 35.5 = 2.7 cc = 2.7 X 10-6
m3
slope of phase boundary in P-T space = 7/2.7 * 106
Pa = 2.59 * 106
PaK-1

Since 1 bar = 105
Pa
Slope = 25.9 bar K-1
Q.54) Five moles of gas A (volume V1) and 3 moles of gas B (volume V2) were kept in separate
containers. These two gases are completely transferred to a new container of volume V.
Assuming isothermal condition, and that the work done is only mechanical, the entropy change
of the system is ____________ Joules K-1
. (R = 8.3 J K-1
mole-1
)
Solution: 43.9
For isothermal expansion of gas, the change in entropy of the system is given by:
∆𝑆 = 𝑛𝑅𝑙𝑛(
𝑉‚
𝑉[
)
V2 = final volume, V1 = initial volume
Since, 𝑉 ∝ 𝑛,
For gas A, entropy change is:
∆𝑆U = 5×8.3×𝑙𝑛
8𝑉
5𝑉
= 24.4 𝐽𝐾`[
For gas B, entropy change is:
∆𝑆X = 3×8.3×𝑙𝑛
8𝑉
3𝑉
= 19.5 𝐽𝐾`[
∆𝑆#Ç#O‡ = ∆𝑆U + ∆𝑆X
∆𝑆#Ç#O‡ = 43.9 𝐽𝐾`[
Q.55) The value of Eh corresponding to the upper limit of natural surface aqueous environment
at pH of 8.0 is _____________ V.
Solution: 0.76V
The Eh

The stability region of water is given by the pourbaix diagram. Below the lower limit, water is
reduced to hydrogen according to the equation:
2H2O + 2e-
-> H2 + 2OH-
Above the upper limit, water is oxidised to equation according to the equation
O2 + 2H2O + 4e-
-> 4OH-
The equation of the straight line marking the upper limit is:
Eh = (1.229 – 0.0591 X pH) volts
We have been given a pH = 8
Therefore,
Eh = (1.229 – 0.0591 X 8) volts
Eh = 0.7562V

Gate 2018 GG Solution (Geology Option)

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