Numericals
The wt.% (correct to two decimal places) of Cu in chalcopyrite (CuFeS2) (atomic weight of Cu=63.55, Fe=55.85, S=32.07) is …………
Formula of chalcopyrite = CuFeS2,
atomic number of chalcopyrite = 63.55 + 55.85 + 64.14
= 183.54
wt% of Cu in chalcopyrite = (atm. weight of Cu in CuFeS2) / (total atomic weight of CuFeS2 )
= 63.55/183.54
=34.62 Ans.
The general formula of an amphibole mineral is A0-1B2C5T8O22(OH)2, where A, B, C and T are cationic sites with different co-ordination numbers as stated as: A=12, B=6-8, C=6, T=4.
The amount of octahedral Al in an amphibole of composition Na0.6Ca2Mg3.0Al3.0Si6.2O22(OH)2 is ……………….
Al can occur in 4 (tetrahedral) as well as 6 (octahedral) coordination number.
That means Al can found in C (6 coordination) and T (4 coordination)
Al and Si total shares 8 tetrahedral sites, out of those Si already takes 6.2 site as given in formula, so remaining tetrahedral Al site = 8-6.2 = 1.8
Total Al site (given 3.0) = Al in tetrahedral + Al in octahedral
3.0 = 1.8 + Al in octahedral
Al in octahedral = 1.2 Ans.
In the block diagram, the net slip (=100 m) is resolved into strike slip (s) and dip slip (d) components. The value (in m, correct to two decimal places) of “s” is …………
Here simple triangle formula can be used –
Cos 300 = s/100
√3/2 = s / 100
s= 50 √3
s = 86.6 m Ans.
The true thickness (t, in m) of bed B in the given diagram is ………..
Here true thickness is t
If (326) is the Miller Index of a crystal face, then the value of x in the corresponding Weiss Parameter of the same face, xa: yb: zc is ……………..
As we know that Miller index and Weiss parameters are inversely proportional
so, Weiss parameter = 1/3 1/2 1/6
to remove denominator multiply it with 6
Weiss parameter = 6/3 6/2 6/6
= 2 3 1
xa : yb : zc = 2 : 3 : 1, x = 2 Ans.
The value of h in the Miller-Bravais Index (4̅1h0) is …………….
When a four number index notation is used. The first three must sum to zero –
-4 + 1 + h = 0
h = 3 Ans.
In an ocean basin, a 4 Ma old oceanic crust lies 40 km away from the ridge axis. The average velocity (in cm/yr) of the oceanic lithosphere is ………………..
Average velocity = distance covered/ time taken
V = 40*100000cm/4*1000000yr
V = 1cm/yr Ans.
An aquifer has a cross sectional area of 1000 m2 and a hydraulic gradient of 0.01. If water is flowing from the aquifer at a rate of 10 m3/sec, the hydraulic conductivity (in m/sec) of the aquifer is …………..
Formula used , q = KA (dh/dl)
where, q = volumetric rate
K = hydraulic conductivity
A = Cross-sectional area of sand pack
dh/dl = hydraulic gradient
so according to formula,
10 = K * 1000*0.01
K = 1 Ans.
According to the mineralogical phase rule, the number of minerals that can coexist at equilibrium in a 8 component chemical system with 2 degrees of freedom is …………
Phase rule , F = C – P + 2
F = degree of freedom
C = Number of component
P = Number of phases
so, 2 = 8 – P + 2
P = 8 Ans.
The grain density (of solids only) and bulk density (solids + voids) of a sandstone sample are 2.7 gm/cm3 and 2.3 gm/cm3, respectively. The total porosity (in %, correct to two decimal places) of the sample is ………..
There is a formula,
Porosity = (1- bulk density/particle density) *100
Porosity = (1- 2.3/2.7)*100
Porosity = (1 – 0.8518) *100
Porosity = 0.1481*100 = 14.81 Ans.
In an undeformed and normal stratigraphic succession, a dolerite dyke was emplaced before deposition of sandstone B. The difference between the maximum ages (in Myr) of deposition of sandstone A and sandstone B is ……..
Here you have to first understand the depositional sequence of the given diagram.
Here first the lower most part Volcanic ash bed is deposited at 132ma than sandstone A is deposited, now dolerite dike emplaced because it is cutting both of the bed but this dyke is older than the sandstone B as it is not cutting sandstone B. at last Volcanic ash bed 2 is deposited at the age of 62ma.
Age difference = 132 -75 = 57 Ans.
A coal seam occurs in a stratigraphic sequence as shown in the figure. If a vertical borehole is drilled at location B, the coal seam will be intersected at a depth (in m) of ………….
Here our bore hole is at 500m elevation and our coal seem intersecting contour at 400m so the bore hole depth is 100m. Ans.
A set of sedimentary rocks A, B and C are affected by a fault F-F. The amount of vertical throw (in m) along the fault is __
As you can see that on the left hand side 500m contour inrsecting and same bed on the left is intersecting at 400m just after the fault. So the vertical throw = 100m Ans.
The retardation of a uniaxial negative mineral of thickness 0.03 mm is 5160 nm in its principal section of indicatrix. If the refractive index corresponding to the E-ray is 1.486, the value of the refractive index (correct to three decimal places) of the O-ray is __
Using formula, Retardation = thickness * birefringence
5160*10-6mm = 0.03 mm * birefringence
birefringence = 5160*10-6/0.03
birefringence = 0.172
As it is an uniaxial negative mineral, so RI of E-ray will be less than RI of O-ray
birefringence = (RI of O-ray) – (RI of E-ray)
0.172 = (RI of O-ray) – 1.486
RI of O-ray = 0.172+1.486
RI of O-ray = 1.658 Ans.
A spherical ore body (diameter = 40m) has 7% metal content and density of 3300 kg/m3 . The reserve (in tonne) of the ore body is –
Volume = (4/3)πr3 = (4/3)*(22/7)*203 = 33520 m
Reserve = volume * density
Reserve = 33520*3300
Reserve = 110616000 kg = 110616 tonne Ans.
From the data shown in the table, the weighted mean size (in micrometer, correct to two decimal places) of the sediment population is ……..
Grain size (micrometer)———— Dry sediment weight (in gram)
4 ——————————————————————–50
20 ——————————————————————-75
40 ——————————————————————125
60 ——————————————————————–50
Here simple mathematical formula will be used ,
that is Σ(grain size*grain weight)/Σ grain weight
Weight mean size = (4*50 + 20*75 + 40*125 + 60*50)/(50 + 75 + 125 + 50)
= 9700/300
= 32.33 Ans.
Consider the schematic isobaric T-X phase diagram in the binary forsterite (Fo)-fayalite (Fa) chemical system. If there is equilibrium crystallization of melt (L), the wt.% of olivine crystallized from a melt of composition “a” at a temperature T2 is …………………
Consider a schematic isobaric ternary phase diagram A-B-C, shown below. The diagram, which is contoured with isopleths of liquidus temperatures (in °C), reveals crystallization behaviors of melt (L) of different compositions during cooling. When a melt of composition “a” lies at a temperature of 1800°C, the variance (or degree of freedom) of the magmatic system is ………….
As it is an isobaric so phase rule will be
F = C – P + 1
at point a phase will be only one that is liquid and components are three
F = 3 – 1 + 1
F = 3 Ans.
An eclogite consists of garnet (60%) and omphacite (40%), where the mineral abundances are in mole %. XMg [=Mg/(Mg+Fe2+)] of garnet and omphacite is 0.50 and 0.75, respectively. The XMg of eclogite is ………………..
Xmg of eclogite = (mole of garnet * 0.6) + (mole of omphacite * 0.75)
Xmg of eclogite = 0.5*0.6 + 0.75*0.4
Xmg of eclogite = 0.3 + 0.3 = 0.6 Ans.
A harzburgite contains pure forsterite and pure enstatite in a molecular ratio of 60:40. The mole % of MgO in the rock is ………………
forsterite (Mg2SiO4)constituent = 2 mole of MgO + 1 mole of SiO2
Enstatite (MgSiO3) constituent =1 mole of MgO + 1mole of SiO2
MgO in forsterite = (2/3)*60 = 40 %
MgO in enstatite = ( 1/2)*40 = 20%
total MgO % in the rock = 40+20 = 60% Ans.
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