The value of φ (phi) of a sediment grain having a diameter of 0.125 mm is ……….. (In integer)
phi (ϕ) = – log2d
we can write it as, d = 2-ϕ
0.125 = 2-ϕ
1/(2*2*2) = 2-ϕ
2-ϕ = 1/8 = 2-3
ϕ = 3 Ans.
The vertical separation of a displaced horizontal stratum along a dip-slip reverse fault is 10 m when measured on a section perpendicular to the fault-strike. If the dip of the fault is 30⁰, the net slip of the fault will be ……………. m. (In integer).
As per the question, the dip is 300, and vertical separation = 10m is drawn in the given diagram.
Where ‘x’ is the net slip
We can simply use right angle triangle formula for calculation.
x = 20m Ans.
The dips of the normal and overturned limbs of a horizontal-overturned antiform are 30⁰ and 70⁰, respectively. The interlimb angle of this fold is ……… degrees. (In integer)
The fold OAB is drawn in the image.
As per question, one dip ∠CAO = 300, Another dip ∠CAB = 700
The interlimb angle is the angle between the two limbs that is ∠OAB
∠OAB = ∠CAB – ∠CAO
∠OAB = 70-30 = 400 Ans.
In a mineral with formula KAl3Si3O10(F0.5OHx), the value of ‘x’ is ………… (Round off to one decimal place)
The given formula is belongs to mica group of members, The general formula is [ XY2-3Z4O10(OH)2]
In the last there must be two atoms of OH, but in the question there is also F [flourine] with 0.5 atoms
toal atoms = atoms of F + atoms of OH
2 = 0.5 + atoms of OH
atoms of OH = 2-0.5 = 1.5 Ans.
The atom percent of Fe in pyrrhotite of composition Fe0.77S is ……….. (Roundoff to two decimal places)
Given, Number of Fe atoms = 0.77
Number of total atoms = 0.77+1 = 1.77
% of Fe in Pyrrotite = (Number of Fe atom/ total atoms)*100
% of Fe in Pyrrotite = (0.77/1.77)*100 = 43.50% Ans.
Consider the univariant metamorphic reaction Albite = Jadeite + Quartz. The minimum number of chemical components required to describe the composition of all the phases is …….. (In integer)
For the univeriant metamorphic reaction F = 1
No. of compoents = C
No. of phases = 3 [albite, jadite, quartz]
According ot Gibbs phase rule , F = C – P + 2
1 = C -3 + 2
C =2 Ans.
The mean flow velocity of water in an open channel having an average depth of 0.2 m, and with Froude Number 4, is ……….. m/s. (Round off to one decimal place) (Use g = 9.8 ms-2)
Froud number = Inertial force / Gravitational force
Froud number = V / √ gd
4 = V/ √ 9.8*0.2
V= 4*√ 1.96
V = 4*1.4 = 5.6 Ans.
An aquifer has a cross-sectional area of 10 m2 and a hydraulic conductivity of 0.25 cm/s. The volume of water that will flow per second through the aquifer for a hydraulic gradient of 0.04 is …………… cm3. (Round off to three decimal places)
Darcy’s law [Q] = KIA
Where, Q = flow volume
K = hydraulic cunductivity
I = hydraulic gradient
A = cross sectional area = 10m2 = 100000cm2
Q = 0.25*0.04*100000 = 1000cm2 Ans.
The geothermal gradient in the continental crust is 0.02 ⁰C/m. If the surface temperature is 25 ⁰C, the temperature at a depth of 18 km from the surface is
………. ⁰C. (In integer)
GIVEN, Geothermal gradient = 0.02 ⁰C/m = 20 ⁰C/km
Temperature increase for 18 km = gradient *depth
Temperature increase for 18 km = 20*18 = 360 ⁰C
Temperature at 18km = surface temperature + temperature increase
Temperature at 18km = 25 + 360 = 385 ⁰C Ans.
The area of a triangular block of a massive orebody is 1500 m2. If the thickness of the orebody is 5 m, 6 m and 7 m at the three corners of the triangular block, and the ore density is 2.5 tons/m3, the estimated ore reserve of the block is ……….. tons. (In integer)
GIVEN that the triangular ore body has different thickness at the three corners. To calculate the total volume we can multiply the area with average thickness.
Average tihickness of the ore body = (5+6+7)/3 = 6m
Volume of the ore body = area *thickness =1500*6
Volume of the ore body = 9000 m3
Ore reserve = volume * density
Ore reserve = 9000 * 2.5 = 22500 tons Ans.
Clinopyroxene crystallizing from a basaltic magma has Sm concentration of 24 ppm. If the clinopyroxene-melt partition coefficient for Sm is 1.2, the concentration of Sm in the basaltic magma will be …………. ppm. (In integer)
GIVEN, partition coefficient = 1.2
Sm concentration in the crystal = 24 ppm
We use the formula, Partition coefficient = Conc. in solid/ Conc. in liquid
1.2 = 24 / Conc. in liquid
Conc in liquid = 24/1.2 = 20 ppm Ans.
The lithostatic pressure at a depth of 36.5 km in the continental crust having an average density of 2800 kg/m3, is ………….. GPa. (Round off to the nearest integer) (Use g = 9.8 m/s2)
GIVEN, depth = 36.5 km = 36500
density = 2800 kg/m3
gravity = 9.8 m/s2
Pressure = density * geavity * height
P = 2800 * 9.8 * 36500
P = 1001560000 Pascal = 1.001 GPa = 1 GPa Ans.
The fraction of 2411𝑁𝑎 atoms remaining after a decay interval of 5.0 hours will be ………….. (Round off to three decimal places) (Use t1/2 = 15.0 hours)
Here we will use two formulas , t1/2 = T/n ——–[1] and y = 1/2n ———[2]
Where, t1/2 = half life,
T = time passed
n = number of half lifes
y = fraction of radioactive material left
number of half lifes, from formula 1, 15 = 5/n
number of half lifes [n] = 5/15 = 0.33 hr
remaining fraction , from formula 2, y = 1/20.33 = 1/1.257 = 0.795 Ans.
The thickness of a dipping coal bed measured along a vertical drill hole is 15 m. If the dip of the coal bed is 30⁰, the orthogonal thickness of the coal bed is
………. m. (Round off to the nearest integer)
Orthogonal thickness is the perpendicular thickness or the true thickness.
sin 600 = x/15
√3/2 = x / 15
x = 15√3/2
x = 12.99 m x= 3m Ans.
The mole fraction of forsterite in olivine with MgO = 29.17 weight %, FeO = 34.65 weight % and SiO2 = 36.18 weight % is ……………. (Round off to two decimal
places) (Use molecular weight, in g/mol, of MgO = 40.31, FeO = 71.85 and SiO2 = 60.00)
A partially saturated soil sample has a volume of 1200 cc. The volume of water present in the sample is 300 cc. The mass of solid in the sample is 1908 g and the particle density is 2.65 g/cc. The porosity (n) of the soil sample is ………….. %. (In integer)
GIVEN, total volume = 1200cc
Particle density = 2.65 g/cc
solid mass or particle mass = 1908 g
Particle/solid volume = mass/density
particle volume = 1908/2.65 = 720 cc
Volume of void = total volume – volume of particle
volume of void = 1200 – 720 = 480 cc
Porosity = [volume of void/ total volume ]*100
porosity = [480/1200]*100 = 40 % Ans.
A rock element during deformation, experienced a pressure change of 5×104 N/m2, due to which its volume changed from 4 cm3 to 3.9 cm3. The bulk modulus of the rock is ……….. ×106 N/m2. (In integer)
Bulk modulus = stress / volumetric strain
stress = 5×104 N/m2
strain = change in volume / initial volume
strain = 0.1 / 4 = 0.025
Bulk modulus = 5×104 / 0.025
Bulk modulus = 200*104 N/m2 = 2*106 N/m2 Ans.
For an anisotropic crystal of thickness 0.04 mm and refractive indices of 1.636 and 1.486 along the slow and fast directions, respectively, the retardation
produced is ………… nm. (In integer)
Retardation = thickness * birefrengence
Thickness = 0.04 mm
Birefrengence = difference in RI = 1.636 – 1.486 = 0.15
Retardation = 0.04*0.15
Retardation = 0.006 mm = 0.006*10-6 nm = 6000 nm Ans.
An orebody contains pyrite and chalcopyrite in the same molar proportions. The percentage concentration of Cu in the ore will be ……………… (Round off to the nearest integer) (Use atomic weight, in g/mol, of Cu = 63.55, Fe = 55.85, S = 32.06)
Atomic weight of pyrite [FeS2]= 55.85 + 32.06*2 = 119.97 g/mol
Atomic weight of chalcopyrite [CuFeS2] = 63.55 + 55.85 + 32.06*2 = 183.52 g/mol
given, atomic weight of Cu = 63.55 g/mol
Total atomic weight of ore body = 119.97 + 183.52 = 303.49 g/mol
The percentage concentration of Cu = [at. weight of copper/ at. weight of ore body]*100
The percentage concentration of Cu = [63.55 / 303.49]*100 = 20.9 % = 21 % Ans.
In the given isobaric binary temperature-composition (T-X) phase diagram involving solids A and B, the fraction of melt remaining at point Q for a magma
having initial composition P will be …………… (Round off to one decimal place)
The horizontal axis of figure is divided into 5 parts representing each dotted line for 20 %
melt fraction = PB /QB = 20/40 = 0.5 Ans.
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