Numericals
The intensity of an earthquake of magnitude 8 on the Richter scale is greater than the intensity of an earthquake of magnitude 5 on the same scale by ………….. times
One point increase in is increase of 10 times of the previous one. Here it is increasing increasing by 3 point so intensity will be 1000 times answer Ans.
Assume: (i) geothermal gradient = 25 °C/km in the crust, (ii) density of the crustal rocks = 3000 kg/m3 and (iii) acceleration due to gravity = 10 m/s2. Based on these values, the lithostatic pressure at a point where temperature is 400 °C will be ……………. MPa
Here is the simple formula of lithostatic pressure = rho*g*h
here h will be calculated with the help of geothermal gradient –
geothermal gradient = temperature at depth h / depth (h)
25 = 400 /h
h = 16 km = 16000m
Lithostatic pressure = 3000*10*16000
Lithostatic pressure = 480000000 Pa
Lithostatic pressure = 480 MPa Ans.
The radii of A+2 and B–ions are 1.12Å and 1.31Å, respectively. The coordination number of A+2 in mineral AB2 is …………
Coordination number can be found by radius ratio rule –
Rcation / Ranion = 1.12/1.31
= 0.854
coordination number will be 8 Ans.
In a sedimentary rock, the diameters of two grains A and B are 1𝜙 and 3𝜙, respectively. The difference in diameters between A and B is…………… mm (rounded off to two decimal places).
phi (ϕ) = – log2d
we can write it as, d = 2-ϕ
dA= 2-1 = 1/2 = 0.5
dB = 2-3 = 1/2*2*2 = 1/8 = 0.125
Difference in diameter = 0.5 – 0.125 = 0.37mm Ans.
A light ray passes from mineral A to mineral B having refractive indices of 1.750 and 1.430, respectively. The limiting value of angle of incidence above which the light ray undergoes total internal reflection is ……………. degree (rounded off to one decimal place)
RIA/RIB = sin(r)/sin(i)
for total internal reflection r = 900
so, 1.75/1.43 = sin(90)/sin(i)
sin (i) = 1.43/1.75 = 0.817
sin(i) = 0.817
i = 54.7 Ans.
Throw and heave of a bed offset by a normal fault are 100 m and 200 m, respectively. The dip of the fault plane is ………………….. degree (rounded off to one decimal place)
given heave = 200m, throw = 100m
fault plane angle = θ
tan(θ) = throw / heave
tan (θ) = 100/200
tan(θ) = 1/2
θ = 26.5 Ans.
The difference between Si/O ratios of K-feldspar and olivine is ……………… (answer in two decimal places).
Si:O in K-feldspar (KAlSi3O8) = 3/8
Si:O in olivine (Mg,Fe,)2SiO4 = 1/4
difference = 3/8 -1/4 = 0.125 Ans.
2nd method
k-feldspar is tectosilicate so Si:O =1:2
olivine is nesosilicate so Si:O = 1:4
difference = 1/2-1/4 = 1/4 = 0.25 Ans.
If a crystal contains 5 faces and 8 edges, the number of vertices in the crystal is …………………
Euler’s formula (F + SA = E +2)
where , F =number of faces
SA = number of solid angles
E = number of edges
so, 5 + SA = 8 +2
SA = 8-5+2 = 5 Ans.
At a temperature of 298.15 Kelvin, the free energy change of a reaction (ΔG0) is 19.737 kCal/mole. If the universal gas constant (R) = 1.98717 Calorie/degree/mole, the log10 of the equilibrium constant K is ……………………. (rounded off to two decimal places).
There is a formula , ΔG0 = -RT*ln(K)
where K is the equilibrium constant
ln(K) can be converted into log base 10 by following equation –
ln(K) = 2.303 *log10(K)
the formula becomes , ΔG0 = -2.303*RT*log(K)
19.737 = -2.303*1.98717*298.15*log(K)
log(K) = -14.46 Ans.
A normal fault displaces a sandstone bed such that the dip-slip and the strike-slip components are 3 m and 4 m, respectively. The net-slip of the displacement is ……………… m
From the diagram we know that strike slip and net slip forms a right angle triangle,
so net-slip = √ [(strike-slip)2 +(dip-slip)2]
net-slip = √ [16+9]
net slip = 5m Ans.
In the given diagram, a 6 km high plateau is supported by a crustal root of thickness r. The system is in isostatic equilibrium as per Airy’s hypothesis of isostasy. Densities of the crust and the mantle are 2800 kg/m3 and 3100 kg/m3, respectively. Assuming the acceleration due to gravity to be same throughout the region, the thickness of the root (r) is ………………. km
At the depth of compensation pressure (rho*g*h) will be equal –
so, pressure below crust = pressure below plateau
(2800*g*10) + (3100*g*r) = 2800*g*(10+r+6)
2800*10 +3100r = 2800*10 + 2800r + 2800*6
300r = 2800*6
r = 56km Ans.
A 50 kg granite boulder gets dislodged from a cliff of height 20 m and undergoes an absolute vertical free fall. If the acceleration due to gravity is 10 m/s2 , the boulder will hit the ground with a velocity of ………………. m/s.
Here static energy of granite boulder will be converted into kinetic energy,
so, mgh = (1/2)mv2
50*10*20 = (1/2)*50*v2
v2 = 400
v = 20 m/sec Ans.
Mass and volume of a fully dried soil sample are 500 g and 250 cm3, respectively. The average density of the particles in the soil sample is 2.5 g/cm3. The void ratio of the soil sample is ……………………%
Given that, total volume of dried soil mass = 250 cm3
mass of the dried sample = 500 g
particle density of the soil = 2.5 g/cc
volume of particles = mass of particle/ density of particle
volume of particles = 500/2.5
volume of particles = 200 cm3
volume of void = 250-200 = 50cm3
void ratio (e) = volume of void / volume of particle
e = 50/200
e = 0.25
e = 25% Ans.
The geological map shows the contact between sandstone and limestone. The two dotted curves are the contours of 400 m and 500 m, respectively. The difference between the dip angles of the contact surface along the AB and AC directions is …………………… degree (rounded off to two decimal places).
Formula is , Dip = vertical interval/ horizontal distance
Dip (AB) = tan-1(100/200) = tan-1(1/2) = 26.56
Dip (AC) = tan-1(100/300) = tan-1(1/3) = 18.43
Required difference = 26.56-18.43 = 8.13 Ans.
A tabular ore body of 9 km2 area and an average thickness of 9 m has a density of 3000 kg/m3. The tonnage (in million tonnes) of the ore body is ……………
Tonnage of any ore body = volume * density
tonnage = (9*1000) * 9*3000
tonnage = 243*109 kg
tonnage = 243 million tons Ans.
Assume that the orbit of the earth is a circle of radius 150×106 km. The gravitational constant and the earth’s orbital velocity are given as 6.7×10-11 Nm2/kg2 and 30×103 m/s, respectively. The calculated mass of the sun is ………………×1030 kg (rounded off to two decimal places).
Now here, Centripetal force will be equal to the gravitational force
Gmems/r2 = meV2/r
where, G = gravitational constant
V = Earth’s Orbital velocity = 30*103m/s
me = mass of earth
ms = mass of sun
r = radius of the circle or orbit = 150*106 km = 150*109 m
so, Gms/r = V2/
ms(6.7*10-11 /150*109) = (30*103)2
ms (6.7*10-11 /150*109) = 900*106
ms = 900*106*150*109/6.7*10-11
ms = 420149.2 *1026
ms = 2.01*1030 Ans.
The difference between XAn and XAb in a feldspar represented by X in the given triangular diagram is …………….. (answer in one decimal place).
The lines in the figure are at 20 interval.
At the bottom right there is complete anorthite is so at the point X it is 0.4
At the bottom left there is complete albite so at the point X it is 0.2
The required difference is = 0.4- 0.2 = 0.20 Ans.
Two vertical wells penetrating a confined aquifer are 200 m apart. The water surface elevations in these wells are 35 m and 40 m above a common reference datum. The discharge per unit area through the aquifer is 0.05 m/day. Using Darcy’s law, the coefficient of permeability is …………………..m/day.
According to the Darcy law, Q/A = KI
Given that , Q/A = 0.05m/day
I = head difference / distance
I = 5/200
So, K = Q/AI
K = 0.05*200/5
K = 2 Ans.
The given P–T diagram shows the relative stability ranges of andalusite, sillimanite and kyanite. The difference in degrees of freedom at points X and Y is ………………..
Using the formula F = C-P+2 (+2 because both T & P are changing)
For point X, F =1-1+2 = 2
For point Y, F =1-3+2 = 0
The difference = 2-0 = 2 Ans.
The core and rim compositions of garnet are (Fe0.75Ca0.90Mn1.35)Al2Si3O12 and (Fe0.90Ca1.35Mn0.75)Al2Si3O12, respectively. The difference in mole fractions of spessartine between the core and the rim is ………………….. (answer in one decimal place).
In one mole of Spessartine (Mn3Al2Si3O12) Mn =3
mole fraction of Spessartine in core = no. of Mn in core/no. of Mn in one mole
mole fraction of Spessartine in core = 1.35/3 = 0.45
mole fraction of Spessartine in rim = no. of Mn in rim/no. of Mn in one mole
mole fraction of Spessartine in rim = 0.75/3 = 0.25
The difference in mole fractions of spessartine between the core and the rim is = 0.45-0.25 = 0.2 Ans.
Views This Month : 5754
Views This Year : 22675
Total views : 166957