Numericals
The amplitude recorded at a station for a magnitude 5 earthquake is x. If another earthquake recorded at the same station has an amplitude of 15x, then the magnitude of this earthquake is ………………(round of to two decimal places)
Earthquake magnitude = log10 (Amplitude)
M2 – M1 = log10 A2 – log10 A1
M2 – 5 = log10 (A2/A1)
M2 -5 = log10(15)
M2 = 1.176 + 5
M2 = 6.176 Ans.
If the intercepts of crystallographic axes are 0.5a : 1b : 0.75c on a crystallographic plane {h k l} the value of ‘l’ is…..(in integer)
The value is calculated by inversing the intercepts of crystallographic axis,
{h, k, l} = 1/0.5, 1/1, 1/0.75
remove denominator by multiplying 3 , {h k l} = 6, 3, 4
value of ‘l’ = 4 Ans.
An ocean wave with a wavelength of 200m approaches the coast. If water depth at the observation point is 75m, the wave velocity is ………… m/s. (Round off to two decimal places)(use g= 10m/s2)
Wave velocity or celerity = √(gλ/2π)
v = √10*200/2*3.14
v = √318.47
v = 17.84 m/s Ans.
A bed with an attitude 0450, 200 SE is rotated 600 clockwise (looking down) about a vertical axis. The strike (in the azimuthal convention following right hand rule) of the rotated bed is ……………..degrees. (In integer)
Solution in the image
A one meter deep and sheet-like waterflow on a sandy beach developed anti-dunes. The minimum velocity of the waterflow was ………………………m/s (round off to two decimal places)( use g=10m/s2)
Anti-dunes are formed when the Froud number is equal to 1
Fr = velocity / √(gd)
1 = v / √10
v = √10
v = 3.16 Ans.
If the angular aperture of a 20X objective is 460 , the numerical aperture of the water immersion objective is …………(round off to two decimal places) (use RI of water =1.33)
use formula, numerical aperture = RI sin(angular aperture/2)
NA = RI sin(AA/2)
NA = 1.33 sin(46/2)
NA = 1.33*0.390
NA = 0.5187 Ans.
A metamorphic rock is composed of grossular garnet (Ca3Al2Si3O12), kyanite (Al2SiO5), anorthite (CaAl2Si2O8) and quartz (SiO2). If these minerals show an univariant reaction relationship, the number of components in this assemblage is …………….. (In integer)
As the reaction is univariant, the degree of freedom = 1
given, mineral phases = 4
there is neither temperature or pressure constant is given so we will use 2 variable in our equation
Using phase rule – F = C – P + 2
1 = C – 4 + 2
C = 3 Ans.
If the dip separation vector on a normal fault plane has an attitude 600 → 0400 and a magnitude of 6 m, the heave on the fault is …………….m. (In integer)
Simply we can use a right angle triangle of heave and through as in the figure –
angle between heave and separation vector = 600
in triangle consider as base than, cos600 = heave/separation
1/2 = heave /6
heave = 3m Ans.
A hillslope with an angle of 400 consists of soil having an internal friction angle of 300. The factor of safety of the hillslope is ………………. (Round off to two decimal places)
Factor of slope/ safety = tan (fr. angle)/tan (hill slope angle)
Factor of slope/ safety = tan300 / tan400
Factor of slope/ safety = 0.577/0.839
Factor of slope/ safety = 0.6877 Ans.
The water table over an area of 1 km2 was lowered by 4 m. If the porosity of rock is 30% and the specific retention is 10%, the change in the groundwater storage is …………….. × 103 m3 . (In integer)
change of ground water storage = volume of water that is removed
As we know that amount of water removed is depends on specific yield, and
specific yield = porosity – specific retension
SPyield = 30-10 = 20%
Change of water volume = area* height* SPyield
Change of water volume = 1*106*4*0.20
= 800*103 m3 Ans.
The 143Nd/144Nd and 147Sm/144Nd ratios of a rock are 0.516 and 0.389, respectively. The rock evolved as a closed system. As per the exact parent-daughter relationship equation, the 143Nd/144Nd ratio of the rock 4.6 × 109 years ago was …………….(Round off to three decimal places) (Use decay constant for 147Sm (λ )= 6.54 × 10-12 y-1)
(eλt -1) = 0.030
The equation is, Dt = D0 + P (eλt -1)
or
(143Nd/144Nd)t = (143Nd/144Nd)0 + (147Sm/144Nd)t * (eλt -1)
0.516 = (143Nd/144Nd)0 +0.389*0.030
(143Nd/144Nd)0 = 0.516 – 0.0116
(143Nd/144Nd)0 = 0.504 Ans.
A longitudinal profile of a river is shown in the figure below. If the average discharge of the river at reach AB is 200 m3/s and increases to 300 m3/s at reach CD, then the stream power from the reach AB to CD will change by a factor of ………………… (Round off to two decimal places) (Use g = 10 m/s2 , ρwater = 1000 kg/m3 )
Stream power = density(ρ)*gravity(g)*discharge(Q)*slope(s)
SP(A-B) = ρgQs(A-B) = ρg*200*100/10000
SP(A-B) = ρg*2
SP(C-D) = ρgQs(C-D) = ρg*300*50/10000
SP(C-D) = ρg*1.5
change in factor = SP(C-D)/SP(A-B) = ρg*1.5/ρg*2
= 0.75 Ans.
An underground vertical dyke is intercepted by an inclined borehole as shown in the figure below. The length of the dyke core intercepted by the borehole is 4 m. If the true thickness of the dyke is 2 m, the inclination of the borehole from the vertical is ………… degrees. (In integer)
A cylindrical copper ore body has a vertical thickness of 45 m and a diameter of 14 m with a density of 2.9 g/cm3. The reserve of the copper ore body is ……………. tons. (In integer)
given, density = 2.9 g/cc = 2900 km/m3
thickness or height of cylinder = 45m
diameter of cylinder = 14m, radius =7m
Volume of cylinder = π r2h
= 3.14*72*45
= 6923.7m3
reserve = volume*density
= 6923.7*2900
= 20078730 kg = 20078.73 tons Ans.
The density of a FCC unit cell is 6.5 g/cm3 . If the mass of a single atom is 60 g/mol, the diagonal length of the face {100} is …………….. Å. (Round off to two decimal places) (Use NA = 6.022 × 1023)
For FCC, a3 = z*M/NA*d
where, a = edge length of unit cell
z = number of atoms
M = atomic mass of element
NA = Avogadro number
d = density of unit cell
a3 = 4*60/6.022*1023 * 6.5
a3 = 6.1313*10-23cm3 = 61.313 Å3
a = 3.94 Å
diagonal = √(a2 + a2)
diagonal = √(2a2)
diagonal = 5.58 Ans.
The following figure shows an isobaric temperature-composition (T-X) phase diagram for the binary system A-B. If ‘P’ is the initial composition of liquid, the amount of liquid that remains in the system when the liquid cools from 1800 0C (point L1) to 15000C (point L2) is ………… %. (In integer)
Amount of liquid = 10/50 = 20%
A water flow transports spherical particles (diameter = 2 mm; density = 3 g/cm3) in suspension mode. If additional particles of density 2 g/cm3 are added into the flow, then the diameter of the particles that can be transported without a change in terminal fall velocity, using Stokes law, is ……….. mm. (Round off to two decimal places) (Use density of water = 1 g/cm3 )
According to the stokes law, fall velocity (V) =2(ρs – ρl)r2g/9μ
If fall velocity is not changing so V1 =V2
where, V1= fall velocity of first particle
V2 = fall velocity of second or new particle
So, 2(ρs1 – ρl)r12g/9μ = 2(ρs2 – ρl)r22g/9μ
r1 = radius of first particle = 2/2 = 1mm = 0.1cm
r2 = radius of second particle
ρs1 = density of first particle = 3g/cc
ρs2 = density of second particle = 2g/cc
ρl = density of water = 1g/cc
(ρs1 – 1)r12 = (ρs2 – 1)r22
(3 – 1)0.12 = (2 – 1)r22
2*0.01 = r22
r22 = 0.02 = 0.141cm
r2 = 1.41mm
diameter = 1.41*2 = 2.82 mm Ans.
If an iron ore body contains 50% hematite (Fe2O3) and 50% magnetite (Fe3O4), then the grade of the iron ore body is ……………..%. (Round off to two decimal places) (Use atomic weight of Fe = 55.85 amu and O = 16 amu).
As ore is consist of half of hematite and half of magnetite
Grade of Iron = (Fe% in hematite)*0.5 + (Fe% in magnetite)*0.5
= (111.7/159.7)*0.5 + (167.55/231.55)*0.5
= 69.94*0.5 + 72.36*0.5
= 71.15 % Ans.
The schematic stereographic projection below shows dip amount and dip direction of three sets of joints (J1, J2 and J3) on a hillslope. If the internal friction angle of the hillslope material is 300, the strike of the potential failure joint plane (in azimuthal convention following right hand rule) is ……………… degrees. (In integer)
J1 is have dip less than the slope and more than the internal frictional angle, it is potential failure
J2 is having dip less than the internal frictional angle, no failure
J3 has opposite dip direction, so no failure plane
The hydraulic conductivity of a 100 cm long cylindrical core is estimated as 1.2 cm/min when hydraulic head difference is 20 cm in an experimental setup. If the effective porosity of the core is 20%, then, assuming steady state Darcy flow, the average interstitial velocity of groundwater through the core is ……………. m/day. (Round off to two decimal places)
Given, hydraulic conductivity = 1.2 cm/min = 8.3*10-6
length (dl) = 100cm, hydraulic difference (dh) = 20cm
dh/dl = 0.2
Discharge (Q) = K(dh/dl)
= 17.28 *0.2
= 3.456
Average Interstitial velocity = Discharge/ eff. porosity
= 3.456/0.2
= 17.28 Ans.
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