Numericals
The ripple symmetry Index (RSI) for the given hypothetical asymmetric ripple is ……….. (round off to two decimal places)
Ripple symmetry index = stoss side / lee side
Ripple symmetry index (RSI) = 10/4 = 2.5 Ans.
Within a fourth drainage basin, the total length of the 1st, 2nd,3rd and 4th order streams are 10.5km, 7.5km, 5.5km and 1.5km, respectively. If the drainage density of the basin is 0.5 km-1, the basin area is …………………….km2 ( In integer)
Formula used = Drainage density (DD) = (total channel length) / (drainage area)
0.5 = 25/ drainage area
drainage area =50 km2 Ans
A soil has a void ratio of 0.5. the total porosity of the soil is …………….(round off to two decimal places)
relationship between the porosity (Φ) and void ratio (e) is – Φ = e/(e+1)
Φ = 0.5/1.5
Φ = 0.33 Ans.
The average unit weight of the uppermost part of the crust is 25000 N/m3. The vertical stress at a depth of 1km would be ………………..MPa. (In integer)
Pressure at a depth = 25000*1000 = 25 *106 Pa = 25 MPa Ans.
The radius of the Earth’s circular orbit around the sun is 149*106 km. The Earth takes 365 days to orbit the sun. The tangential velocity of the Earth is ………………………………. km/hr. (π = 3.14) (round off to one decimal place)
Total time = 365*24 = 8760 hr
Tangential velocity = angular velocity * radius
Tangential velocity = (2π/time)* radius
= 2*3.14*149*106/8760
= 106817.3 Ans.
A bore hole inclined at 600 to the horizontal pierces a vertical basaltic dyke of uniform thickness. If the length of the basaltic drill core along the core axis is 12m, the thickness of the dyke is ………….m. (In integer)
A P-ray arrives at the mantle-core boundary at an angle 250 with respect to the normal. At what angle to the normal does it enter the core ? (P – wave velocity in the lower mantle is 13.7 km/s and outer core is 8.1 km/s)(round off to two decimal places)
Here we can use Snell’s law, sin(i)/sin(r) = v1/v2
sin 250 / sin (r) = 13.7/8.1
0.422/sin (r) = 1.691
sin(r) = 0.249
r = 14.410 Ans.
The mass of the Earth is 80 times that of the Moon while the radius of the Earth is four times that of the Moon. The surface gravity of the Earth is …………………………times that of the moon ? (In integer)
Here we will use the gravity formula, g = G*M/R2
gravity at Moon (gm) = G*Mm /Rm2
gravity at Earth (ge) = G*Me / Re2 = G*80*Mm/ (4Rm)2
= (80/16)*G*Mm /Rm2
Now, ge/gm = 80/16 = 5
ge = 5*gm Ans.
A hypothetical rock contains the assemblage kyanite, sillimanite and quartz. The variance (degree of freedom) of the assemblage is …………………………( In integer)
Here one thing remember that although phases are three; kyanite, sillimanite and quartz
but we have only two component that is, SiO2 and Al2SiO5
Using the phase rule , F = C-P+2
F = 2-3+2 = 1 Ans.
The cut off grade of copper is 0.45 wt%. A mine has 1 million tonne of waste with a grade of 0.25 wt%. The mine also has stock of high grade ore with a grade of 1.8 wt%. How much of this high grade ore (in million tonne) must be blended with the waste to sell the blended ore at a grade of 0.5 wt%. ? (round off to three decimal places)
Here we can use the simple formula –
Average grade = [(grade 1)* (volume 1) + (grade 2)*(volume 2)]/ total volume
let’s assume that x million tone of 1.8wt% ore is blended to make the waste of 0.5wt% ore
0.5 = (0.25*1 + 1.8*x)/(1+x)
0.5 + 0.5x = 0.25 + 1.8x
1.3x = 0.5 – 0.25
x = 0.25/1.3
x = 0.192 million tons Ans.
The maximum and minimum principal stresses in a zone of active normal faulting are 28 MPa and 8MPa. The fault plane strikes N300E and dips 600 towards SE. Considering Anderson’s theory of faulting, the normal stress on the fault plane is …………………MPa. (In integer)
Given that the maximum principal stress (σ1) = 18MPa
Minimum principal stress (σ3) = 8 MPa
dip of fault plane = 600
According to the Andersen theory,
Normal stress (σn) = (σ1 + σ3)/2 + {(σ1 – σ3)/2}*cos2θ
= 18 + 10*cos 1200
= 18-5
= 13 Ans.
A granitic starts sliding on a slope (inclination of 300 with the horizontal) under the effect of gravity only, along the true direction of inclination of the slope and hits the ground in 4sec. Considering zero cohesion during sliding, the vertical height of the point (with respect to the ground) from where the block was dislodged is …………………..m. (g = 10m/s2)( In integer)
Here we can use the conservation of energy formula that is, total potential energy is being converted to the kinetic energy .
(1/2)*m*v2 = m*g*h
(1/2) *20*20*m = m*10*h (v= 20m/s)
200 = 10h
h= 20m Ans.
A cylindrical soil sample is encased in an open-ended inclined tube with a diameter of 100mm. There is a constant supply of water from the upper end of the sample and the outflow from the other end is collected in a beaker. The average amount of water collected is 1000mm3 every 10 second. The average outflow velocity is …………………….. mm/sec. (π = 3.14) ( round off to three decimal places)
There is a simple formula of velocity that is amount of outgoing water per unit area at unit second.
given, diameter = 100mm, radius =50mm
Q = 1000 mm3/10 sec = 100 mm3/sec
velocity (V) = Q/A
V = 100 / πr2
V = 100/3.14*2500
V = 0.012 Ans.
Using Airy’s hypothesis, calculate the thickness of the root beneath a 4km high mountain in isostatic equilibrium with a 40km thick continental crust of density 2800 kg/m3 and a mantle of density 3300 kg/m3. Express your answer in km. (round off to one decimal place)
Pressure at level of compensation will be equal
40*2800 + r*3300 = (44+r)*2800
40*2800 + 3300r = 2800r + 44*2800
500r = 4*2800
r = 22.4km Ans.
Given atomic weights of Cu, Fe and S as 63.55, 55.85 and 32.10 respectively, find out the weight of copper (in grams) metal in an ore (no associated gangue) of 1 kg weight constituting of bornite, chalcopyrite and chalcocite present in weight fractions of 0.4, 0.4 and .2 respectively. (round of to one decimal place)
Cu in the ore = Cu in Bornite + Cu in Chalcopyrite + Cu in Chalcocite
Bornite ——- Cu5FeS4 ———– At. wt. 502 —————- amount of Cu per kg ore (253.187)
Chalcopyrite –CuFeS2 ———— At. wt. 183.6 ————- amount of Cu per kg ore (138.453)
Chalcosite —- Cu2S —————- At. wt. 159.2 ————- amount of Cu per kg ore (159.67)
Total Cu = 551.31 Ans.
An ore body defined by a 300m * 300m area is shown in the figure in which the drill hole locations on equally spaced square grid are marked (number 1-16) The average thickness of the ore body at the 4 interior points is 10.8 m, at the 4 corners is 11.0 m and at the remaining 8 boundary locations is 10.5 m, respectively. The corresponding average grades are 1.5, 1.9 and 1.8 wt%, respectively. calculate the average grade (in wt%) of the full ore body using the included area method. (round of to two decimal places)
Length and width of the grids are given, depth is calculated by calculating average of the associated 4 bore holes –
In the same way average grades are also calculated.
Average grade = (weight of copper / total volume)*100
total volume = V1 +4*V2 + 4*V3 (V1, V2, V3 are column of blocks with 10.8m, 10.7 and 10.65m thickness)
Total volume = (100*100*10.8) + (4*100*100*10.7) + (4*100*100*10.65)
Total volume = 108000 + 428000 + 426000 = 962000m3
total weight of copper = (V1*G1) + (V2*G2) + (V3*G3) ,
where G1,G2,G3 are grades of blocks with 10.8m,10.7m,10.65m thickness}
Total weight of copper = 108000*1.5/100 + 428000*1.75/100 + 426000*1.65/100
Total weight of copper = 1620 + 7490 + 7029 = 16139 kg
Average grade = (16139/962000)*100 = 0.01677*100 =1.67 % Ans.
The 87Sr/86Sr ratio of a 1000Ma granite was measured as 0.8001. If its 87Rb/86Sr ratio is 2.499, what was the Sr isotopic ration of the source at the time of derivation of the granite ? (decay constant of 87Rb = 1.39*10-11 yr-1) (round off to three places of decimals)
The Rb- Sr dating equation is – (87Sr/86Sr)t = (87Sr/86Sr)o + (87Rb/86Sr)t (eλt -1)
(87Sr/86Sr)o = (87Sr/86Sr)t – (87Rb/86Sr)t (eλt -1)
(87Sr/86Sr)o = 0.8001 – 2.499 (e1.39*(10^-11)*10^9)
(87Sr/86Sr)o = 0.765 Ans.
The coefficient of permeability of two aquifers -1 and 2, are 60m/day and 40 m/day, respectively. Their saturated thickness are 30m and 15m respectively. Assuming steady state Darcian flow, the transmissivity of aquifer 1 is …………….. times that of aquifer2. ( In integer)
As we know transmissivity = K*b
T1/T2 = K1*b1/K2*b2
T1/T2 = 60*30/40*15
T1 = 3*T2
The transmissivity of aquifer 1 is 3 times the transmissivity of aquifer 2. Ans.
Assume that 218Po, with a half life of 138 days, is in secular equilibrium with 238U whose half life is 4.5*109y. How many grams of 218Po will be present for each gram of 238U in the mineral ? Express your answer in logarithm (to the base 10) (round of to two decimal places)
Secular Equilibrium : λ1N1 = λ2N2 or Nu/Np = λp/λu
as λ = log 2/t1/2
So, according to secular equilibrium equation ,
Np/Nu = (log2/4.5*109*365)/(log 2/138)
Np/Nu =138/ 4.5*109*365 (log 2 will be canceled out)
Np/Nu = -10.07 (as log base 10) Ans.
The figure below is an isobaric binary temperature-composition (T-X) plot. what amount (in%) of the equilibrium melting of rock R will generate a melt of composition L ? (round off to one decimal place)
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