The bulk density and water content of a soil are 1800 kg/m3 and 18%, respectively. The dry density of the soil calculated from the given information is ______________ kg/m3.

The dry density of the soil is 1525.42 kg/m3.

The decay constant of a radioactive isotope is 1.21 × 10−4 year−1. The half-life of the isotope is _________________ years.

The half-life of the isotope is 5727 years.

Solution to GATE 2023 Geology GG1 Numerical Problem

Q: In a seismic reflection survey over a two-layered Earth model having densities and seismic velocities ρ1=2000 kg/m3, V1=1800 m/s for the first layer and ρ2=3000 kg/m3, V2=2100 m/s for the second layer, the normal incidence P-wave reflection coefficient is ______________.

The normal incidence P-wave reflection coefficient is 0.272.

Q: The resistivity of a rock, 100% saturated with water of resistivity 0.25 Ωm, is 60 Ωm. Assuming tortuosity and cementation exponents to be 1 and 2, respectively, the porosity of the rock is __________ (in %).

The porosity of the rock is 6.45%.

Q: The amount of Fe in a sample of 25 g of pyrrhotite (FeS) is ____________ g. (Atomic wt. of Fe = 55.85 and S = 32.06)

The amount of Fe in the sample is 15.87 g.

Q: The rate of spreading about a symmetric spreading center at the middle of a 4000 km wide sea is 40 mm/year. The spreading began ________ million years before present.

The spreading began 100 million years ago.

Q: A vertical aerial photograph is obtained over flat terrain with a 30 cm focal-length camera lens from an altitude of 18288 m. If the width of a dolerite dyke on this vertical photograph is 2 mm, its actual width on the terrain is ____________ m.

The actual width of the dolerite dyke is 121.22 m.

Q: In a drainage basin, the number of the 1st, 2nd, 3rd, 4th and 5th order streams are 240, 40, 8, 2 and 1, respectively. The average of all calculated bifurcation ratios is ____________.

The average bifurcation ratio is 4.25.

Q: A sandstone follows Mohr-Coulomb failure criterion. If the uniaxial compressive strength and the angle of the internal friction of the sandstone are 7 MPa and 30°, respectively, the calculated cohesion of the rock is _______________ MPa.

The calculated cohesion of the rock is 2.01 MPa.

The bulk density and water content of a soil are 1800 kg/m³ and 18%, respectively. The dry density of the soil calculated from the given information is ______________ kg/m³. [round off to 2 decimal places]

Given:

Bulk density (ρ_b): 1800 kg/m³
Water content (w): 18%

Objective: Calculate the dry density (ρ_d) of the soil.

Formula for Dry Density:

The dry density (ρ_d) can be determined using the formula:

ρ_d = ρ_b / (1 + w)

Where:

ρ_d = Dry Density
ρ_b = Bulk Density
w = Water Content (in decimal form)

Step-by-Step Calculation:

1. Convert the Water Content from Percentage to Decimal:

w = 18 / 100 = 0.18

1. Calculate the Dry Density:

ρ_d = ρ_b / (1 + w)

ρ_d = 1800 / (1 + 0.18)

Calculate the denominator:

1 + 0.18 = 1.18

Divide the bulk density by this value:

ρ_d = 1800 / 1.18

Perform the division to find the dry density:

ρ_d ≈ 1525.42 kg/m³

Answer:

The dry density of the soil is approximately 1525.42 kg/m³, rounded to two decimal places.

In a seismic reflection survey over a two-layered Earth model having densities and seismic velocities ρ₁=2000 kg/m³, V₁=1800 m/s for the first layer and ρ₂=3000 kg/m³, V₂=2100 m/s for the second layer, the normal incidence P-wave reflection coefficient is ______________. [round off to 3 decimal places]

Given:

Density of the first layer (ρ₁): 2000 kg/m³
Seismic velocity of the first layer (V₁): 1800 m/s
Density of the second layer (ρ₂): 3000 kg/m³
Seismic velocity of the second layer (V₂): 2100 m/s

Objective: Calculate the normal incidence P-wave reflection coefficient (R) between the two layers.

Formula for Normal Incidence P-Wave Reflection Coefficient:

The reflection coefficient (R) at normal incidence is given by:

R = (ρ₂ V₂ - ρ₁ V₁) / (ρ₂ V₂ + ρ₁ V₁)

Step-by-Step Calculation:

1. Substitute the given values into the formula:

R = (3000 × 2100 - 2000 × 1800) / (3000 × 2100 + 2000 × 1800)

1. Perform the multiplications:

3000 × 2100 = 6300000

2000 × 1800 = 3600000

1. Calculate the numerator:

Numerator = 6300000 - 3600000 = 2700000

1. Calculate the denominator:

Denominator = 6300000 + 3600000 = 9900000

1. Compute the reflection coefficient:

R = 2700000 / 9900000

R ≈ 0.273

Answer:

The normal incidence P-wave reflection coefficient between the two layers is approximately 0.273, rounded to three decimal places.

The resistivity of a rock, 100% saturated with water of resistivity 0.25 Ωm, is 60 Ωm. Assuming tortuosity and cementation exponents to be 1 and 2, respectively, the porosity of the rock is __________ (in %). [round off to 2 decimal places]

Given:

Resistivity of the saturated rock (R_t): 60 Ωm
Resistivity of water (R_w): 0.25 Ωm
Tortuosity exponent (a): 1
Cementation exponent (m): 2

Objective: Calculate the porosity of the rock.

Formula for Porosity:

The porosity (φ) can be calculated using the Archie’s equation for saturated rocks:

R_t = R_w / φ^m

Rearrange the formula to solve for porosity (φ):

φ = (R_w / R_t)^1/m

Step-by-Step Calculation:

1. Substitute the given values into the formula:

φ = (0.25 / 60)^1/2

1. Calculate the fraction inside the parentheses:

0.25 / 60 ≈ 0.004167

1. Calculate the porosity:

φ = (0.004167)^0.5

φ ≈ 0.064

1. Convert porosity to percentage:

φ = 0.064 × 100 ≈ 6.38%

Answer:

The porosity of the rock is approximately 6.38%, rounded to two decimal places.

Let us consider that a student misses cancelling the self-potential between potential electrodes before injecting current into the subsurface, in a Wenner electrical resistivity survey using DC resistivity meter over a horizontally stratified Earth. In direct and reverse modes of measurement (when current flows from C1 to C2 and C2 to C1, respectively) with the same magnitude of current flow, the potential differences recorded are +158 mV and −214 mV, respectively. The self-potential between the potential electrodes before injecting current was _________ mV. [in integer]

Given:

Potential difference in direct mode (V_+direct): +158 mV
Potential difference in reverse mode (V_−reverse): −214 mV

Objective: Determine the self-potential (SP) between the potential electrodes before injecting current.

Formula and Calculation:

The self-potential (SP) can be calculated using the following relationship:

SP = (V_+direct + V_−reverse) / 2

Step-by-Step Calculation:

1. Substitute the given values into the formula:

SP = (158 + (-214)) / 2

Calculate the sum:

158 + (-214) = 158 - 214 = -56

Divide by 2 to find the self-potential:

SP = -56 / 2 = -28

Answer:

The self-potential between the potential electrodes before injecting current was -28 mV.

The amount of Fe in a sample of 25 g of pyrrhotite (FeS) is ____________ g. (Atomic wt. of Fe = 55.85 and S = 32.06) [round off to 2 decimal places]

Given:

Sample weight: 25 g
Atomic weight of Fe: 55.85
Atomic weight of S: 32.06

Objective: Determine the amount of Fe in the 25 g sample of pyrrhotite (FeS).

Formula and Calculation:

The amount of Fe can be calculated using the following steps:

1. Calculate the molar mass of FeS:

Molar mass of FeS = Atomic weight of Fe + Atomic weight of S

Molar mass of FeS = 55.85 + 32.06 = 87.91 g/mol

1. Calculate the fraction of Fe in FeS:

Fraction of Fe = Atomic weight of Fe / Molar mass of FeS

Fraction of Fe = 55.85 / 87.91 ≈ 0.635

1. Calculate the amount of Fe in the 25 g sample:

Amount of Fe = Fraction of Fe × Sample weight

Amount of Fe = 0.635 × 25

Amount of Fe ≈ 15.88 g

Answer:

The amount of Fe in a sample of 25 g of pyrrhotite (FeS) is approximately 15.88 g, rounded to two decimal places.

The rate of spreading about a symmetric spreading center at the middle of a 4000 km wide sea is 40 mm/year. The spreading began ________ million years before present. [in integer]

Given:

Rate of spreading: 40 mm/year
Width of the sea (distance between the spreading centers): 4000 km

Objective: Determine how many million years ago the spreading began.

Formula and Calculation:

The total time (T) in years required for the spreading to cover the given width can be calculated using the formula:

T = D / (2 × R)

where:

D = Distance between the spreading centers (4000 km)
R = Rate of spreading (40 mm/year)

Convert the width from kilometers to millimeters:

D = 4000 km × 1,000,000 mm/km = 4,000,000,000 mm

Calculate the time required in years:

T = 4,000,000,000 mm / (2 × 40 mm/year)

T = 4,000,000,000 / 80 = 50,000,000 years

Note: The calculated time of 50 million years represents the time for one side of the symmetric spreading center. Since the distance given is for the full width (spreading in both directions from the center), the time for the full width would be:

Total time = 2 × 50,000,000 years = 100,000,000 years

Convert the time into million years:

Total time = 100,000,000 years / 1,000,000 = 100 million years

Answer:

The spreading began 100 million years before present.

A vertical aerial photograph is obtained over flat terrain with a 30 cm focal-length camera lens from an altitude of 18288 m. If the width of a dolerite dyke on this vertical photograph is 2 mm, its actual width on the terrain is ____________ m. [round off to 2 decimal places]

Given:

Focal length of camera (f): 30 cm
Altitude (H): 18288 m
Width of the dolerite dyke on the photograph (w_photo): 2 mm

Objective: Determine the actual width of the dolerite dyke on the terrain.

Formula and Calculation:

The actual width of the dyke (w_actual) can be determined using the formula:

w_actual = w_photo × (H / f)

where:

w_photo = Width of the dolerite dyke on the photograph
H = Altitude of the camera
f = Focal length of the camera

Step-by-Step Calculation:

1. Convert the focal length from cm to meters:

f = 30 cm × 0.01 m/cm = 0.30 m

1. Convert the width of the dyke from mm to meters:

w_photo = 2 mm × 0.001 m/mm = 0.002 m

1. Calculate the actual width of the dyke:

w_actual = 0.002 m × (18288 m / 0.30 m)

w_actual = 0.002 × 60960

w_actual = 121.92 m

Answer:

The actual width of the dolerite dyke on the terrain is approximately 121.92 m, rounded to two decimal places.

The decay constant of a radioactive isotope is 1.21 × 10⁻⁴ year⁻¹. The half-life of the isotope is _________________ years. [round off to nearest integer]

Given:

Decay constant (λ): 1.21 × 10⁻⁴ year⁻¹

Objective: Determine the half-life (T_1/2) of the radioactive isotope.

Formula and Calculation:

The half-life of a radioactive isotope can be calculated using the formula:

T_1/2 = ln(2) / λ

where:

ln(2) is the natural logarithm of 2 (approximately 0.693)
λ is the decay constant

Step-by-Step Calculation:

1. Substitute the given decay constant into the formula:

T_1/2 = 0.693 / (1.21 × 10⁻⁴)

1. Calculate the half-life:

T_1/2 = 0.693 / 1.21 × 10⁻⁴

T_1/2 = 0.693 / 0.000121 = 5735.54 years

1. Round off to the nearest integer:

T_1/2 = 5736 years

Answer:

The half-life of the isotope is approximately 5736 years, rounded to the nearest integer.

In a drainage basin, the number of the 1st, 2nd, 3rd, 4th, and 5th order streams are 240, 40, 8, 2, and 1, respectively. The average of all calculated bifurcation ratios is ____________. [round off to 2 decimal places]

Given:

Number of 1st order streams (N₁): 240
Number of 2nd order streams (N₂): 40
Number of 3rd order streams (N₃): 8
Number of 4th order streams (N₄): 2
Number of 5th order streams (N₅): 1

Objective: Determine the average bifurcation ratio.

Formula and Calculation:

The bifurcation ratio (R_i) between two consecutive orders of streams is calculated as:

R_i = N_i / N_i+1

To find the average bifurcation ratio, calculate the bifurcation ratios between each pair of consecutive orders and then average them.

Step-by-Step Calculation:

Calculate the bifurcation ratios:

R₁ = N₁ / N₂ = 240 / 40 = 6.00

R₂ = N₂ / N₃ = 40 / 8 = 5.00

R₃ = N₃ / N₄ = 8 / 2 = 4.00

R₄ = N₄ / N₅ = 2 / 1 = 2.00

Average R = (R₁ + R₂ + R₃ + R₄) / 4

Average R = (6.00 + 5.00 + 4.00 + 2.00) / 4 = 17.00 / 4 = 4.25

Answer:

The average of all calculated bifurcation ratios is approximately 4.25, rounded to two decimal places.

A sandstone follows Mohr-Coulomb failure criterion. If the uniaxial compressive strength and the angle of the internal friction of the sandstone are 7 MPa and 30°, respectively, the calculated cohesion of the rock is _______________ MPa. [round off to 2 decimal places]

Given:

Uniaxial compressive strength (σ_c): 7 MPa
Angle of internal friction (φ): 30°

Objective: Determine the cohesion (c) of the sandstone.

Formula and Calculation:

The Mohr-Coulomb failure criterion relates the uniaxial compressive strength, cohesion, and angle of internal friction with the formula:

σ_c = 2c / (1 - sin(φ))

where:

σ_c = Uniaxial compressive strength
c = Cohesion
φ = Angle of internal friction

Rearrange the formula to solve for cohesion (c):

c = σ_c × (1 - sin(φ)) / 2

Step-by-Step Calculation:

1. Calculate the sine of the angle φ:

sin(30°) = 0.5

1. Substitute the values into the formula:

c = 7 MPa × (1 - 0.5) / 2

c = 7 MPa × 0.5 / 2

c = 3.5 MPa / 2 = 1.75 MPa

Answer:

The calculated cohesion of the sandstone is 1.75 MPa, rounded to two decimal places.

At a certain depth in the crust, the maximum and minimum principal compressive stresses are 150 MPa and 75 MPa, respectively, which lead to normal faulting. If the average density of the crust is 2700 kg/m³, the crustal depth of fracture initiation according to Anderson’s theory of faulting is ________________ km. (g = 10 m/s²) [round off to one decimal place]

Given:

Maximum principal compressive stress (σ₁): 150 MPa
Minimum principal compressive stress (σ₃): 75 MPa
Density of the crust (ρ): 2700 kg/m³
Acceleration due to gravity (g): 10 m/s²

Objective: Determine the crustal depth of fracture initiation according to Anderson’s theory of faulting.

Formula and Calculation:

According to Anderson’s theory of faulting, normal faulting occurs when the vertical stress is greater than the horizontal stress. The depth at which this happens can be calculated by using the relationship between the stress and depth:

In a normal faulting regime, the vertical stress (σ_v) at depth (D) is given by:

σ_v = ρ × g × D

The difference between the vertical stress and the minimum principal stress (σ₃) should be equal to the difference between the maximum principal stress (σ₁) and the minimum principal stress (σ₃):

σ_v - σ₃ = σ₁ - σ₃

Rearrange to solve for depth (D):

D = (σ₁ - σ₃ + σ₃) / (ρ × g)

Step-by-Step Calculation:

1. Substitute the given values into the formula:

D = (150 MPa - 75 MPa + 75 MPa) / (2700 kg/m³ × 10 m/s²)

1. Perform the calculation:

D = 150 MPa / (2700 kg/m³ × 10 m/s²)

D = 150 × 10⁶ Pa / (27000 kg/m³ × m/s²)

D = 150 × 10⁶ / 27000

D ≈ 5555.56 m

D ≈ 5.56 km

Answer:

The crustal depth of fracture initiation according to Anderson’s theory of faulting is approximately 5.6 km, rounded to one decimal place.

A cylindrical soil sample of 10 cm diameter is tested in a constant-head permeameter. A volume of 250 cm³ of water is collected in 5 minutes when the constant-head difference between tapping points 15 cm apart is 5 cm. Considering Darcy flow, the absolute value of the coefficient of permeability in cm/s is ____________. (π = 3.14) [round off to 3 decimal places]

Given:

Diameter of the cylindrical soil sample: 10 cm
Volume of water collected (V): 250 cm³
Time (t): 5 minutes (300 seconds)
Constant-head difference (Δh): 5 cm
Length of the soil sample (L): 15 cm

Objective: Determine the coefficient of permeability (\(k\)) in cm/s.

Formula and Calculation:

The coefficient of permeability can be calculated using Darcy’s Law:

$$Q = k \times A \times \frac{\Delta h}{L \times t}$$

Where:

Q = Volume of water collected (250 cm³)
A = Cross-sectional area of the soil sample
Δh = Head difference (5 cm)
L = Length of the soil sample (15 cm)
t = Time (300 seconds)

Rearranging to solve for \(k\):

$$k = \frac{Q \times L}{A \times \Delta h \times t}$$

Step-by-Step Calculation:

Calculate the cross-sectional area (\(A\)):
Substitute the values into the formula for \(k\):
Perform the calculation:

Answer:

The absolute value of the coefficient of permeability is approximately 0.032 cm/s, rounded to three decimal places.

The bulk density and water content of a soil are 1800 kg/m3 and 18%, respectively. The dry density of the soil calculated from the given information is ______________ kg/m3. [round off to 2 decimal places]

The resistivity of a rock, 100% saturated with water of resistivity 0.25 Ωm, is 60 Ωm. Assuming tortuosity and cementation exponents to be 1 and 2, respectively, the porosity of the rock is __________ (in %). [round off to 2 decimal places]

The amount of Fe in a sample of 25 g of pyrrhotite (FeS) is ____________ g. (Atomic wt. of Fe = 55.85 and S = 32.06) [round off to 2 decimal places]

The rate of spreading about a symmetric spreading center at the middle of a 4000 km wide sea is 40 mm/year. The spreading began ________ million years before present. [in integer]

A vertical aerial photograph is obtained over flat terrain with a 30 cm focal-length camera lens from an altitude of 18288 m. If the width of a dolerite dyke on this vertical photograph is 2 mm, its actual width on the terrain is ____________ m. [round off to 2 decimal places]

The decay constant of a radioactive isotope is 1.21 × 10−4 year−1. The half-life of the isotope is _________________ years. [round off to nearest integer]

In a drainage basin, the number of the 1st, 2nd, 3rd, 4th, and 5th order streams are 240, 40, 8, 2, and 1, respectively. The average of all calculated bifurcation ratios is ____________. [round off to 2 decimal places]

A sandstone follows Mohr-Coulomb failure criterion. If the uniaxial compressive strength and the angle of the internal friction of the sandstone are 7 MPa and 30°, respectively, the calculated cohesion of the rock is _______________ MPa. [round off to 2 decimal places]

Responses

The bulk density and water content of a soil are 1800 kg/m³ and 18%, respectively. The dry density of the soil calculated from the given information is ______________ kg/m³. [round off to 2 decimal places]

The decay constant of a radioactive isotope is 1.21 × 10⁻⁴ year⁻¹. The half-life of the isotope is _________________ years. [round off to nearest integer]