Applied Geology

Step 1: Analyzing the types of dams.
- (A) Gravity dams are rigid concrete or masonry structures that resist water pressure primarily through their own weight.
- (B) Arch dams are typically built on very strong, stable foundations, as they rely on the strength of the rock to support the structure.
- (C) A buttress dam has a watertight upstream face (typically a sloped concrete slab) supported by downstream buttresses.
- (D) Earth dams typically have a sloped surface, not a concrete wall. Their structure is not curved in the manner described.

Step 2: Conclusion.
Answer: (A) and (C) are correct

Given a slope with an inclination of 30° and a granite block sliding down under gravity, we need to find the vertical height from which the block was dislodged. Considering gravity and neglecting friction and cohesion, we use the following approach to solve the problem.

Since the inclination is 30°, the acceleration of the block along the slope is given by:

The block hits the ground in 4 seconds. We use the kinematic equation for the distance traveled along the slope:

Assuming initial velocity , we have:

Next, we calculate the vertical height from which the block was dislodged:

The computed vertical height is 20 m.

The Ripple Symmetry Index (RSI) is defined as the ratio of the stoss side length to the lee side length of a ripple. In the given figure, the stoss side length is 10 cm and the lee side length is 4 cm.
RSI is calculated as follows:

RSI = stoss side length / lee side length

Plugging in the given values, we have:

RSI = 10 / 4 = 2.5

The calculated RSI value is 2.5

To find the total porosity of soil with a given void ratio, we can use the relationship between void ratio (e) and porosity (n), expressed by the formula:

n = e / (1 + e)

Given the void ratio e is 0.5, substitute this value into the formula:

n = 0.5 / (1 + 0.5) = 0.5 / 1.5

Perform the division to find the total porosity:

n = 0.3333...

Rounding to two decimal places, the porosity is:

n = 0.33

The vertical stress at a certain depth in the Earth's crust can be calculated using the formula for lithostatic pressure: σ = ρgh, where σ is the vertical stress, ρ is the density or unit weight of the crust, g is the acceleration due to gravity, and h is the depth.

Given:

• Unit weight ρ = 25000 N/m³
• Depth h = 1 km = 1000 m

Acceleration due to gravity, g, is approximately 9.81 m/s². However, since the unit weight ρ already includes gravity, it is not needed explicitly in our calculation for this particular scenario.

We directly use the unit weight (which is force per unit volume) at the given depth:

σ = ρ × h = 25000 N/m³ × 1000 m

σ = 25000000 N/m²

Now, convert the stress from N/m² to MPa (since 1 MPa = 1,000,000 N/m²):

σ = 25000000 N/m² ÷ 1,000,000 = 25 MPa

The vertical stress at a depth of 1 km in the Earth's crust is 25 MPa

To determine the angle at which a P-wave enters the core, we apply Snell's Law, which relates the angles of incidence and refraction to the velocities of the wave in the two media.

The formula is: where is the angle with respect to the normal in the mantle, is the P-wave velocity in the mantle, is the P-wave velocity in the core, and is the angle with respect to the normal in the core, which we need to find.

Rearranging for , we get: .

Substituting the values: .

Calculating : .

Thus, the P-wave enters the core at an angle of 14.35° with respect to the normal.

To determine the average outflow velocity of water from the cylindrical soil sample, we can use the formula for velocity, v, which is given by:
v = Q/A
where Q is the volumetric flow rate and A is the cross-sectional area of the tube.
First, calculate the cross-sectional area, A, of the tube. The formula for the area of a circle is:
A = π(d/2)², where d is the diameter of the tube.
Here, d = 100 mm.
Thus, A = 3.14 × (100/2)² = 3.14 × 50² = 3.14 × 2500 = 7850 mm².
Now, calculate the volumetric flow rate, Q. Given that 1000 mm³ of water is collected every 10 seconds, Q = 1000 mm³ per 10 sec = 100 mm³/sec.
Substitute Q and A into the velocity formula:
v = 100/7850
v ≈ 0.012738.
Rounded to three decimal places, the average outflow velocity is approximately 0.013 mm/sec.

Airy's hypothesis states that isostatic compensation occurs due to vertical movements of crustal columns of varying thicknesses, all of which have the same density ( ), but float on a higher-density, uniform substratum (mantle, ).

The pressure at the compensation depth ( ) must be equal for two columns: the Mountain Column ( ) and the Normal Column ( ).

Height of the Mountain ( ):

Normal Crustal Thickness ( ):

Crustal Density ( ):

Mantle Density ( ):

Root Thickness ( ): The unknown thickness of the crustal root extending below the normal crustal thickness. (This is the required value).

The pressure exerted by a column of rock is , where is density, is gravity, and is thickness. Since cancels out, we equate the mass per unit area.

$ h T r X C T+r T+r T r P_M = P_N \implies M_M = M_N 0 = (\rho_m - \rho_c)r r=0 \text{Correct Airy Balance} T + r h r h r h \rho_c r \rho_m - \rho_c \text{2. Calculate the Root Thickness } r r \text{3. Substitute Values} h = 4 \text{ km} \rho_c = 2800 \text{ kg/m}^3 \rho_m = 3300 \text{ kg/m}^3 $

To find the initial ⁸⁷Sr/⁸⁶Sr ratio of the source from which the granite was derived, we will use the following formula:

R_t = R₀ + (e^λt - 1) (⁸⁷Rb/⁸⁶Sr)

Where:

• R_t = Measured ⁸⁷Sr/⁸⁶Sr ratio today = 0.8001
• R₀ = Initial ⁸⁷Sr/⁸⁶Sr ratio we are solving for
• ⁸⁷Rb/⁸⁶Sr = 2.499
• λ = Decay constant for ⁸⁷Rb = 1.39 × 10^-11 yr^-1
• t = 1000 Ma = 1000 × 10⁶ years

Step 1: Calculate e^λt
e^λt = e^{(1.39 × 10-11 yr-1 × 1000 × 106 yr)} = e^0.0139 ≈ 1.0140
Step 2: Substitute known values and solve for R₀
0.8001 = R₀ + (1.0140 - 1) × 2.499
0.8001 = R₀ + 0.0350
R₀ = 0.8001 - 0.0350 = 0.7651

Upon rounding R₀ to three decimal places, we find it to be 0.765.

To solve the problem, we need to determine the transmissivity of each aquifer and then compare them. Transmissivity ( ) is calculated as the product of the coefficient of permeability ( ) and the saturated thickness ( ) of the aquifer. Thus, the formula is:

Calculate the transmissivity for Aquifer 1:

Next, calculate the transmissivity for Aquifer 2:

Now, determine how many times larger the transmissivity of aquifer 1 is compared to aquifer 2 by dividing by :

The result is 3, which indicates that the transmissivity of aquifer 1 is 3 times that of aquifer 2.

In secular equilibrium, the decay rate of the parent equals the decay rate of the daughter:

$ P \text{Uranium-238, } {}^{238}\text{U} D \text{Polonium-218, } {}^{218}\text{Po} N \lambda \lambda t_{1/2} \text{Given Values} t_{1/2, D} ({}^{218}\text{Po}) = 138 \text{ days} t_{1/2, P} ({}^{238}\text{U}) = 4.5 \times 10^9 \text{ years} \text{2. Calculate the Mass Ratio} \text{mass}_D / \text{mass}_P N m M N_A \left(\frac{N_D}{N_P} = \frac{t_{1/2, D}}{t_{1/2, P}}\right) \text{Substitute Values} \text{Mass Numbers (Molar Mass in grams):} M_D = 218 \text{ g/mol} M_P = 238 \text{ g/mol} \text{Half-lives (convert to the same unit, days):} {}^{218}\text{Po} {}^{238}\text{U} m_P = 1 \text{ g} \text{3. Express the Answer in Logarithm} \log_{10} \log(A \times B) = \log(A) + \log(B) $

Given:
Angular aperture of objective:
Refractive index of water:

Step 1 — Recall the formula for numerical aperture (NA):

Step 2 — Plug in the values:
Answer (rounded to two decimals):

Given:
Area of aquifer:
Lowering of water table:
Porosity:
Specific retention:
Step 1 — Compute the specific yield:
For unconfined aquifer, the change in groundwater storage per unit area is
Step 2 — Compute the total change in storage:
Answer:

Given:
Present ratios for the rock:
Decay constant: . Time interval: . (Assume a closed system.)
Step 1 — Parent → daughter mass-balance
If and are the parent (Sm) and daughter (Nd) amounts initially, then after time : The term in brackets is the amount of parent decayed into daughter. Divide everything by the stable isotope (which does not change). Using present (now) values and rearranging gives the standard relation (derivation: and the decayed amount is .) So the initial (ancient) ratio is
Step 2 — compute

Step 3 — plug numbers in

Answer (rounded to three decimals):

Given:
Water depth . Use . Antidunes occur when flow is at least critical (Froude number ). The minimum velocity corresponds to the critical condition .
Froude number: Set so .
Compute: Rounded to two decimals:

To determine the diameter of particles that can still be transported without changing terminal fall velocity, we apply Stokes' Law for the terminal velocity (v) given by:
v = (2/9) * (r²(ρ_p - ρ_w)/η) * g
where ρ_p is the density of the particle, ρ_w is the density of the water, η is the dynamic viscosity of the fluid, g is the acceleration due to gravity (assumed constant), and r is the radius of the particle.

We need the new particle diameter for the same terminal velocity with ρ_p = 2 g/cm³.
Since v remains constant and g, η, and ρ_w do not change, equate velocities:
((2/9) * (r₁²(ρ₁ - ρ_w))) = ((2/9) * (r₂²(ρ₂ - ρ_w)))
where ρ₁, ρ₂ are densities, and r₁, r₂ the respective radii.

Given:

• ρ₁ = 3 g/cm³
• ρ₂ = 2 g/cm³
• d₁ = 2 mm => r₁ = 1 mm (since diameter = 2*radius)

By inserting known values:
(1² * (3 - 1)) = (r₂² * (2 - 1))
Solving: 1² * 2 = r₂² * 1
Thus, r₂² = 2 → r₂ = √2 mm.
So, the diameter d₂ is 2√2 mm.
Approximating: d₂ = 2.83 mm.

To find the average interstitial velocity of groundwater through the core, we start by using the formula for interstitial velocity, :



Where:
= hydraulic conductivity = 1.2 cm/min
= hydraulic gradient = = 0.2 (since head difference is 20 cm and length of core is 100 cm)
= effective porosity = 0.2

Now, converting hydraulic conductivity from cm/min to m/day:

(m/day) = 1.2 cm/min × 60 min/hour × 24 hour/day × 0.01 m/cm = 17.28 m/day

Substitute the values into the formula for interstitial velocity:



This confirms that the calculated interstitial velocity is 17.28 m/day,

Neutron-Rich Unstable Nuclides and β⁻ Decay

Neutron Excess and Instability

Neutron-rich unstable nuclides have an excess of neutrons, making them unstable. To achieve stability, they undergo β⁻ (negatron) decay.

β⁻ Decay Process

In β⁻ decay, a neutron ( ) is converted into a proton ( ), emitting an electron ( ) and an antineutrino ( ):

Explanation

• The neutron transforms into a proton, increasing the proton-to-neutron ratio.
• An electron ( ) is emitted as a beta particle.
• An antineutrino ( ) is released to conserve lepton number.

Effect on Atomic Number

Since a neutron is converted into a proton, the atomic number ( ) of the nuclide increases by 1, but the mass number ( ) remains unchanged.

Example of β⁻ Decay

Consider the beta decay of carbon-14:

Here, carbon-14 ( ) transforms into nitrogen-14 ( ), increasing its atomic number while keeping the mass number constant.

Conclusion

β⁻ decay is a fundamental nuclear process that allows neutron-rich nuclides to move toward a stable configuration by converting excess neutrons into protons.

Calculating the Bulk Modulus

Step 1: Understanding the Formula

The bulk modulus is defined by the formula:

Where:

• is the pressure change.
• is the change in volume.
• V is the initial volume.

Step 2: Given Values

Given:

• Initial volume
• Final volume

Step 3: Calculate the Change in Volume

The change in volume is calculated as:

Step 4: Calculate the Volumetric Strain

The volumetric strain is calculated as:

Step 5: Calculate the Bulk Modulus

Now, we can calculate the bulk modulus by substituting the values into the formula:

Final Answer:

The bulk modulus of the rock is: 2 × 10⁶ N/m².

Calculating the Porosity of the Soil Sample

Step 1: Calculate the Volume of Voids

The formula to calculate porosity ( ) is given by:

Where:

• is the volume of voids (air + water).
• is the total volume of the sample.

To calculate the volume of voids, we use the following equation:

Given:

• V_total = 1200 cc
• Mass of solids = 1908 g
• Particle density = 2.65 g/cc

First, calculate the volume of solids:

Now, calculate the volume of voids:

Step 2: Calculate the Porosity

Now, we can calculate the porosity using the formula:

Final Answer:

The porosity of the soil sample is: 40%.

Calculating the Fraction of Atoms Remaining After a Given Time

Step 1: Understanding the Formula

The fraction of atoms remaining after a given time can be calculated using the radioactive decay formula:

Where:

• is the fraction of remaining atoms.
• t is the time elapsed (5.0 hours).
• t₁/₂ is the half-life of the isotope (15.0 hours).

Step 2: Substituting the Given Values

Substitute the given values into the formula:

Final Answer:

The fraction of ²⁴11Na atoms remaining is approximately: 0.796 (rounded to three decimal places).

Calculating the Lithostatic Pressure

Step 1: Understanding the Formula

The lithostatic pressure (P) can be calculated using the formula:

Where:

• P is the lithostatic pressure.
• is the density of the rock (2800 kg/m³).
• g is the acceleration due to gravity (9.8 m/s²).
• h is the depth (36.5 km = 36,500 m).

Step 2: Substituting the Given Values

Substituting the given values into the formula:

Step 3: Convert Pascals to Gigapascals (GPa)

To convert from Pascals to Gigapascals (GPa), divide by :

Final Answer:

The lithostatic pressure is: 1 GPa.

Calculating the Geothermal Temperature

Step 1: Understanding the Formula

The geothermal temperature can be calculated using the formula:

Where:

• T_surface is the surface temperature (25°C).
• Gradient is the geothermal gradient (0.02°C/m).
• Depth is the depth (18 km = 18,000 meters).

Step 2: Substitute the Given Values

Substitute the given values into the formula:

Now, calculate the geothermal temperature:

Final Answer:

The temperature at a depth of 18 km is: 385°C (rounded to the nearest integer).

Calculating the Volume of Water Flowing Through the Aquifer Using Darcy's Law

Step 1: Understanding the Formula

Darcy's Law is given by the equation:

Where:

• Q is the volume of water per second (discharge).
• K is the hydraulic conductivity (0.25 cm/s).
• A is the cross-sectional area (10 m²).
• i is the hydraulic gradient (0.04).

Step 2: Convert the Cross-Sectional Area

The given cross-sectional area is in square meters (m²). We need to convert it to square centimeters (cm²):

Step 3: Substitute the Values into the Formula

Substitute the given values into Darcy's Law:

Now, calculate the volume of water flowing per second:

Final Answer:

The volume of water flowing per second is: 1000 cm³/s (rounded to three decimal places).

Calculating the Mean Flow Velocity using the Froude Number

Step 1: Understanding the Formula

The Froude Number (Fr) is given by the equation:

Where:

• V is the mean flow velocity in m/s.
• g is the acceleration due to gravity (9.8 m/s²).
• D is the depth of the flow (0.2 m).

Step 2: Rearranging the Equation to Solve for V

Rearranging the equation to solve for the mean flow velocity (V):

Step 3: Substituting the Given Values

Substituting the given values into the equation:

First, calculate .

Now, take the square root:

Thus, the mean flow velocity (V) is:

Final Answer:

The mean flow velocity is: 5.6 m/s (rounded to one decimal place).

Darcy’s Law and Groundwater Flow

Understanding Darcy’s Law

Darcy’s Law quantifies groundwater flow as a function of:

• Permeability (K): The ability of a porous medium to transmit fluid.
• Hydraulic Gradient (I): The change in hydraulic head per unit distance.
• Cross-Sectional Area (A): The area through which groundwater flows.

Mathematically, Darcy’s Law is expressed as:

where:

• = Discharge (volume of water flowing per unit time)
• = Hydraulic conductivity (permeability)
• = Hydraulic gradient
• = Cross-sectional area

Primary Quantification

Although Darcy’s Law considers permeability, hydraulic gradient, and cross-sectional area, its primary purpose is to quantify the volume of water flowing per unit time in a porous medium.

Conclusion:

Darcy’s Law is fundamental in hydrogeology, helping estimate groundwater movement and manage water resources effectively.

Step 1: Understanding the Concept of Saturated Thickness in an Unconfined Aquifer

An unconfined aquifer is one where the upper boundary is defined by the water table. The saturated thickness is the vertical distance between the water table (upper boundary) and the underlying confining layer (lower boundary).

Step 2: Explanation of Options

• (A): Incorrect. This defines the depth of the unsaturated zone, not the saturated thickness.
• (B): Correct. Saturated thickness is the vertical distance between the water table and the confining layer beneath.
• (C): Incorrect. This refers to an arbitrary reference (mean sea level), not the aquifer’s boundaries.
• (D): Incorrect. This would include both saturated and unsaturated zones, which is not the definition of saturated thickness.

Final Answer:

Hence, the correct option is: (B).

Step 1: Identifying the events of interest.
From the given table, we focus on the years when a "Large + GLOF" event occurred

From the table, the "Large + GLOF" events occurred in the years 1658, 1872, and 2023. These are the relevant events to calculate the probability.

Step 2: Total number of events in the last 400 years.
The total number of events from 1625 to 2023 is 11.

Step 3: Number of events with "Large + GLOF".
There are 3 events where "Large + GLOF" occurred (1658, 1872, 2023).

Step 4: Probability calculation.
The probability of a large flood accompanied by a GLOF in 2025 is the ratio of "Large + GLOF" events to the total number of events: Step 5: Adjust for the year range.
Since this is over a 400-year period, the probability per year is:
Final Answer: