Suppose a mountain at location A is in isostatic equilibrium with a column at location B, which is at sea-level, as shown in the figure. The height of the mountain is 4 km and the thickness of the crust at B is 1 km. Given that the densities of crust and mantle are 2700 kg/m\(^3\) and 3300 kg/m\(^3\), respectively, the thickness of the mountain root (r1) is km. (Answer in integer)
Step 1: Understand the principle of isostatic equilibrium.
Isostatic equilibrium implies that the pressure at a certain depth (compensation depth) due to the weight of the overlying material is the same for adjacent columns.
Step 2: Consider the excess mass due to the mountain.
The mountain at location A has an excess height of 4 km above the normal crustal thickness at sea level (location B). This excess mass per unit area is given by the height of the mountain multiplied by the density of the crust: \[ {Excess mass} = h_{mountain} \times \rho_c = 4 { km} \times 2700 { kg/m}^3 \] Step 3: Consider the buoyant force provided by the mountain root.
The mountain root (r1) is composed of crustal material that displaces mantle material. The buoyant force per unit area provided by the root is due to the density difference between the mantle and the crust multiplied by the thickness of the root: \[ {Buoyant force (due to root)} = r1 \times (\rho_m - \rho_c) = r1 \times (3300 - 2700) { kg/m}^3 \] Step 4: Equate the excess mass and the buoyant force for isostatic equilibrium.
For the mountain to be in isostatic equilibrium, the excess mass must be balanced by the buoyant force provided by the root: \[ h_{mountain} \times \rho_c = r1 \times (\rho_m - \rho_c) \] \[ 4 \times 2700 = r1 \times (3300 - 2700) \] \[ 10800 = r1 \times 600 \] Step 5: Solve for the thickness of the mountain root (r1). \[ r1 = \frac{10800}{600} = 18 { km} \] The thickness of the mountain root (r1) is 18 km.
Consider a medium of uniform resistivity with a pair of source and sink electrodes separated by a distance \( L \), as shown in the figure. The fraction of the input current \( (I) \) that flows horizontally \( (I_x) \) across the median plane between depths \( z_1 = \frac{L}{2} \) and \( z_2 = \frac{L\sqrt{3}}{2} \), is given by \( \frac{I_x}{I} = \frac{L}{\pi} \int_{z_1}^{z_2} \frac{dz}{(L^2/4 + z^2)} \). The value of \( \frac{I_x}{I} \) is equal to 
A magma having density of 2900 kg m\(^-3\) just reaches the surface through a two-layered crust as shown in the figure below. Assuming isostatic equilibrium, its depth of melting is .......... km. (Round off to one decimal place) 
While doing Bayesian inference, consider estimating the posterior distribution of the model parameter (m), given data (d). Assume that Prior and Likelihood are proportional to Gaussian functions given by \[ {Prior} \propto \exp(-0.5(m - 1)^2) \] \[ {Likelihood} \propto \exp(-0.5(m - 3)^2) \] 
The mean of the posterior distribution is (Answer in integer)
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