Question

The lithostatic pressure at the base of a 35 km thick continental crust of average density of 2.8 g/cc is................ \(\times 10^8\) Pa.

Hint:

Lithostatic pressure can be calculated using the formula \( P = \rho \cdot g \cdot h \), where \( \rho \) is the density, \( g \) is the gravitational acceleration, and \( h \) is the height of the rock column.

Answer 1

Correct Answer: 9.8

Step 1. Lithostatic pressure is the pressure exerted by a column of rock. It is calculated using the formula:
\[ P = \rho \cdot g \cdot h \]
Where:
- \( \rho \) is the density of the rock (in kg/m\(^3\))
- \( g \) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\))
- \( h \) is the height of the rock column (in meters)
Given: - \( \rho = 2.8 \, \text{g/cc} = 2800 \, \text{kg/m}^3 \) - \( h = 35 \, \text{km} = 35 \times 10^3 \, \text{m} \) - \( g = 9.8 \, \text{m/s}^2 \)

Step 2. Now, substituting the values into the formula:
\[ P = 2800 \times 9.8 \times 35 \times 10^3 \]
\[ P = 9.8 \times 10^8 \, \text{Pa} \] Thus, the lithostatic pressure at the base of the continental crust is \( 9.8 \times 10^8 \) Pa.
\[ \boxed{9.8 \times 10^8 \, \text{Pa}} \]