GATE-2018 (24) :Wyllie – Time average equation

CSIR-NET 2015

79. If the average density of rocks in the crust is 2750 kg/m^3  and the value of 'g' (acceleration due to gravity) is 10 m/s^2, then what is the value of litho-static stress at a depth of 35km in the crust?

A) 962.5 MPa        
B) 96.25 MPa            
C) 9.625 MPa                 
D) 9625.0 MPa

Solution:

Litho-static stress (or) litho-static pressure (p) =

 Force/area-----------(1)

Force = mass*acceleration

F=mg-------(2)

P=mg/A-----------(3)

mass(m) = density*volume = ρ*v ------(4)

equation (4) substitute in equation (3)

P=ρ * v * g / A = ρ * h³ * g / h² = ρ * h*g----------(5)

density (ρ) = 2750 kg/m³

depth (or) height (h) = 35 Km

acceleration due to gravity (g) =10m/s²

These values substitution in equation (5),

P= ρ * h*g

  =2750 * 35000 * 10 kg/m³ * m * m/s²

  = 962,500,000 Kg/ms²

  =962,500,000 Pa

1Mega pascal = 10,00,000 pascal

Therefore, litho-static stress (or) P = 962.5MPa