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

 CSIR-NET JUNE 2019

A 1km thick elevated land mass of density 2.7 gm/cc is associated with a free anomaly which is half the Bouguer anomaly. If the density contrast at the crust mantle boundry is 0.3 gm/cc what would be the thickness of the root?

 

(Thank you so much Pawan Singh, BHU)

Solution:

Bouguer anomaly:

$\triangle g_{BA}=g_{m}+(\triangle g_{FC }-\triangle g_{BC}+ \triangle g_{T} +\triangle g_{tide})-g_{n}$-----------(1)

Free air anomaly:

$\triangle g_{FA}=g_{m}+(\triangle g_{FC}+\triangle g_{T}+\triangle g_{Tide})-g_{n}$---------(2)

$equation \left(2-1\right)$

$\triangle g_{FA}-\triangle g_{BA}=\triangle g_{BC}$

$\triangle g_{FA}=\triangle g_{BC}+\triangle g_{BA}----(3)$

$\triangle g_{FA}$---Free air anomaly

$\triangle g_{BA}$----Bouguer anomaly

$\triangle g_{BC}$----Bouguer correction

From above solution

              $\triangle g_{FA}=\frac{1}{2}\triangle g_{BA}---4$

             $ \triangle g_{BA}=2 \triangle g_{FA} $

             Substitute equation 4 in equation 3

$ \frac{1}{2}\triangle g_{BA}=\triangle g_{BC}+\triangle g_{BA}$

$-\frac{1}{2}\triangle g_{BA}=\triangle g_{BC} $

$ -\frac{1}{2}\triangle g_{BA}=2\pi G\triangle\rho t$

$ \triangle g_{BA}=-2(2\pi G\triangle \rho t)$

$\triangle \rho=2700 \frac{kg}{m^{3}}$,thickness =1000m

$ \triangle g_{BA}=-2(0.0419*10^{-3}*2700*1000)$

                         $ =-2(113.13)$

       $\triangle g_{BA}=226.26mgal$

Bouguer anomaly indicating directly root zone. After making the free air and bouguer plate corrections there remains a bouguer anomaly due to block that represents the root zone of a mountain range.

Let the thickness of the root zone is

      $ \triangle g_{BA} =2\pi G\triangle \rho t$

       $226.26=2\pi G(300)*t$

       $226.26=0.0419*10^{-3}(300)*t$

       $\frac{226.26}{0.0419*10^{-3}*300}=t$

       t=18000m

       t=18km.