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

 

110)  Gravity anomalies of values 0.1 mgals and 0.2 mgals are located at points separated by 300 m and 200 m respectively along a profile across an anomalous body resembling a horizontal circular cylinder. The depth to the center of cylinder is 

(Special thanks to Y.Suresh Sir, GSI)

(Thanks to Chandrasekhar, ANU)

Solution:

The Gravity anomaly for a Horizontal cylinder is

$\triangledown g=\frac{2Gmz}{(x^{2}+z^{2})}$

where,

z$\rightarrow$ Depth to the centre of the cylinder

x$\rightarrow$Distance in profile

m$\rightarrow$ mass of the cylinder

$\triangledown g$ Gravity anomaly

 

Given that

 At $x_{1}=300 m \triangledown g_{1}=0.1 mgal$

 

$x_{2}=200 m \triangledown g_{2}=0.2 mgal$

Therefore by substituting the given values

$\triangledown g=\frac{2Gmz}{(x^{2}+z^{2})}$

 

$0.1=\frac{2Gmz}{((300)^{2}+z^{2})}$---------(1)

 

$0.2=\frac{2Gmz}{((200)^{2}+z^{2})}$---------(2)

 

$\frac{(1)}{(2)}\Rightarrow \frac{0.1}{0.2}=\frac{2Gmz}{((300)^{2}+z^{2})}\times \frac{((200)^{2}+z^{2})}{2Gmz}$

 

$\Rightarrow\frac{1}{2}=\frac{((200)^{2}+z^{2})}{((300)^{2}+z^{2})}$

 

$\Rightarrow 1\times ((300)^{2}+z^{2})=2\times((200)^{2}+z^{2})$

 

$\Rightarrow 9\times 10^{4}+z^{2}=8\times 10^{4}+2z^{2}$

 

$\Rightarrow z^{2}=1\times 10^{4}$

 

$z=100m$