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

csir net(dec)-2018

107) A 2.0 km thick elevated land mass of density 2.7g/cc is associated with a Bouger anomaly of -176g/cc. The free air anamoly is (assume that 2πG = 42 mgals/km/g/cc). 

a) -51 mgals 

b)50 mgals

c) -85 mgals 

d) 85 mgals

(Thanks to Suresh Sir ANU)

(Thanks to Neeraja AU)

Sol:

$$\triangle g_B=g_m+(\triangle g_{FA}-\triangle g_{BP}+\triangle g_T+\triangle g_{tide})-g_n-(1)$$

$$\triangle g_F=g_m+(\triangle g_{FA}+\triangle g_T+\triangle g_{tide})-g_n-(2)$$

 Bouger anamoly $(\triangle g_B)$ - Free  air anamoly$(\triangle g_F)$ =$-\triangle g_{BP} $

$$\triangle g_B-\triangle g_F=-\triangle g_{BP}-(3a)$$

$$\triangle g_{BP}\rightarrow Bouger plate correction$$

$$\triangle g_{BP}=2\pi G\triangle\rho t-(4)$$

$$\triangle g_B-\triangle g_F=-2\pi G\triangle\rho t-(5)$$

$$\triangle\rho=2.7g/cc$$

$$\triangle g_B=-176mgals$$

$$2\pi G=42mgal/km/g/cc$$

$$\triangle g_B-\triangle g_F=-2\pi G\triangle\rho t-(5)$$

$$-176-\triangle g_F=-42\times2.7\times2$$

$$-176-\triangle g_F=-226.8$$

$$\triangle g_F=-226.8+176$$

$$-\triangle g_F=-50.8$$

$$\therefore \triangle g_F=50mgal$$

$$\therefore  Free air anamoly( \triangle g_F)=50mgal$$

$g_m\rightarrow $  measured  gravity  value

$g_n\rightarrow $ normal  gravity  value

$g_{FA}\rightarrow $  Free air  correction

$g_{BP}\rightarrow$  Bouger  plate  correction

$g_T\rightarrow $  terrain  correction

$g_{tide}{\rightarrow   tidal  correction}$