1. Gravitational acceleration (gravity) is commonly expressed in units of milligals (mGal). $1 Gal=1cm/s^2=0.01 m/s^2$ $1mGal=10^{-3}Gal=10^{-3} cm/s^2=10^{-5} m/s^2$ References:  1. Fundamentals of Geophysics by William Lowrie. 2. Whole Earth Geophysics by J. Lilliwe ..read more

 96. A Thick Sedimentary formation, in Which the seismic wave Velocity V Increases with depth Z from $V_0$ at the surface to $V_1$ at the bottom according to the relation $V_1=V_0e^{\lambda z}$ . the Two way reflection travel time at a point closest to the short point is ….. (Special Thanks to Pawan Singh) (Thanks to Ashok, AKNU) Solution: Given that $V=V_0e^{\lambda z}$ At  Z  =  0          $V = V_0$     ( surface ) At   Z =  Z     &nb ..read more

 24.  A seismic reflection survey is carried out over a 1500 m thick horizontal layer with a P-wave velocity of 2000 m/s. The travel time of a reflected wave at a surface detector placed 1000m from a surface source is _______ milliseconds. (Thanks to Pragnath, AU) Solution: Given that $V_{p}=2000 m/s=2 Km/s~$;     d( depth)=1500 m=1.5 Km;       x(offset distance)=1000m=1 Km. $T^{2}=\frac{X^{2}}{V^{2}}+(\frac{2D}{V})^2$ $T^{2}=\frac{1^{2}}{2^{2}}+(\frac{2(1.5)}{2})^2$ $T^{2}=\frac{1}{4}+(\frac{3}{4})^2$ $T^{2}=0.25+(1.5)^2$ $T^{2}=0.25 ..read more

 22. Skin Depth in homogeneous media of resistivity $\rho_{1}$ and $\rho_{2}$ are 100 m and 200 m respectively, at 1000Hz frequency. The ratio $\frac{\rho_{1}}{\rho_{2}}$ will be __________.  (Thanks to Pragnath, AU) Solution: Skin depth formula  $D = 503.8 * \sqrt{(\frac{\rho}{f})}$ Given that,  $D_{1}= 100m~,D_{2}= 200 m$;         $f=1000~Hz$ $\frac{D_{1} = 503.8 * \sqrt{(\frac{\rho_{1}}{f})}}{D_{2} = 503.8 * \sqrt{(\frac{\rho_{2}}{f})}}$ $\frac{100}{200}=\frac{503.8 * \sqrt{(\frac{\rho_{1}}{1000})}}{ 503.8 * \sqrt{(\frac{\rho_{2}}{1000 ..read more

 21. A Mid-oceanic-Ridge has symmetric magnetic anomalies about the ridge axis as shown below. Using the information given in the figure, the average relative velocity between the Plates A and b is calculated to be ________ cm/year. (Thanks to Pragnath, AU) Solution: We know that $Velocity=\frac{Distance}{Time}$ Given that,  Distance=50 Km= 5*10^6 cm Time= 1 million year= 10^6 years $Velocity=\frac{5*10^6}{10^6}$ Relative velocity of the ridge is  =10 cm/year   If you fixed one side of the ridge then, the Absolute Velocity is= 5cm/year ..read more

 22.The transmission coefficient for the vertically incident seismic wave at the interface between layer 1 and layer 2 is? Layer1= $V_{1}=1000 m/s,~~~~\rho_{1}=1200kg/m^{3}$ Layer2= $V_{2}=1500 m/s,~~~~\rho_{2}=1300kg/m^{3}$ $V_{1}~V_{2}=P-wave~velocity,~~~~\rho_{1},\rho_{2}=densities$ (Thanks to Pragnath, AU) Solution:         we know that Transmission coefficient $(T~C)=\frac{2(V_{1}*\rho_{1})}{(V_{2}*\rho_{2})+(V_{1}*\rho_{1})}----(1)$ Substituting above values in equation (1) $T~C=\frac{2(1000*1200)}{(1500*1300)+(1000*1200)}$ $T~C=\frac{(24*10 ..read more

 116.What will be the approximate skin depth of electromagnetic waves of frequency 1 kHz in a conductor having conductivity 100 S/m ? a) 16 meters                b). 1.6 meters              c). 016 meters             d). 0.016 meter (Thanks to Pragnath, AU) Solution: -  Skin depth formula $D = 503.8 * \sqrt{(\frac{\rho}{f})}$ Given that (conductivity)$ ~\sigma= 100 S/m~$     &n ..read more

 104.The Laplace Transform of unit step function is:  (a) 1          (b) $\frac{1}{s}$        (c) $\frac{1}{s^{2}}$ (d) $\frac{1}{s^{3}}$   (Thanks to Pragnath, AU)   Solution: - The Laplace transform is a mathematical tool which is used to convert the differential equation in time domain into the algebraic equations in the frequency domain.  if x(t) is a time-domain function, then its Laplace transform is defined as $X_{(s)} = L[u(t)] = \int_{0}^{\infty}~u(t)e^{-st}~dt$   Th ..read more

 97. In a seismic reflection survey, the inferred upper and lower layer velocities are 2 km/s and 6 km/s, respectively. The depth of the refractor is 0.5 km. What will be the crossover distance? a) 2828m            b) 1414m            c) 707m              d) 353.5m  (Thanks to Pragnath, AU) Solution:  $X_{~crossover}=2Z*\sqrt{~\frac{V_{2}+V_{1}}{V_{2}-V_{1}}}$   Given that $Z=0.5km$    ..read more

 96. The foldage of a 2D common depth point (CDP) reflection profiling conducted using a 72-channel digital recording unit with a geophone interval of 50 m and shot interval of 100 m is a) 18          b) 36          c) 72          d) 144 (Thanks to Pragnath, AU) Solution: Foldage:- A measure of the redundancy of common midpoint per bin in seismic data. $Foldage=\frac{1}{2}*[Total~receivers~*\frac{Receiver~interval}{Shot~interval}]$   Given that ..read more