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CSIR-NET 2016 June

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94. The primary (P) and secondary (S) waves from a shallow focus earthquake reached a seismological observatory at 8:30:04 Hrs and 8:31:16 Hrs respectively. If the velocities of the P and S waves are in the ratio 1.6:1, then the time of occurrence of the earthquake would be

A) 8:27:16 Hrs
B) 8:27:32 Hrs
C) 8:28:04 Hrs
D) 8:28:52 Hrs

Solution:




Travel time of P-wave  from earthquake to seismometer  t_p =8:30:04hr

                              =30,604sec
  

Travel time of S-wave from earthquake to seismometer t_s = 8:31:16 hrs

                            = 30,676sec
 

{v_p}:{v_s}= {1.6}:{1}


v_p =1.6 velocity of the P-wave


v_s = 1 velocity of the S-wave


Distance can be calculated by using below this formula

t_{s}-t_{p}=D(\frac{1}{v_s}-\frac{1}{v_p})

D=\frac{t_s-t_p}{(\frac{1}{v_s}-\frac{1}{v_p})}

D=\frac{72}{(\frac{1}{1}-\frac{1}{1.6})}

D=\frac{72}{(\frac{1.6-1}{1.6})}

D=\frac{72*1.6}{0.6}

D=192 km



The relation between the distance, velocity of the P-wave and origin time of earthquake (t_0)

D=V_p (t_p-t_0)

t_0= (t_p-\frac{D}{V_p})

t_0= (30,604-\frac{192}{1.6})

t_0= (30,604-120)

t_0= 30,484 sec

Convert 30,484 sec we will get 8:28:04 Hrs

Reference :http://mjgeophysics.blogspot.com/


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