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CSIR NET- JUNE 2019

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CSIR NET- JUNE 2019


97) A seismic source at a depth of 20m generates a seismic wave which is reflected and recorded at the surface at an offset distance of 30m, Assuming that the horizontal reflector lies at a depth of 30m and the P-wave velocity of the medium is 2.5 km/s, the travel time for the reflected wave is ( in milli seconds).


A) 15

B) 20

C) 25

D) 30
(Thanks to Chandrasekhar, ANU)


Solution:

 We need to calculate the travel time from SO to OG.

Where given Offset distance (x) = 30m
Depth of the reflector(z) = 30m
P-wave velocity(Vp)= 2.5 km/s

From Pythagoras theorem, 

From triangle SOB  SO^2 =SB^2 +BO^2---(1)

SB is the half of the offset distance =15 m


OB = 10 m

SO^2 =15^2 +10^2

SO^2 =225 +100

SO^2 =335

SO= 18.3

Travel time from SO is t_{SO} =\frac{Distance}{Velocity}

t_{SO} =\frac{18.3}{2500}

t_{SO} =0.0073 sec----(2)



From triangle AOG  OG^2 =AG^2 +OA^2

OG^2 =15^2 +30^2

OG^2 =225 +900

OG^2 =1125

OG^2 =15^2 +30^2

OG^2 =33.54---(3)

Travel time from OG is t_{OG} =\frac{Distance}{Velocity}

 t_{OG} =\frac{33.54}{2500}

 t_{OG} =0.0134--(4)


Total travel time is t_{SO} +t_{OG} =0.0073+0.0134 s

t_{SG}= 0.0207 s

t_{SG}= 20.7 ms







 







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12 comments

  1. EXACT ANSWER IS 20 milli second
    this is not the correct way to solve the question

    ReplyDelete
    Replies
    1. Same thing i already posted below this post..If you know the solution please share in this blog.

      Delete
  2. you can easily solve this by taking the mirror image of down going wave about reflector.

    ReplyDelete
    Replies
    1. This problem is not possible with mirror image..please check it once..

      Delete
    2. Plz solve 2019 JUNE NET No 98 problem.

      Delete
    3. http://gpsurya.blogspot.com/2020/05/csir-net-2019-june.html

      Delete
    4. If you like this blog please follow me.....

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  3. Nice explanation....keep it up

    ReplyDelete

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