NEET Class neet Answered

Would you like to explain me this one question please

Asked by arpita030903 | 23 Jul, 2021, 11:43: PM

Expert Answer

In a dip circle , apparant dip angle θ is related to true dip angle δ as given below

begin mathsize 14px style tan theta space equals space fraction numerator tan delta over denominator cos alpha end fraction end style

............................(1)

where α is anle made by plane of dip circle with magnetic meridian.

In first case , apparant dip angle is 30^o

Hence , we get from eqn.(1) ,

begin mathsize 14px style cos alpha space equals space fraction numerator tan delta over denominator tan 30 end fraction space equals space square root of 3 space tan delta space end style

...........................(2)

In second case , plane of dip circle is rotated by 90^o and apparaent dip angle is 45^o

Hence , we get from eqn.(1) ,

begin mathsize 14px style tan 45 space equals space fraction numerator tan delta over denominator cos left parenthesis 90 minus alpha right parenthesis end fraction space equals fraction numerator tan delta over denominator sin alpha end fraction space end style

...........................(3)

Hence , we get , sinα = tanδ ..................... (4)

If we eliminate α in eqn.(2) and (4) , using sin² α + cos² α = 1 ,

then we get , 3 tan² δ + tan² δ = 1

From above expression, we get , tan δ = 1/2

Hence , true dip angle δ = tan^-1( 1/2 )

Answered by Thiyagarajan K | 24 Jul, 2021, 11:17: PM

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