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At an angle of 30^(@) to the magnetic m...

At an angle of `30^(@)` to the magnetic meridian, the apparent dip is `45^(@)`. Find the true dip:

A

`"tan"^(-1)(1)/(sqrt(3))`

B

`tan^(-1)sqrt(3)`

C

`"tan"^(-1)(2)/(sqrt(3))`

D

`"tan"^(-1)(2)/(sqrt(3))`

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Knowledge Check

  • At 45^(@) to the magnetic meridian the apparent dip is 60^(@) . The true dip is

    A
    `tan^(-1)sqrt(3)`
    B
    `tan^(-1)"1/sqrt(3)`
    C
    `tan^(-1)"sqrt(3)/2`
    D
    `tan^(-1)"sqrt(1/6)`

  • If a magnet is suspended at an angle of 30^(@) to the magnetic meridián, the dip needle makes an angle of 45^(@) with the horizontal. The real dip is

    A
    `tan^(-1)(sqrt3"/"2)`
    B
    `tan^(-1)(sqrt3)`
    C
    `tan^(-1)(sqrt3"/"sqrt2)`
    D
    `tan^(-1)(2"/"sqrt3)`

  • The real angle of dip, if a magnet is suspended at an angle of 30^(@) to the magnetic meridian and the dip needle makes an angle of 45^(@) with horizontal, is:

    A
    `tan^(-1)(sqrt(3)/2)`
    B
    `tan^(-1)(sqrt(3))`
    C
    `tan^(-1)(sqrt(3)/sqrt(2))`
    D
    `tan^(-1)(2/sqrt(3))`

    • Similar Questions

    Explore conceptually related problems

    In a vertical plane p_(1) making angle 30^(@) with magnetic meridian , apparent angle of dip is 60^(@) . The apparent angle of dip in vertical plane p_(2) , which is perpendicular to plane p_(1) is

    The angle of dip at the magnetic equator is

    The angle of dip at the magnetic equator is

    If a dip needle is suspended at an angle 30° to the magnetic meridian, it makes an angle of 45^@ with the horizontal. The real dip is

    The angle of dip at the magnetic pole is

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