At a certain place,the horizontal component of earth's magnetic field is $\sqrt{3}$ times the vertical component. The angle of dip at that place is....$^o$

  • A
    $60$
  • B
    $45$
  • C
    $90$
  • D
    $30$

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If ${\phi_1}$ and ${\phi_2}$ are the angles of dip observed in two vertical planes at right angles to each other and ${\phi}$ is the true angle of dip,then:

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$A$ dip circle is adjusted so that its needle moves freely in the magnetic meridian. In this position,the angle of dip is $40^{\circ}$. Now the dip circle is rotated so that the plane in which the needle moves makes an angle of $30^{\circ}$ with the magnetic meridian. In this position,the needle will dip by an angle:

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