A dip circle is set such that the dip needle moves freely in the magnetic meridian. In this position, the angle of dip is 39 $^{0}$ . Now the dip circle is rotated so that the plane in which the needle moves makes an angle of 30 $^{0}$ with the magnetic meridian. In this position ,the needle will dip by an angle :

exactly 39

^{0}

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^{0}

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more than 39

^{0}

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less than 39

^{0}

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Solution

The correct option is B more than 39 $^{0}$
The relation between true dip and the apparent dip is given by
$t a n δ_{a} = \frac{t a n δ_{t}}{c o s θ}$
$t a n δ_{a} = t a n 39^{o} \times 1.15$
Therefore $δ_{a}$ i.e , the apparent dip angle value increases

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Similar questions

Q. A dip of circle is so set that it moves freely in the magnetic meridian. In this position, the angle of dip is

40^{0}

. Now, the dip circle is rotated so that the plane in which the needle moves makes an angle of

30^{0}

with the magnetic meridian. In this position, the needle will dip by an angle of

Q. 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

Q. The needle of the dip circle of a place stay at

30^{0}

. The dip circle is inclined at an angle towards

30^{0}

with the magnetic meridian . The true dip at the place is:

The real angle of dip, if a magnet is suspended at an angle of $30^{0}$ to the magnetic meridian and the dip needle makes an angle of 45 $^{0}$ with horizontal is :

Q. A magnetic needle is free to rotate in vertical plane. Initially plane of rotation of needle is perpendicular to magnetic meridian and needle is in vertical position (i.e. angle of dip is

90^{\circ}

). Now if plane of rotation of needle is rotated by

60^{\circ}

about a vertical axis and angle of dip measured in this position is found to be

30^{\circ}

, then true angle of dip at this place is

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The correct option is B more than 390The relation between true dip and the apparent dip is given bytanδa=tanδtcosθtanδa=tan39o×1.15Therefore δa i.e , the apparent dip angle value increases

The correct option is B more than 39 $^{0}$
The relation between true dip and the apparent dip is given by
$t a n δ_{a} = \frac{t a n δ_{t}}{c o s θ}$
$t a n δ_{a} = t a n 39^{o} \times 1.15$
Therefore $δ_{a}$ i.e , the apparent dip angle value increases