Question

# A dip needle lies initially in the magnetic meridian when it shows an angle of dip $\theta$ at a place. The dip circle is rotated through an angle $x$ in the horizontal plane and then it shows an angle of dip $\theta \text{'}$. Then $\left(\frac{\mathrm{tan}\theta \text{'}}{\mathrm{tan}\theta }\right)$ is

A

$\frac{1}{\mathrm{cos}x}$

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B

$\frac{1}{\mathrm{sin}x}$

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C

$\frac{1}{\mathrm{tan}x}$

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D

$\mathrm{cos}x$

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Solution

## The correct option is A $\frac{1}{\mathrm{cos}x}$Step 1: Given dataA dip needle lies initially in the magnetic meridian when it shows an angle of dip $\theta$ at a place. The dip circle is rotated through an angle $x$ in the horizontal plane and then it shows an angle of dip $\theta \text{'}$.Step 2: Dip needleA dip needle, also known as a dip circle, is a compass with a pivot that allows it to move in a plane that contains the earth's magnetic field. The magnetic inclination, commonly known as the dip angle, is the angle created by the earth's magnetic field with respect to the horizontal. Step 3: Calculating the value of $\left(\frac{\mathrm{tan}\theta \text{'}}{\mathrm{tan}\theta }\right)$The dip needle is originally in the plane of the magnetic meridian, with a dip angle of. The plane in question is the plane $ABCD$, and the dip angle, or the angle between the horizontal ${B}_{H}$ and the earth's magnetic field ${B}_{V}$, is as follows: $\mathrm{tan}\theta =\frac{{B}_{V}}{{B}_{H}}$Now, the dip circle is rotated through an angle $x$ in the horizontal plane, changing the dip angle to $\theta \text{'}$ .Let this new plane be $CDEF$. Here, there is a component of the horizontal acting in this place, in addition to the earth’s magnetic field. Therefore, $\mathrm{tan}\theta \text{'}=\frac{{B}_{V}}{{B}_{H}\mathrm{cos}\left(x\right)}$Dividing the two equations, we get$\frac{\mathrm{tan}\theta \text{'}}{\mathrm{tan}\theta }=\frac{\left(\frac{{B}_{V}}{{B}_{H}\mathrm{cos}\left(x\right)}\right)}{\left(\frac{{B}_{V}}{{B}_{H}}\right)}\phantom{\rule{0ex}{0ex}}\frac{\mathrm{tan}\theta \text{'}}{\mathrm{tan}\theta }=\frac{1}{\mathrm{cos}\left(x\right)}$Hence, option A is the correct option.

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