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# The correct value of dip angle at a place is ${45^\circ }$ . If the dip circle is rotated by ${45^\circ }$ out of the meridian, then the tangent of the angle of apparent dip at the place is:(A) $1$ (B) $0.5$ (C) $\dfrac{1}{{\sqrt 2 }}$ (D) $\sqrt 2$  Verified
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Hint: The problem is grounded on using the formulae that relate the magnetic dip and the magnetic declination. The actual and the apparent dip values are calculated and then compared to obtain the expression in terms of the apparent dip. Lastly, the given values are substituted in the acquired expression to find the required value.

Formula used:
$\Rightarrow \tan \rho = \dfrac{V}{H}$

Complete step by step solution
As per the given data, we have the data as follows.
The true value of the dip is,
$\rho = {45^\circ }$
The magnetic dip is the consequence of the propensity of a magnet to align itself with the lines of the magnetic field. A dip needle deals with the inclination or the dip of the Earth’s magnetic field.
The magnetic declination is,
$\theta = {45^\circ }$
The magnetic declination is the angle on the horizontal plane between the magnetic north (the direction of the earth’s magnetic field lines) and the true north, that is, the geographical North Pole.
First, let us acquire the expression for the true dip.
So, we have,
$\tan \rho = \dfrac{V}{H}$ ............ $\left( 1 \right)$
where $V$ is the vertical component and $H$ is the horizontal component.
Now let us find the expression for the apparent dip.
So, we have,
$\tan \rho ' = \dfrac{V}{{H\cos \theta }}$ ............ $\left( 2 \right)$
where $V$ is the vertical component and $H\cos \theta$ is the horizontal component, because of the inclination.
Now compare the equations $\left( 1 \right)$ and $\left( 2 \right)$ to obtain the expression in terms of the apparent dip.
So, we get,
$\tan \rho ' = \dfrac{{\tan \rho }}{{\cos \theta }}$
On Substituting the values in the above equation, so we get,
$\tan \rho ' = \dfrac{{\tan {{45}^\circ }}}{{\cos {{45}^\circ }}}$
Substitute the values of the angles.
$\tan \rho ' = \dfrac{1}{{\dfrac{1}{{\sqrt 2 }}}}$
$\Rightarrow \tan \rho ' = \sqrt 2$
Therefore, the tangent of the angle of apparent dip at the place is $\sqrt 2$ .
So the correct option is (B) $\sqrt 2$ .

At the magnetic equator, the magnetic dip measures ${0^\circ }$ , and at each of the magnetic poles, the magnetic dip measures ${90^\circ }$ .