Answer

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**Hint:**A Dip at a certain area or place is determined by a dip circle. It has a needle which is capable of vertical rotation in an axis which is horizontal. The angle which is made by the needle with the horizontal axis is called apparent dip.

**Complete step by step solution:**

Find the true dip

\[B_H' = {B_H}\cos \theta = {B_H}\cos 45\];

Put the value of cos45 in the above equation and solve;

\[ \Rightarrow B_H' = \dfrac{{{B_H}}}{{\sqrt 2 }}\];

The apparent dip is related to the vertical magnetic field $Bv$ upon the horizontal magnetic field $B_H'$.

$\tan \delta ' = \dfrac{{Bv}}{{B_H'}}$;

Put the value of $B_H'$in the above equation:

$ \Rightarrow \tan \delta ' = \dfrac{{Bv}}{{\dfrac{{{B_H}}}{{\sqrt 2 }}}}$;

Simplify the equation:

$ \Rightarrow \tan \delta ' = \dfrac{1}{{\sqrt 2 }}\dfrac{{Bv}}{{{B_H}}}$;

$ \Rightarrow \tan \delta ' = \sqrt 2 \tan \delta $; ….$\left( {\delta = \dfrac{{{B_v}}}{{{B_H}}}} \right)$

Here the apparent dip is given by$\delta ' = 30^\circ $:

$ \Rightarrow \tan 30 = \sqrt 2 \tan \delta $;

$ \Rightarrow 0.57 = \sqrt 2 \tan \delta $;

Take the root value to the LHS:

$ \Rightarrow \dfrac{{0.57}}{{\sqrt 2 }} = \tan \delta $;

The value for the true dip is:

${\tan ^{ - 1}}\dfrac{1}{{\sqrt 6 }} = \delta $;

**Hence, Option (D) is correct.**

The true dip is \[{\tan ^{ - 1}}\dfrac{1}{{\sqrt 6 }}\].

**Additional Information:**

When a dip circle in a certain plane of scale comes to rest when the needle is in the magnetic meridian and is in the direction of Earth’s magnetic field. The angle which is made by the needle with an axis which is horizontal is called True Dip.

**Note:**Here we have to establish a relation between true dip and apparent dip but before that make a relation between the horizontal magnetic field$\left( {B_H'} \right)$, Vertical magnetic field $\left( {B_v'} \right)$ and the apparent dip. The true dip is given by $\left( {\delta = \dfrac{{{B_v}}}{{{B_H}}}} \right)$. Solve the equations by using trigonometric properties and find out the true dip.

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