In India,the magnetic declination at Delhi is . . . . . . .

At a certain place,the angle of dip is $30^{\circ}$ and the horizontal component of the Earth's magnetic field is $0.50 \text{ Oersted}$. The Earth's total magnetic field is:

The dip angle in a vertical plane at an angle of $\cos^{-1} \left( \frac{1}{\sqrt{2}} \right)$ from the magnetic meridian is $60^\circ$. Find the actual dip angle at that place.

The angle of dip at a place on the earth gives:

$A$ 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:

The angles of dip are $30^{\circ}$ and $45^{\circ}$ at two different places. The ratio of the horizontal components of the Earth's magnetic field at these places will be: