The angle of dip at a certain place on earth is $60^{\circ}$ and the magnitude of earth's horizontal component of magnetic field is $0.26 \, G$. The magnetic field at the place on earth is.....$G$

The angle of dip is $90^{\circ}$ at:

At an angle of $30^{\circ}$ to the magnetic meridian,the apparent dip is $45^{\circ}$. Find the true dip.

$A$ 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:

If $\theta_1$ and $\theta_2$ are the apparent angles of dip observed in two vertical planes at right angles to each other,then the true angle of dip $\theta$ is given by:

$A$ compass needle oscillates $20$ times per minute at a place where the dip is $30^{\circ}$ and $30$ times per minute where the dip is $60^{\circ}$. The ratio of total magnetic field due to the earth at the two places respectively is $\frac{4}{\sqrt{x}}$. The value of $x$ is $............$.