$A$ dip circle is at a right angle to the magnetic meridian. What will be the apparent dip in degrees?

$A$ magnet is suspended in such a way that it oscillates in the horizontal plane. It makes $20$ oscillations per minute at a place where the dip angle is $30^{\circ}$ and $15$ oscillations per minute at a place where the dip angle is $60^{\circ}$. The ratio of the total Earth's magnetic field at the two places is:

$A$ 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 $\theta'$. Then $\frac{\tan \theta'}{\tan \theta}$ is

In the magnetic meridian of a certain place,the horizontal component of the earth's magnetic field is $0.26 \ G$ and the dip angle is $60^{\circ}$. What is the magnetic field of the earth at this location (in $G$)?

The value of the horizontal component of the earth's magnetic field and the angle of dip are $1.8 \times 10^{-5} \, Wb/m^2$ and $30^{\circ}$ respectively at a certain place. The total intensity of the earth's magnetic field at that place will be:

Magnetic meridian is a