$A$ compass needle oscillates $20$ times per minute at a place where the dip is $45^{\circ}$ and the magnetic field is $B_1$. The same needle oscillates $30$ times per minute at a place where the dip is $30^{\circ}$ and magnetic field is $B_2$. Then $B_1: B_2$ is

$A$ bar magnet is placed in the magnetic meridian with its south pole pointing towards the geographic north. The neutral point is at a distance of $20 \, cm$ from the center of the magnet. If the horizontal component of the Earth's magnetic field is $B_H = 0.3 \times 10^{-4} \, Wb/m^2$,what is the magnetic moment of the magnet?

The time period of a magnet is $T$. If it is divided into four equal parts along its axis and perpendicular to its axis as shown,then the time period for each part will be

With a standard rectangular bar magnet, the time period of a vibration magnetometer is $4 \,s$. The bar magnet is cut parallel to its length into four equal pieces. The time period of the vibration magnetometer when one piece is used (in seconds) (bar magnet breadth is small) is

$A$ short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14 \; cm$ from the centre of the magnet. The earth's magnetic field at the place is $0.36 \; G$ and the angle of dip is zero. If the bar magnet is turned around by $180^o$,where will the new null points (in $cm$) be located?

$A$ vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is ${2^{5/4}} \ s$. One of the magnets is removed and if the other magnet oscillates in the same field,then the time period in seconds is