$A$ magnet oscillating in a horizontal plane has a time period of $2 \ s$ at a place where the angle of dip is $30^o$ and $3 \ s$ at another place where the angle of dip is $60^o$. The ratio of resultant magnetic fields at the two places is

$A$ deflection magnetometer is adjusted and a magnet of magnetic moment $M$ is placed on it in the usual manner and the observed deflection is $\theta$. The period of oscillation of the needle before settling of the deflection is $T$. When the magnet is removed,the period of oscillation of the needle is $T_0$ before settling to $0^{\circ}-0^{\circ}$. If the earth's horizontal magnetic field is $B_H$,the relation between $T$ and $T_0$ is

$A$ magnet is placed horizontally on the ground with its north pole pointing towards the geographic north pole of the Earth. Where is the neutral point obtained?

In a vibration magnetometer,the time period of a bar magnet oscillating in the horizontal component of the Earth's magnetic field is $2 \ s$. When a magnet is brought near and parallel to it,the time period reduces to $1 \ s$. The ratio $H/F$ of the horizontal component $H$ and the field $F$ due to the magnet will be:

$A$ bar magnet is placed on a table along the east-west direction in the Earth's magnetic field. The number of neutral points,where the magnetic field is zero,is

$A$ magnet of magnetic moment $M$ oscillating freely in earth's horizontal magnetic field makes $n$ oscillations per minute. If the magnetic moment is quadrupled and the earth's field is doubled,the number of oscillations made per minute would be