The period of oscillations of a magnetic needle in a magnetic field is $1.0 \, s$. If the length of the needle is halved by cutting it,the time period will be.....$s$.

$A$ 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?

$A$ short bar magnet having a magnetic moment $4 \text{ Am}^2$,placed in a vibrating magnetometer,vibrates with a time period of $8 \text{ s}$. Another short bar magnet having a magnetic moment $8 \text{ Am}^2$ vibrates with a time period of $6 \text{ s}$. If the moment of inertia of the second magnet is $9 \times 10^{-2} \text{ kg-m}^2$,the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field).

Two magnets are held together in a vibration magnetometer and are allowed to oscillate in the Earth's magnetic field. With like poles of magnets together,$12$ oscillations per minute are made,but for unlike poles together,only $4$ oscillations per minute are executed. The ratio of their magnetic moments is

Two bar magnets having the same geometry with magnetic moments $M$ and $2M$ are first placed in such a way that their similar poles are on the same side,and their time period of oscillation is $T_{1}$. Now,the polarity of one of the magnets is reversed,and the time period of oscillation becomes $T_{2}$. Then:

Two magnets of same size and mass make respectively $10$ and $15$ oscillations per minute at a certain place. The ratio of their magnetic moments is