$A$ magnetic needle suspended horizontally by an unspun silk fibre,oscillates in the horizontal plane because of the restoring force originating mainly from

$A$ 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).

At a certain place, a magnet makes $30$ oscillations per minute. At another place where the magnetic field is double, its time period will be:

The time period of oscillation of a magnet in a vibration magnetometer is $1.5 \ s$. The time period of oscillation of another magnet similar in size,shape,and mass but having one-fourth magnetic moment than that of the first magnet,oscillating at the same place,will be.....$s$.

Two small magnets have their masses and lengths in the ratio $1:2$. The maximum torques experienced by them in a uniform magnetic field are the same. For small oscillations, the ratio of their time periods is

$A$ vibration magnetometer placed in the magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2 \, s$ in the earth's horizontal magnetic field of $24 \, \mu T$. When a horizontal field of $18 \, \mu T$ is produced opposite to the earth's field by placing a current-carrying wire,the new time period of the magnet will be....$s$