To compare magnetic moments of two magnets by vibration magnetometer,the 'sum and difference method' is better because:

$A$ short bar magnet having 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.)

$A$ magnet freely suspended in a vibration magnetometer makes $10$ oscillations per minute at a place $A$ and $20$ oscillations per minute at a place $B$. If the horizontal component of earth's magnetic field at $A$ is $36 \times 10^{-6} \ T$,then its value at $B$ is

The period of oscillation of a magnet in a vibration magnetometer is $2 \, s$. The period of oscillation of a magnet whose magnetic moment is four times that of the first magnet is ... $s$.

The period of oscillation of a vibration magnetometer depends on which of the following factors? (where $I$ is the moment of inertia of the magnet about the axis of suspension,$M$ is the magnetic moment of the magnet,and $H$ is the external magnetic field.)

$A$ thin rectangular magnet suspended freely has a period of oscillation equal to $T$. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T'$,then the ratio $\frac{T'}{T}$ is