The strength of the magnetic field in which the magnet of a vibration magnetometer is oscillating is increased $4$ times its original value. The frequency of oscillation would then become

$A$ 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 $T'$,the ratio $T'/T$ is :-

Two short magnets having magnetic moments in the ratio $27 : 8$,when placed on opposite sides of a deflection magnetometer,produce no deflection. If the distance of the weaker magnet is $0.12 \ m$ from the centre of the deflection magnetometer,the distance of the stronger magnet from the centre is.......$m$

$A$ 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:

In the sum and difference method of a vibration magnetometer,the time period is greater if:

$A$ 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).