Twists of suspension fibre should be removed in a vibration magnetometer so that:

$A$ 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

When a bar magnet is freely suspended for oscillation,its time period is $T$. If it is cut into $3$ equal parts perpendicular to its length,then the time period of oscillation for each part will be

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \, s$. The magnet is cut along its length into three equal parts and these three parts are then placed on each other with their like poles together. The time period of this combination will be

$A$ short bar magnet is placed in the magnetic meridian of the earth with its north pole pointing towards the geographic north. Neutral points are found at a distance of $30 \, cm$ from the magnet on the East-West line, drawn through the center of the magnet. The magnetic moment of the magnet in $A \cdot m^2$ is close to: (Given $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$ and $B_H = 3.6 \times 10^{-5} \, T$)

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