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

Two bar magnets $A$ and $B$ are placed one over the other and are allowed to vibrate in a vibration magnetometer. They make $20$ oscillations per minute when the similar poles of $A$ and $B$ are on the same side, while they make $15$ oscillations per minute when their opposite poles lie on the same side. If $M_A$ and $M_B$ are the magnetic moments of $A$ and $B$ and if $M_A > M_B$, the ratio of $M_A$ and $M_B$ is

Write the equation of the periodic time for an oscillating bar magnet in a uniform magnetic field.

$A$ bar magnet of length $10 \, cm$ is kept with its north $(N)$-pole pointing north. $A$ neutral point is formed at a distance of $15 \, cm$ from each pole. Given the horizontal component of the earth's magnetic field is $0.4 \, Gauss$, the pole strength of the magnet is: (in $ \, A-m$)

$A$ thin magnetic iron rod of length $30 \ cm$ is suspended in a uniform magnetic field. Its time period of oscillation is $4 \ s$. It is broken into three equal parts. The time period in seconds of oscillation of one part when suspended in the same magnetic field is

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