$A$ magnet makes $40$ oscillations per minute at a place having a magnetic intensity of $0.1 \times 10^{-5} \, T$. At another place, it takes $2.5 \, s$ to complete one oscillation. The value of the Earth's horizontal magnetic field at that place is:

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ (about its center,perpendicular to its length) is cut into two equal pieces,perpendicular to its length. Let $T$ be the period of oscillation of the original magnet about an axis through its midpoint,perpendicular to its length,in a uniform magnetic field $B$. What would be the period $T'$ for each piece?

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

Two different magnets are tied together and allowed to vibrate in a horizontal plane. When their like poles are joined,the time period of oscillation is $5 \, s$,and when their unlike poles are joined,the time period of oscillation is $15 \, s$. The ratio of their magnetic moments is

$A$ bar magnet is oscillating in the Earth's magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled?

$A$ 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....$s$