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

The moment of inertia of a magnetic needle is $40 \, g \cdot cm^2$ and it has a time period of $3 \, s$ in the Earth's horizontal magnetic field of $3.6 \times 10^{-5} \, Wb/m^2$. Its magnetic moment will be (in $A \cdot m^2$):

$A$ bar magnet of moment of inertia $49 \times 10^{-2} \,kg-m^2$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \,T$. The time period of vibration is $8.8 \,s$. The magnetic moment of the bar magnet is (in $\,A-m^2$)

$A$ 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....$s$

$A$ bar magnet is placed in the north-south direction with its north pole pointing towards the geographic north. In which direction from the center of the magnet will the points of zero magnetic field (neutral points) be located?

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 ... $s$.