$A$ 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:

$A$ 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....$s$

The time period of a freely suspended magnet is $2 \ s$. If it is broken in length into two equal parts and one part is suspended in the same way,then its time period will be.....$s$.

$A$ 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).

Two mutually perpendicular uniform magnetic fields $F$ and $H$ act on a magnetic needle. The needle makes an angle of $60^\circ$ with the field $F$. Find the ratio $\frac{F}{H}$.