The time period of a vibration magnetometer is $T_0$. Its magnet is replaced by another magnet whose moment of inertia is $3$ times and magnetic moment is $1/3$ of the initial magnet. The time period now will be

$A$ short bar magnet having 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.)

$A$ 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 $\,A-m^2$)

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

In the figure,the magnetic needle has a magnetic moment $m = 6.7 \times 10^{-2} \; A \cdot m^2$ and a moment of inertia $I = 7.5 \times 10^{-6} \; kg \cdot m^2$. It performs $10$ complete oscillations in $6.70 \; s$. What is the magnitude of the magnetic field $B$ (in $; T$)?

$A$ magnetic needle has a magnetic moment of $5 \times 10^{-2} \text{ A m}^2$ and a moment of inertia of $8 \times 10^{-6} \text{ kg m}^2$. It has a period of oscillation of $2 \text{ s}$ in a magnetic field $\vec{B}$. The magnitude of the magnetic field is approximately: