$A$ magnetic needle of magnetic moment $6.7 \times 10^{-2} \ Am^2$ and moment of inertia $7.5 \times 10^{-6} \ kgm^2$ is performing simple harmonic oscillations in a magnetic field of $0.01 \ T$. Time taken for $10$ complete oscillations is.....$s$

When the $N$-pole of a bar magnet points towards the geographic south and the $S$-pole points towards the geographic north,the null points are located at the:

When two magnetic moments are compared using the equal distance method,the deflections produced are $45^\circ$ and $30^\circ$. If the lengths of the magnets are in the ratio $1:2$,the ratio of their pole strengths is:

The time period of a thin bar magnet in Earth's magnetic field is $T$. If the magnet is cut into two equal parts perpendicular to its length,the time period of each part in the same field 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 $\,A-m^2$)