$A$ 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.....$s$

$A$ 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?

$A$ magnet oscillating in a horizontal plane has a time period of $2 \ s$ at a place where the angle of dip is $30^o$ and $3 \ s$ at another place where the angle of dip is $60^o$. The ratio of resultant magnetic fields at the two places is

$A$ thin rectangular magnet suspended freely has a period of oscillation equal to $T$. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T'$,then the ratio $\frac{T'}{T}$ is

$A$ short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14 \; cm$ from the centre of the magnet. The earth's magnetic field at the place is $0.36 \; G$ and the angle of dip is zero. What is the total magnetic field (in $G$) on the normal bisector of the magnet at the same distance as the null-point (i.e.,$14 \; cm$) from the centre of the magnet? (At null points,field due to a magnet is equal and opposite to the horizontal component of earth's magnetic 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 $\,A-m^2$)