With a standard rectangular bar magnet of length $l$, breadth $b$ $(b \ll l)$ and magnetic moment $M$, the time period of the magnet in a vibration magnetometer is $4 \ s$. If the magnet is cut normal to its length into four equal pieces, the time period (in seconds) with one of the pieces is

The period of oscillation of a bar magnet at a place is $2 \,s$. At the same place, the period of oscillation of another identical bar magnet whose magnetic moment is $4$ times that of the first magnet is: (in $\,s$)

The magnetic needle of a vibration magnetometer makes $12$ oscillations per minute in the horizontal component of earth's magnetic field. When an external short bar magnet is placed at some distance along the axis of the needle in the same line, it makes $15$ oscillations per minute. If the poles of the bar magnet are interchanged, the number of oscillations it makes per minute is

$A$ short magnetic needle is pivoted in a uniform magnetic field of strength $1 \, T$. When another magnetic field of strength $\sqrt{3} \, T$ is applied to the needle in a perpendicular direction,the needle deflects through an angle $\theta$. The value of $\theta$ is: (in $^o$)

$A$ thin magnetic iron rod of length $30 \ cm$ is suspended in a uniform magnetic field. Its time period of oscillation is $4 \ s$. It is broken into three equal parts. The time period in seconds of oscillation of one part when suspended in the same magnetic field is

In a uniform magnetic field,a magnetic needle has a magnetic moment $9.85 \times 10^{-2} \, A \cdot m^{2}$ and a moment of inertia $5 \times 10^{-6} \, kg \cdot m^{2}$. If it performs $10$ complete oscillations in $5 \, s$,then the magnitude of the magnetic field is $.... \, mT$. [ Take $\pi^{2} = 9.85$ ]