Two magnets are held together in a vibration magnetometer and are allowed to oscillate in the Earth's magnetic field. With like poles of magnets together,$12$ oscillations per minute are made,but for unlike poles together,only $4$ oscillations per minute are executed. The ratio of their magnetic moments is

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.....$s$.

In a uniform magnetic field of $0.049 \ T$, a magnetic needle performs $20$ complete oscillations in $5 \ s$. The moment of inertia of the needle is $9.8 \times 10^{-5} \ kg \ m^2$. If the magnitude of the magnetic moment of the needle is $x \times 10^{-5} \ A \ m^2$, then the value of '$x$' is: (in $\pi^2$)

$A$ magnetic needle is made to vibrate in a uniform magnetic field $H$. Its time period is $T$. If it vibrates in a field of intensity $4H$,its time period will be:

$A$ magnetic needle oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60^{\circ}$. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian,its period will be

$A$ 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. If the bar magnet is turned around by $180^o$,where will the new null points (in $cm$) be located?