$A$ bar magnet $A$ of magnetic moment $M_A$ is found to oscillate at a frequency twice that of magnet $B$ of magnetic moment $M_B$ when placed in a vibrating magnetometer. We may say that

Two bar magnets having the same geometry with magnetic moments $M$ and $2M$ are first placed in such a way that their similar poles are on the same side,and their time period of oscillation is $T_{1}$. Now,the polarity of one of the magnets is reversed,and the time period of oscillation becomes $T_{2}$. Then:

$A$ short magnet of magnetic moment $6.75 \, Am^2$ produces a neutral point on its axis. If the horizontal component of the Earth's magnetic field is $5 \times 10^{-5} \, Wb/m^2$, then the distance of the neutral point from the center of the magnet is.....$cm$.

Two bar magnets $A$ and $B$ are placed one over the other and are allowed to vibrate in a vibration magnetometer. They make $20$ oscillations per minute when the similar poles of $A$ and $B$ are on the same side, while they make $15$ oscillations per minute when their opposite poles lie on the same side. If $M_A$ and $M_B$ are the magnetic moments of $A$ and $B$ and if $M_A > M_B$, the ratio of $M_A$ and $M_B$ 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.)

$A$ magnet freely suspended in a vibration magnetometer makes $40$ oscillations per minute at place $A$ and $20$ oscillations per minute at a place $B$. If the horizontal component of earth's magnetic field at $A$ is $36 \times 10^{-6} \ T$, then its value at $B$ is