$A$ magnet freely suspended in a vibration magnetometer makes $10$ oscillations per minute at a 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

$A$ 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.)

With a standard rectangular bar magnet, the time period of a vibration magnetometer is $4 \,s$. The bar magnet is cut parallel to its length into four equal pieces. The time period of the vibration magnetometer when one piece is used (in seconds) (bar magnet breadth is small) is

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:

$A$ bar magnet of $10 \,cm$ length is kept with its north $(N)$-pole pointing north. $A$ neutral point is formed at a distance of $15 \,cm$ from each pole. Given the horizontal component of Earth's magnetic field is $0.4 \,Gauss$, the pole strength of the magnet is: (in $\,A-m$)

$A$ vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is ${2^{5/4}} \ s$. One of the magnets is removed and if the other magnet oscillates in the same field,then the time period in seconds is