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 bar magnet having a magnetic moment $4 \text{ Am}^2$,placed in a vibrating magnetometer,vibrates with a time period of $8 \text{ s}$. Another short bar magnet having a magnetic moment $8 \text{ Am}^2$ vibrates with a time period of $6 \text{ s}$. If the moment of inertia of the second magnet is $9 \times 10^{-2} \text{ kg-m}^2$,the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field).

$A$ dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be $2$ seconds. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be $2$ seconds. Then the angle of dip is.....$^o$

The unit of the reduction factor of a tangent galvanometer is

$A$ magnet is suspended horizontally in the earth's magnetic field. When it is displaced and then released,it oscillates in a horizontal plane with a period $T$. If a piece of wood of the same moment of inertia (about the axis of rotation) as the magnet is attached to the magnet,what would the new period of oscillation of the system become?

The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by $36\%$. By doing this,the periodic time of the magnetometer will