$A$ 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:

$A$ compass needle oscillates $20$ times per minute at a place where the dip is $45^{\circ}$ and the magnetic field is $B_1$. The same needle oscillates $30$ times per minute at a place where the dip is $30^{\circ}$ and magnetic field is $B_2$. Then $B_1: B_2$ is

$A$ short bar magnet having 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.)

The time period of a magnet is $T$. If it is divided into four equal parts along its axis and perpendicular to its axis as shown,then the time period for each part will be

Two tangent galvanometers,which are identical except in their number of turns,are connected in parallel. The ratio of their resistances of the coils is $1 : 3$. If the deflections in the two tangent galvanometers are $30^{\circ}$ and $60^{\circ}$ respectively,then the ratio of their number of turns is

$A$ certain amount of current when flowing in a properly set tangent galvanometer,produces a deflection of $45^{\circ}$. If the current is reduced by a factor of $\sqrt{3}$,the deflection would: