The moment of inertia of a magnetic needle is $40 \, g \cdot cm^2$ and it has a time period of $3 \, s$ in the Earth's horizontal magnetic field of $3.6 \times 10^{-5} \, Wb/m^2$. Its magnetic moment will be (in $A \cdot m^2$):

When a bar magnet is freely suspended for oscillation,its time period is $T$. If it is cut into $3$ equal parts perpendicular to its length,then the time period of oscillation for each part will be

Two tangent galvanometers having coils of the same radius are connected in series. $A$ current flowing in them produces deflections of $60^{\circ}$ and $45^{\circ}$ respectively. The ratio of the number of turns in the coils is

Time period of a freely suspended magnet does not depend upon

At two places $A$ and $B$,using a vibration magnetometer,a magnet vibrates in a horizontal plane. Its respective periodic times are $2 \ s$ and $3 \ s$. At these places,the Earth's horizontal components are $H_A$ and $H_B$ respectively. Then the ratio between $H_A$ and $H_B$ will be:

$A$ compass needle of an oscillation magnetometer oscillates $20$ times per minute at a place $P$ where the angle of dip is $30^{\circ}$. The number of oscillations per minute becomes $10$ at another place $Q$ where the angle of dip is $60^{\circ}$. The ratio of the total magnetic field at the two places $(B_{Q}: B_{P})$ is: