In the sum and difference method of a vibration magnetometer,the time period is greater if:

Write the equation of the periodic time for an oscillating bar magnet in a uniform magnetic field.

$A$ bar magnet of moment of inertia $49 \times 10^{-2} \,kg-m^2$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \,T$. The time period of vibration is $8.8 \,s$. The magnetic moment of the bar magnet is (in $\,A-m^2$)

$A$ 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

The time period of oscillation of a bar magnet suspended horizontally along the magnetic meridian is $T_0$. If this magnet is replaced by another magnet of the same size and pole strength but with double the mass,the new time period 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