The number of turns and radius of cross-section of the coil of a tangent galvanometer are doubled. The reduction factor $K$ will be

$A$ vibration magnetometer placed in the magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2 \, s$ in the earth's horizontal magnetic field of $24 \, \mu T$. When a horizontal field of $18 \, \mu T$ is produced opposite to the earth's field by placing a current-carrying wire,the new time period of the magnet will be....$s$

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

$A$ 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 $\,A-m^2$)

Two tangent galvanometer coils of the same radius are connected in series. The current flowing produces deflections of $60^o$ and $45^o$. The ratio of the number of turns in the coils is:

Which curve best represents the relationship between current $i$ and deflection $\theta$ in a tangent galvanometer?