$A$ dip circle lies initially in the magnetic meridian,it shows an angle of dip $\delta$ at a place. The dip circle is rotated through an angle $\alpha$ in the horizontal plane and then it shows an angle of dip $\delta^{\prime}$. Hence $\frac{\tan \delta^{\prime}}{\tan \delta}$ is

$A$ short magnetic needle is pivoted in a uniform magnetic field of strength $1 \, T$. When another magnetic field of strength $\sqrt{3} \, T$ is applied to the needle in a perpendicular direction,the needle deflects through an angle $\theta$. The value of $\theta$ is: (in $^o$)

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

$A$ short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14 \; cm$ from the centre of the magnet. The earth's magnetic field at the place is $0.36 \; G$ and the angle of dip is zero. If the bar magnet is turned around by $180^o$,where will the new null points (in $cm$) be located?

The time period of oscillation of a magnet in a vibration magnetometer is $1.5 \ s$. The time period of oscillation of another magnet similar in size,shape,and mass but having one-fourth magnetic moment than that of the first magnet,oscillating at the same place,will be.....$s$.

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