The time period in a vibration magnetometer will be infinity at

$A$ 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 $^o$)

Two short magnets having magnetic moments in the ratio $27 : 8$,when placed on opposite sides of a deflection magnetometer,produce no deflection. If the distance of the weaker magnet is $0.12 \ m$ from the centre of the deflection magnetometer,the distance of the stronger magnet from the centre is.......$m$

Magnets $A$ and $B$ are geometrically similar,but the magnetic moment of $A$ is twice that of $B$. If $T_1$ and $T_2$ are the time periods of oscillation when their like poles and unlike poles are kept together respectively,then the ratio $\frac{T_1}{T_2}$ will be:

$A$ 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

$A$ very small magnet is placed in the magnetic meridian with its south pole pointing north. The null point is obtained $20 \, cm$ away from the centre of the magnet. If the earth's magnetic field (horizontal component) at this point is $0.3 \, gauss$,the magnetic moment of the magnet is: