The tangent galvanometer,when connected in series with a standard resistance,can be used as:

$A$ 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. What is the total magnetic field (in $G$) on the normal bisector of the magnet at the same distance as the null-point (i.e.,$14 \; cm$) from the centre of the magnet? (At null points,field due to a magnet is equal and opposite to the horizontal component of earth's magnetic field.)

$A$ magnet is suspended horizontally in the Earth's magnetic field. When it is made to oscillate in a horizontal plane,its time period is $T$. If a wooden block of the same moment of inertia as the magnet is attached to it,what will be the new time period of the system?

$A$ current of $\left(\frac{2}{\sqrt{3}}\right) \text{ A}$ produces a deflection of $60^{\circ}$ in a tangent galvanometer. The reduction factor is:

Two magnets $A$ and $B$ are identical in mass,length,and breadth but have different magnetic moments. In a vibration magnetometer,if the time period of $B$ is twice the time period of $A$,then the ratio of the magnetic moments $M_A/M_B$ of the magnets will be:

$A$ short magnet oscillates with a time period $0.1 \, s$ at a place where the horizontal magnetic field is $24 \, \mu T$. $A$ downward current of $18 \, A$ is established in a vertical wire kept at a distance of $20 \, cm$ east of the magnet. The new time period of oscillations of the magnet is (in $s$)