If the number of turns and the radius of the cross-section of the coil of a tangent galvanometer are both doubled,then the reduction factor $K$ will become:

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:

$A$ bar magnet of $10 \,cm$ length is kept with its north $(N)$-pole pointing north. $A$ neutral point is formed at a distance of $15 \,cm$ from each pole. Given the horizontal component of Earth's magnetic field is $0.4 \,Gauss$, the pole strength of the magnet is: (in $\,A-m$)

Two bar magnets having the same geometry with magnetic moments $M$ and $2M$ are first placed in such a way that their similar poles are on the same side,and their time period of oscillation is $T_{1}$. Now,the polarity of one of the magnets is reversed,and the time period of oscillation becomes $T_{2}$. Then:

$A$ 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.)

How many neutral points will be obtained when a bar magnet is kept with its magnetic moment parallel to the Earth's magnetic field?