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

$A$ magnetic dipole is placed horizontally with the north pole pointing towards north. The horizontal component of Earth's magnetic field is $20 \mu T$. If the neutral point is found at a distance of $20 \ cm$ in the plane bisecting the dipole, then the magnetic moment of the dipole is (Assume $\mu_0 = 4 \pi \times 10^{-7} \text{ S.I. units}$) (in $\text{ A m}^2$)

The moment of inertia of a magnetic needle is $40 \, g \cdot cm^2$ and it has a time period of $3 \, s$ in the Earth's horizontal magnetic field of $3.6 \times 10^{-5} \, Wb/m^2$. Its magnetic moment will be (in $A \cdot m^2$):

$A$ short bar magnet having a magnetic moment $4 \text{ Am}^2$,placed in a vibrating magnetometer,vibrates with a time period of $8 \text{ s}$. Another short bar magnet having a magnetic moment $8 \text{ Am}^2$ vibrates with a time period of $6 \text{ s}$. If the moment of inertia of the second magnet is $9 \times 10^{-2} \text{ kg-m}^2$,the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field).

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

When two magnetic moments are compared using the equal distance method,the deflections produced are $45^\circ$ and $30^\circ$. If the lengths of the magnets are in the ratio $1:2$,the ratio of their pole strengths is: