If there is no torsion in the suspension thread, then the time period of a magnet executing $SHM$ is

$A$ magnet of magnetic moment $20$ $C.G.S.$ units is freely suspended in a uniform magnetic field of intensity $0.3$ $C.G.S.$ units. The amount of work done in deflecting it by an angle of $30^{\circ}$ in $C.G.S.$ units is:

$A$ short bar magnet of magnetic moment $0.4\,JT^{-1}$ is placed in a uniform magnetic field of $0.16\,T$. The magnet is in stable equilibrium when the potential energy is

$A$ magnet of magnetic moment $4 \, A-m^2$ is held in a uniform magnetic field $5 \times 10^{-4} \, T$ with the magnetic moment vector making an angle $30^{\circ}$ with the field. Calculate the work done in increasing the angle from $30^{\circ}$ to $45^{\circ}$.

The magnetic potential at a point on the axial line of a bar magnet of dipole moment $M$ is $V$. What is the magnetic potential due to a bar magnet of dipole moment $\frac{M}{4}$ at the same point?

The distance of two points on the axis of a magnet from its centre is $10 \, cm$ and $20 \, cm$ respectively. The ratio of magnetic intensity at these points is $12.5 : 1$. The length of the magnet will be......$cm$.