The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{\circ}$ from the magnetic meridian is $n$ times the corresponding work done to turn it through an angle of $60^{\circ}$. The value of $n$ is:

The magnetic moment of a bar magnet is $0.5 \text{ A m}^2$. It is suspended in a uniform magnetic field of $8 \times 10^{-2} \text{ T}$. The work done in rotating it from its most stable to most unstable position is:

The magnetic moment of a short magnet is $16 \ Am^2$. The magnetic induction at a point $20 \ cm$ away from its midpoint on $(i)$ axial point and $(ii)$ equatorial point respectively,will be:

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ is freely suspended such that the magnetic axial line is in the direction of the magnetic meridian. If the magnet is displaced by a very small angle $\theta$, the angular acceleration is (Magnetic induction of earth's horizontal field $= B_H$)

$A$ short bar magnet of magnetic moment $2.5 \text{ Am}^2$ is kept in a uniform magnetic field of $4 \times 10^{-5} \text{ T}$. The work done in moving the magnet from its most stable position to its most unstable position is:

Points $A$ and $B$ are situated along the extended axis of a $2 \ cm$ long bar magnet at distances $x$ and $2x$ $cm$ respectively from the nearer pole. The ratio of the magnetic field at $A$ and $B$ will be: