Two bar magnets oscillate in a horizontal plane in Earth's magnetic field with time periods of $3\,s$ and $4\,s$ respectively. If their moments of inertia are in the ratio of $3: 2$,then the ratio of their magnetic moments will be.

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \, s$. The magnet is cut along its length into three equal parts and these three parts are then placed on each other with their like poles together. The time period of this combination will be

$A$ short magnetic needle is pivoted in a uniform magnetic field of induction $1 \text{ T}$. Now, simultaneously another magnetic field of induction $\sqrt{3} \text{ T}$ is applied at right angles to the first field; the needle deflects through an angle $\theta$ whose value is (in $^{\circ}$)

$A$ short bar magnet with its north pole facing north forms a neutral point at $P$ in the horizontal plane. If the magnet is rotated by $90^o$ in the horizontal plane,the net magnetic induction at $P$ is (Horizontal component of earth's magnetic field = $B_H$)

$A$ magnet makes $20$ and $15$ oscillations in $1$ minute at two places where the angles of dip are $30^{\circ}$ and $60^{\circ}$ respectively. What is the ratio of the total magnetic fields at these two places?

$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ (about its center,perpendicular to its length) is cut into two equal pieces,perpendicular to its length. Let $T$ be the period of oscillation of the original magnet about an axis through its midpoint,perpendicular to its length,in a uniform magnetic field $B$. What would be the period $T'$ for each piece?