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.

$A$ bar magnet is placed on a table along the east-west direction in the Earth's magnetic field. The number of neutral points,where the magnetic field is zero,is

$A$ magnet of magnetic moment $M$ oscillating freely in earth's horizontal magnetic field makes $n$ oscillations per minute. If the magnetic moment is quadrupled and the earth's field is doubled,the number of oscillations made per minute would be

$A$ magnet freely suspended in a vibration magnetometer makes $40$ oscillations per minute at place $A$ and $20$ oscillations per minute at a place $B$. If the horizontal component of earth's magnetic field at $A$ is $36 \times 10^{-6} \ T$, then its value at $B$ is

$A$ short bar magnet is placed in the magnetic meridian of the earth with its north pole pointing towards the geographic north. Neutral points are found at a distance of $30 \, cm$ from the magnet on the East-West line, drawn through the center of the magnet. The magnetic moment of the magnet in $A \cdot m^2$ is close to: (Given $\frac{\mu_0}{4\pi} = 10^{-7} \, T \cdot m/A$ and $B_H = 3.6 \times 10^{-5} \, T$)

Two magnets of same size and mass make respectively $10$ and $15$ oscillations per minute at a certain place. The ratio of their magnetic moments is