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.

In a uniform magnetic field of $0.049 \ T$, a magnetic needle performs $20$ complete oscillations in $5 \ s$. The moment of inertia of the needle is $9.8 \times 10^{-5} \ kg \ m^2$. If the magnitude of the magnetic moment of the needle is $x \times 10^{-5} \ A \ m^2$, then the value of '$x$' is: (in $\pi^2$)

$A$ 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

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

The length $l$ 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 into three equal parts of length $l/3$ each. If these parts are placed on each other with their like poles together,then the time period of this combination is

$A$ magnetic needle has a magnetic moment of $5 \times 10^{-2} \text{ A m}^2$ and a moment of inertia of $8 \times 10^{-6} \text{ kg m}^2$. It has a period of oscillation of $2 \text{ s}$ in a magnetic field $\vec{B}$. The magnitude of the magnetic field is approximately: