$A$ 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 having magnetic moment $4 \text{ Am}^2$, placed in a vibrating magnetometer, vibrates with a time period of $8 \text{ s}$. Another short bar magnet having a magnetic moment $8 \text{ Am}^2$ vibrates with a time period of $6 \text{ s}$. If the moment of inertia of the second magnet is $9 \times 10^{-2} \text{ kg m}^2$, the moment of inertia of the first magnet is (assume that both magnets are kept in the same uniform magnetic induction field.)

$A$ bar magnet of length $14 \, cm$ is placed in the magnetic meridian with its north pole pointing towards the geographic north pole. $A$ neutral point is obtained at a distance of $18 \, cm$ from the center of the magnet. If $B_{H} = 0.4 \, G$,the magnetic moment of the magnet is $\left(1 \, G = 10^{-4} \, T\right)$.

At two places $A$ and $B$,using a vibration magnetometer,a magnet vibrates in a horizontal plane. Its respective periodic times are $2 \ s$ and $3 \ s$. At these places,the Earth's horizontal components are $H_A$ and $H_B$ respectively. Then the ratio between $H_A$ and $H_B$ will be:

$A$ bar magnet is oscillating in the Earth's magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled?

$A$ bar magnet of moment of inertia $49 \times 10^{-2} \,kg-m^2$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \,T$. The time period of vibration is $8.8 \,s$. The magnetic moment of the bar magnet is (in $\,A-m^2$)