$A$ 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 $^{\circ}$)

$A$ magnet makes $40$ oscillations per minute at a place having magnetic field intensity of $0.1 \times 10^{-5} \,T$. At another place,it takes $2.5 \,s$ to complete one vibration. The value of earth's horizontal field at that place is:

Two different magnets are tied together and allowed to vibrate in a horizontal plane. When their like poles are joined,the time period of oscillation is $5 \, s$,and when their unlike poles are joined,the time period of oscillation is $15 \, s$. The ratio of their magnetic moments is

$A$ dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be $2$ seconds. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be $2$ seconds. Then the angle of dip is.....$^o$

$A$ short bar magnet having a 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).

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: