$A$ short magnet of magnetic moment $6.75 \, Am^2$ produces a neutral point on its axis. If the horizontal component of the Earth's magnetic field is $5 \times 10^{-5} \, Wb/m^2$, then the distance of the neutral point from the center of the magnet is.....$cm$.

$A$ deflection magnetometer is adjusted and a magnet of magnetic moment $M$ is placed on it in the usual manner and the observed deflection is $\theta$. The period of oscillation of the needle before settling of the deflection is $T$. When the magnet is removed, the period of oscillation of the needle is $T_0$ before settling to $0^{\circ}-0^{\circ}$. If the earth's horizontal magnetic field is $B_H$, the relation between $T$ and $T_0$ is

$A$ bar magnet has a magnetic moment equal to $5 \times 10^{-5} \, \text{Wb} \cdot \text{m}$. It is suspended in a magnetic field which has a magnetic induction $B$ equal to $8\pi \times 10^{-4} \, \text{T}$. The magnet vibrates with a period of vibration equal to $15 \, \text{s}$. The moment of inertia of the magnet is:

$A$ 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?

$A$ vibration magnetometer works on the principle of

Two magnets $A$ and $B$ are identical in mass,length,and breadth but have different magnetic moments. In a vibration magnetometer,if the time period of $B$ is twice the time period of $A$,then the ratio of the magnetic moments $M_A/M_B$ of the magnets will be: