In a uniform magnetic field,a magnetic needle has a magnetic moment $9.85 \times 10^{-2} \, A \cdot m^{2}$ and a moment of inertia $5 \times 10^{-6} \, kg \cdot m^{2}$. If it performs $10$ complete oscillations in $5 \, s$,then the magnitude of the magnetic field is $.... \, mT$. [ Take $\pi^{2} = 9.85$ ]

$A$ 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 $^{\circ}$)

Two bar magnets $A$ and $B$ are geometrically similar but the magnetic moment of $A$ is twice that of $B$. $T_1$ is the time period of oscillation when their like poles are kept together. When unlike poles are kept together,the time period of oscillation is $T_2$. The ratio $T_1: T_2$ will be

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:

$A$ vibration magnetometer consists of two identical bar magnets placed one over the other such that they are mutually perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is $4 \, s$. If one of the magnets is taken away,find the period of oscillation of the other in the same field (in $s$):-

The time period of a vibration magnetometer is $T_0$. Its magnet is replaced by another magnet whose moment of inertia is $3$ times and magnetic moment is $1/3$ of the initial magnet. The time period now will be