$A$ thin rectangular magnet suspended freely has a period of oscillation equal to $T$. Now it is broken into two equal halves (each having half of the original length) and one piece is made to oscillate freely in the same field. If its period of oscillation is $T'$,then the ratio $\frac{T'}{T}$ is

$A$ magnet oscillating in a horizontal plane has a time period of $2 \ s$ at a place where the angle of dip is $30^o$ and $3 \ s$ at another place where the angle of dip is $60^o$. The ratio of resultant magnetic fields at the two places is

$A$ 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 $s$):-

$A$ tangent galvanometer has a coil with $50$ turns and a radius equal to $4$ $cm$. $A$ current of $0.1$ $A$ is passing through it. The plane of the coil is set parallel to the earth's magnetic meridian. If the value of the earth's horizontal component of the magnetic field is $7 \times 10^{-5}$ $T$ and $\mu_0 = 4\pi \times 10^{-7}$ $T \cdot m/A$,then the deflection in the galvanometer needle will be.....$^o$

Two small magnets have their masses and lengths in the ratio $1:2$. The maximum torques experienced by them in a uniform magnetic field are the same. For small oscillations, the ratio of their time periods is

$A$ 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).