In a vibration magnetometer,the time period of a bar magnet oscillating in the horizontal component of the Earth's magnetic field is $2 \ s$. When a magnet is brought near and parallel to it,the time period reduces to $1 \ s$. The ratio $H/F$ of the horizontal component $H$ and the field $F$ due to the magnet will be:

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $T$. The magnet is cut along its length into six parts and these parts are then placed together as shown in the figure. The time period of this combination will be

$A$ bar magnet is in simple harmonic motion with time period $T$ in an Earth's magnetic field. If its mass is increased by $9$ times,the time period becomes:

$A$ magnet of magnetic moment $M$ oscillating freely in earth's horizontal magnetic field makes $n$ oscillations per minute. If the magnetic moment is quadrupled and the earth's field is doubled,the number of oscillations made per minute would be

$A$ magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of the wire is rotated through $180^o$ to deflect the magnet by $30^o$ from the magnetic meridian. When this magnet is replaced by another magnet,the upper end of the wire is rotated through $270^o$ to deflect the magnet $30^o$ from the magnetic meridian. The ratio of the magnetic moments of the magnets 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 $s$):-