At a certain place, a magnet makes $30$ oscillations per minute. At another place where the magnetic field is double, its time period will be:

Twists of suspension fibre should be removed in a vibration magnetometer so that:

Magnets $A$ and $B$ are geometrically similar,but the magnetic moment of $A$ is twice that of $B$. If $T_1$ and $T_2$ are the time periods of oscillation when their like poles and unlike poles are kept together respectively,then the ratio $\frac{T_1}{T_2}$ will be:

$A$ 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:

$A$ bar magnet of length $10 \, cm$ is kept with its north $(N)$-pole pointing north. $A$ neutral point is formed at a distance of $15 \, cm$ from each pole. Given the horizontal component of the earth's magnetic field is $0.4 \, Gauss$, the pole strength of the magnet is: (in $ \, A-m$)

The period of oscillations of a magnetic needle in a magnetic field is $1.0 \, s$. If the length of the needle is halved by cutting it,the time period will be.....$s$.