$A$ bar magnet is oscillating in the earth's magnetic field with time period $T.$ If its mass is increased four times,then its time period will be

$A$ short magnet oscillates with a frequency $10 \, Hz$ where the horizontal component of Earth's magnetic field is $12 \, \mu T$. $A$ downward current of $15 \, A$ is established in a vertical wire placed $20 \, cm$ west of the magnet. The new frequency of the magnet is (in $ \, Hz$)

$A$ bar magnet of moment of inertia $49 \times 10^{-2} \,kg-m^2$ vibrates in a magnetic field of induction $0.5 \times 10^{-4} \,T$. The time period of vibration is $8.8 \,s$. The magnetic moment of the bar magnet is (in $\,A-m^2$)

$A$ bar magnet of $10 \,cm$ length 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 Earth's magnetic field is $0.4 \,Gauss$, the pole strength of the magnet is: (in $\,A-m$)

The time period of a magnet is $T$. If it is divided into four equal parts along its axis and perpendicular to its axis as shown,then the time period for each part will be

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: