$A$ 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:

The length of a magnet is large compared to its width and breadth. The time period of its oscillation in a vibration magnetometer is $2 \, s$. The magnet is cut along its length into three equal parts and these three parts are then placed on each other with their like poles together. The time period of this combination 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:

$A$ vibration magnetometer placed in the magnetic meridian has a small bar magnet. The magnet executes oscillations with a time period of $2 \, s$ in the earth's horizontal magnetic field of $24 \, \mu T$. When a horizontal field of $18 \, \mu T$ is produced opposite to the earth's field by placing a current-carrying wire,the new time period of the magnet will be....$s$

$A$ tangent galvanometer shows a deflection of $30^{\circ}$ under the influence of the horizontal component of the Earth's magnetic field,which is $0.34 \times 10^{-4} \, T$. What is the magnetic field of the coil?

If a brass bar is placed on a vibrating magnet, then its time period