$A$ 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:

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:

$A$ tangent galvanometer has a coil of $25$ turns and a radius of $15\, cm$. The horizontal component of the earth's magnetic field is $3 \times 10^{-5}\, T$. The current required to produce a deflection of $45^{\circ}$ in it is....$A$

With a standard rectangular bar magnet, the time period of a vibration magnetometer is $4 \,s$. The bar magnet is cut parallel to its length into four equal pieces. The time period of the vibration magnetometer when one piece is used (in seconds) (bar magnet breadth is small) is

$A$ 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:

$A$ 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).