When a bar magnet is freely suspended for oscillation,its time period is $T$. If it is cut into $3$ equal parts perpendicular to its length,then the time period of oscillation for each part will be

In a uniform magnetic field of $0.049 \ T$, a magnetic needle performs $20$ complete oscillations in $5 \ s$. The moment of inertia of the needle is $9.8 \times 10^{-5} \ kg \ m^2$. If the magnitude of the magnetic moment of the needle is $x \times 10^{-5} \ A \ m^2$, then the value of '$x$' is: (in $\pi^2$)

$A$ vibration magnetometer consists of two identical bar magnets placed one over the other such that they are perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is $2^{5/4} \ s$. One of the magnets is removed and if the other magnet oscillates in the same field, then the time period in seconds is:

The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by $19\%$. By doing this,the periodic time of the magnetometer will

The moment of inertia of a magnetic needle is $40 \, g \cdot cm^2$ and it has a time period of $3 \, s$ in the Earth's horizontal magnetic field of $3.6 \times 10^{-5} \, Wb/m^2$. Its magnetic moment will be (in $A \cdot m^2$):

Two identical bar magnets are held perpendicular to each other with a certain separation,as shown below. The area around the magnets is divided into four zones. Given that there is a neutral point,it is located in: