$A$ magnet is suspended horizontally in the Earth's magnetic field. When it is made to oscillate in a horizontal plane,its time period is $T$. If a wooden block of the same moment of inertia as the magnet is attached to it,what will be the new time period of the system?

Two magnets of same size and mass make respectively $10$ and $15$ oscillations per minute at a certain place. The ratio of their magnetic moments is

$A$ deflection magnetometer is adjusted and a magnet of magnetic moment $M$ is placed on it in the usual manner and the observed deflection is $\theta$. The period of oscillation of the needle before settling of the deflection is $T$. When the magnet is removed,the period of oscillation of the needle is $T_0$ before settling to $0^{\circ}-0^{\circ}$. If the earth's horizontal magnetic field is $B_H$,the relation between $T$ and $T_0$ is

The time period of a magnet is $2 \, s$. The horizontal component of the Earth's magnetic field is $H$. When an external magnetic field $F$ is applied in the same direction,the new time period becomes $1 \, s$. What is the ratio $H/F$?

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 shows a deflection of $45^{\circ}$ when a $10 \, mA$ current passes through it. If the horizontal component of the Earth's magnetic field is $3.6 \times 10^{-5} \, T$ and the radius of the coil is $10 \, cm$,find the number of turns in the coil.