$A$ 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

Two mutually perpendicular uniform magnetic fields $F$ and $H$ act on a magnetic needle. The needle makes an angle of $60^\circ$ with the field $F$. Find the ratio $\frac{F}{H}$.

In a vibration magnetometer,the time period of a bar magnet oscillating in the horizontal component of the Earth's magnetic field is $2 \ s$. When a magnet is brought near and parallel to it,the time period reduces to $1 \ s$. The ratio $H/F$ of the horizontal component $H$ and the field $F$ due to the magnet will be:

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 with $50$ turns and a radius equal to $4$ $cm$. $A$ current of $0.1$ $A$ is passing through it. The plane of the coil is set parallel to the earth's magnetic meridian. If the value of the earth's horizontal component of the magnetic field is $7 \times 10^{-5}$ $T$ and $\mu_0 = 4\pi \times 10^{-7}$ $T \cdot m/A$,then the deflection in the galvanometer needle will be.....$^o$

$A$ magnetic needle suspended horizontally by an unspun silk fibre,oscillates in the horizontal plane because of the restoring force originating mainly from