$A$ current of $\left(\frac{2}{\sqrt{3}}\right) \text{ A}$ produces a deflection of $60^{\circ}$ in a tangent galvanometer. The reduction factor is:

$A$ bar magnet has a magnetic moment equal to $5 \times 10^{-5} \, \text{Wb} \cdot \text{m}$. It is suspended in a magnetic field which has a magnetic induction $B$ equal to $8\pi \times 10^{-4} \, \text{T}$. The magnet vibrates with a period of vibration equal to $15 \, \text{s}$. The moment of inertia of the magnet is:

$A$ vibration magnetometer is placed at the south pole; then its time period will be

$A$ magnetic needle oscillates in a horizontal plane with a period $T$ at a place where the angle of dip is $60^{\circ}$. When the same needle is made to oscillate in a vertical plane coinciding with the magnetic meridian,its period will be

$A$ compass needle oscillates $20$ times per minute at a place where the dip is $45^{\circ}$ and the magnetic field is $B_1$. The same needle oscillates $30$ times per minute at a place where the dip is $30^{\circ}$ and magnetic field is $B_2$. Then $B_1: B_2$ is

$A$ 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.....$^o$