The vertical component of the earth's magnetic field is $$6 \times 10^{-5} \mathrm{~T}$$ at any place where the angle of dip is $$37^{\circ}$$. The earth's resultant magnetic field at that place will be $$\left(\right.$$Given $$\left.\tan 37^{\circ}=\frac{3}{4}\right)$$

A compass needle of oscillation magnetometer oscillates 20 times per minute at a place $$\mathrm{P}$$ of $$\operatorname{dip} 30^{\circ}$$. The number of oscillations per minute become 10 at another place $$\mathrm{Q}$$ of $$60^{\circ}$$ dip. The ratio of the total magnetic field at the two places $$\left(B_{Q}: B_{P}\right)$$ is :

Two bar magnets oscillate in a horizontal plane in earth's magnetic field with time periods of $$3 \mathrm{~s}$$ and $$4 \mathrm{~s}$$ respectively. If their moments of inertia are in the ratio of $$3: 2$$, then the ratio of their magnetic moments will be:

A magnet hung at $$45^{\circ}$$ with magnetic meridian makes an angle of $$60^{\circ}$$ with the horizontal. The actual value of the angle of dip is -