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 -

An electron with energy 0.1 keV moves at right angle to the earth's magnetic field of 1 $$\times$$ 10^$$-$$4 Wbm^$$-$$2. The frequency of revolution of the electron will be :

(Take mass of electron = 9.0 $$\times$$ 10^$$-$$31 kg)