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

$A$ magnetic needle of magnetic moment $6.7 \times 10^{-2} \ Am^2$ and moment of inertia $7.5 \times 10^{-6} \ kgm^2$ is performing simple harmonic oscillations in a magnetic field of $0.01 \ T$. Time taken for $10$ complete oscillations is.....$s$

How many neutral points will be obtained when a bar magnet is kept with its magnetic moment parallel to the Earth's magnetic field?

Two tangent galvanometers $A$ and $B$ have coils of radii $8 \text{ cm}$ and $16 \text{ cm}$ respectively and have a resistance of $8 \Omega$ each. They are connected in parallel with a cell of emf $4 \text{ V}$ and negligible internal resistance. The deflections produced in the tangent galvanometers $A$ and $B$ are $30^{\circ}$ and $60^{\circ}$, respectively. If $A$ has $2$ turns, then $B$ must have: (in $turns$)

With a standard rectangular bar magnet, the time period of a vibration magnetometer is $4 \,s$. The bar magnet is cut parallel to its length into four equal pieces. The time period of the vibration magnetometer when one piece is used (in seconds) (bar magnet breadth is small) is

$A$ bar magnet of length $10 \, cm$ is kept with its north $(N)$-pole pointing north. $A$ neutral point is formed at a distance of $15 \, cm$ from each pole. Given the horizontal component of the earth's magnetic field is $0.4 \, Gauss$, the pole strength of the magnet is: (in $ \, A-m$)