$A$ magnetic needle of negligible breadth and thickness compared to its length oscillates in a horizontal plane with a period $T$. What is the period of oscillation of each part obtained after breaking the magnet into $n$ equal parts perpendicular to its length?

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 $turns$)

$A$ magnet freely suspended in a vibration magnetometer makes $40$ oscillations per minute at place $A$ and $20$ oscillations per minute at a place $B$. If the horizontal component of earth's magnetic field at $A$ is $36 \times 10^{-6} \ T$, then its value at $B$ is

$A$ short magnet oscillates with a time period $0.1 \, s$ at a place where the horizontal magnetic field is $24 \, \mu T$. $A$ downward current of $18 \, A$ is established in a vertical wire $20 \, cm$ east of the magnet. Find the new time period of the oscillator. (in $, s$)

The frequency of oscillation of a small magnet in a magnetic field of induction $B$ is $n$. If the frequency of oscillations of the same magnet in a field of induction $X$ falls to $n/3$, the value of $X$ is

$A$ vibration magnetometer consists of two identical bar magnets placed one over the other such that they are mutually perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is $4 \, s$. If one of the magnets is taken away,find the period of oscillation of the other in the same field (in $s$):-