$A$ bar magnet of magnetic moment $M$ and moment of inertia $I$ (about its center,perpendicular to its length) is cut into two equal pieces,perpendicular to its length. Let $T$ be the period of oscillation of the original magnet about an axis through its midpoint,perpendicular to its length,in a uniform magnetic field $B$. What would be the period $T'$ for each piece?

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

Two small magnets have their masses and lengths in the ratio $1:2$. The maximum torques experienced by them in a uniform magnetic field are the same. For small oscillations,the ratio of their time periods is

$A$ dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be $2$ seconds. The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be $2$ seconds. Then the angle of dip is.....$^o$

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