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) is (bar magnet breadth is small)
Answer: (C) $4$
Concept. $T = 2\pi\sqrt{\dfrac{I}{MB}}$. Cutting parallel to the length changes both $I$ and $M$.
Step 1. Moment of inertia of a piece. Each piece has $\dfrac{1}{4}$ of the mass and the same length; with small breadth $I \approx \dfrac{mL^2}{12}$, so $I' = \dfrac{I}{4}$.
Step 2. Magnetic moment of a piece. The pole strength is shared among the four pieces and the length is unchanged, so $M' = \dfrac{M}{4}$.
Step 3. New period. $\dfrac{I'}{M'} = \dfrac{I}{M}$, so the period is unchanged: $T' = 4$ s.
Answer. $4$ s (Option C).
Solution by Sreeraj P, M.Sc Physics