Vibration Magnetometer question from JEE Main 2003 (AIEEE)

Answer: (b)

1. $T = 2\pi\sqrt{\dfrac{I}{MB_H}}$ with $I = \dfrac{m l^2}{12}$ for a thin magnet.
2. Each half: mass $\dfrac{m}{2}$, length $\dfrac{l}{2}$, so $I' = \dfrac{(m/2)(l/2)^2}{12} = \dfrac{I}{8}$ and $M' = \dfrac{M}{2}$.
3. $\dfrac{T'}{T} = \sqrt{\dfrac{I'}{I}\cdot\dfrac{M}{M'}} = \sqrt{\dfrac{1}{8}\times 2} = \dfrac{1}{2}$
4. So the answer is (b) $\dfrac{1}{2}$.