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(B) The time period of a magnet in a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$.Here,$I = \frac{ml^2}{12}$ is the moment of in

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$2/3 \ s$

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(B) The time period of a magnet in a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$.Here,$I = \frac{ml^2}{12}$ is the moment of inertia and $M = m_p l$ is the magnetic moment,where $m_p$ is the pole strength.Substituting these,$T = 2 \pi \sqrt{\frac{ml^2}{12 m_p l B}} = 2 \pi \sqrt{\frac{ml}{12 m_p B}}$.Since the mass $m$ is also proportional to length $l$,we have $m \propto l$,so $T \propto \sqrt{\frac{l^2}{m_p}} \propto \frac{l}{\sqrt{m_p}}$.When the magnet is cut into three equal parts,the length of each part becomes $l' = l/3$,and the pole strength $m_p$ remains the same for each part.When these three parts are stacked with like poles together,the new length is $l' = l/3$ and the new pole strength is $M'_{p} = 3m_p$.The new moment of inertia $I'$ for the combination is $3 	imes \frac{(m/3)(l/3)^2}{12} = \frac{ml^2}{108}$.The new magnetic moment $M'$ is $3m_p 	imes (l/3) = m_p l = M$.The new time period $T' = 2 \pi \sqrt{\frac{I'}{M'B}} = 2 \pi \sqrt{\frac{ml^2/108}{M B}} = \frac{1}{\sqrt{9}} T = \frac{T}{3}$.Given $T = 2 \ s$,the new time period $T' = 2/3 \ s$.

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