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(B) The time period of a magnet in a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is t

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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 $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.Initially,$T = 2 \, s = 2\pi \sqrt{\frac{I}{MB}}$.When the magnet is cut into three equal parts along its length,for each part,the mass $m' = m/3$ and length $l' = l/3$.The moment of inertia of each part about the center is $I' = \frac{1}{12} m' (l')^2 = \frac{1}{12} (m/3) (l/3)^2 = \frac{1}{27} I$.The magnetic moment of each part is $M' = M/3$.When three such parts are placed on each other,the total moment of inertia $I_s = 3 	imes I' = 3 	imes (I/27) = I/9$.The total magnetic moment $M_s = 3 	imes M' = 3 	imes (M/3) = M$.The new time period $T_s = 2\pi \sqrt{\frac{I_s}{M_s B}} = 2\pi \sqrt{\frac{I/9}{MB}} = \frac{1}{3} 	imes 2\pi \sqrt{\frac{I}{MB}} = \frac{T}{3}$.Substituting $T = 2 \, s$,we get $T_s = 2/3 \, s$.

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