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

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$4$

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(C) The time period of 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.When the bar magnet is cut parallel to its length into four equal pieces, the mass of each piece becomes $m' = \frac{m}{4}$.The new moment of inertia of one piece about the axis of rotation is $I' = \frac{I}{4}$.The new magnetic moment of one piece is $M' = \frac{M}{4}$.Substituting these into the formula for the new time period $T'$:$T' = 2 \pi \sqrt{\frac{I'}{M' B}} = 2 \pi \sqrt{\frac{I/4}{(M/4) B}} = 2 \pi \sqrt{\frac{I}{MB}} = T$.Therefore, the new time period $T' = 4 \,s$.

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

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