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(C) The time period of a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$.When a bar magnet is cut parallel to its length into $4$ e

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$4$

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

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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