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

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(C) The time period $T$ of a magnet in a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MH}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $H$ is the horizontal component of the Earth's magnetic field.For the original magnet: $I_1 = \frac{ml^2}{12}$ and $M_1 = M$.When the magnet is cut into four equal pieces normal to its length,each piece has length $l' = l/4$ and mass $m' = m/4$.The new moment of inertia is $I_2 = \frac{m'(l')^2}{12} = \frac{(m/4)(l/4)^2}{12} = \frac{ml^2}{12 	imes 4 	imes 16} = \frac{I_1}{64}$.The new magnetic moment is $M_2 = M/4$.Using the ratio: $\frac{T_2}{T_1} = \sqrt{\frac{I_2}{I_1} \cdot \frac{M_1}{M_2}} = \sqrt{\frac{I_1/64}{I_1} \cdot \frac{M}{M/4}} = \sqrt{\frac{1}{64} \cdot 4} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.Given $T_1 = 4 \ s$,then $T_2 = T_1 	imes \frac{1}{4} = 4 	imes \frac{1}{4} = 1 \ s$.

With a standard rectangular bar magnet of length $l$,breadth $b$ $(b \ll l)$,and magnetic moment $M$,the time period of the magnet in a vibration magnetometer is $4 \ s$. If the magnet is cut normal to its length into four equal pieces,the time period (in seconds) with one of the pieces is:

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