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(C) The time period 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 th

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(C) The time period 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: $I_1 = \frac{m l^2}{12}$ and $M_1 = M$.When the magnet is cut into four equal pieces normal to its length, each piece has mass $m' = \frac{m}{4}$ and length $l' = \frac{l}{4}$.The new moment of inertia is $I_2 = \frac{m' (l')^2}{12} = \frac{(m/4) (l/4)^2}{12} = \frac{m l^2}{12 	imes 4 	imes 16} = \frac{I_1}{64}$.The new magnetic moment is $M_2 = \frac{M}{4}$.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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