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(C) The time period $T$ of a magnet in a vibration magnetometer is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of i

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$4$

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(C) The time period $T$ of a magnet in a vibration magnetometer is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the Earth's magnetic field.Since the magnets are identical in mass,length,and breadth,their moment of inertia $I$ is the same.Thus,$T \propto \frac{1}{\sqrt{M}}$,which implies $T^2 \propto \frac{1}{M}$ or $M \propto \frac{1}{T^2}$.Given that the time period of $B$ is twice the time period of $A$,we have $T_B = 2T_A$.Therefore,the ratio of the magnetic moments is $\frac{M_A}{M_B} = \left(\frac{T_B}{T_A}\right)^2 = \left(\frac{2T_A}{T_A}\right)^2 = 2^2 = 4$.Hence,the ratio $M_A/M_B$ is $4$.

Two magnets $A$ and $B$ are identical in mass,length,and breadth but have different magnetic moments. In a vibration magnetometer,if the time period of $B$ is twice the time period of $A$,then the ratio of the magnetic moments $M_A/M_B$ of the magnets will be:

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