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(C) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$.Initially,two identical magnets of magnetic moment $M$ an

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(C) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$.Initially,two identical magnets of magnetic moment $M$ and moment of inertia $I$ are placed perpendicular to each other.The resultant magnetic moment is $M_1 = \sqrt{M^2 + M^2} = M\sqrt{2}$.The total moment of inertia is $I_1 = I + I = 2I$.The initial time period is $T_1 = 2\pi \sqrt{\frac{2I}{M\sqrt{2} B_H}} = 2\pi \sqrt{\frac{\sqrt{2}I}{MB_H}} = 2^{1/4} \cdot 2\pi \sqrt{\frac{I}{MB_H}}$.Given $T_1 = 2^{5/4} \ s$,we have $2^{5/4} = 2^{1/4} \cdot T_0$,where $T_0 = 2\pi \sqrt{\frac{I}{MB_H}}$ is the time period of a single magnet.Thus,$T_0 = \frac{2^{5/4}}{2^{1/4}} = 2^1 = 2 \ s$.When one magnet is removed,the time period of the remaining magnet is $T_2 = T_0 = 2 \ s$.

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