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(C) The time period of a magnet in a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{MH}}$.In the first case, two identical magnets are p

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(C) The time period of a magnet in a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{MH}}$.In the first case, two identical magnets are placed perpendicular to each other, so the total moment of inertia is $I_{total} = I + I = 2I$ and the resultant magnetic moment is $M' = \sqrt{M^2 + M^2} = M\sqrt{2}$.Thus, $T_1 = 2 \pi \sqrt{\frac{2I}{M\sqrt{2}H}} = 2 \pi \sqrt{\frac{\sqrt{2}I}{MH}}$.Given $T_1 = 2^{5/4} \ s$, we have $2^{5/4} = 2 \pi \sqrt{\frac{\sqrt{2}I}{MH}}$ ... $(i)$.When one magnet is removed, the time period $T_2$ is $T_2 = 2 \pi \sqrt{\frac{I}{MH}}$ ... (ii).Dividing equation $(i)$ by (ii):$\frac{2^{5/4}}{T_2} = \frac{2 \pi \sqrt{\frac{\sqrt{2}I}{MH}}}{2 \pi \sqrt{\frac{I}{MH}}} = \sqrt{\sqrt{2}} = 2^{1/4}$.Therefore, $T_2 = \frac{2^{5/4}}{2^{1/4}} = 2^{5/4 - 1/4} = 2^1 = 2 \ s$.

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