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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 pla

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$2$

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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,$T_1 = 2\pi \sqrt{\frac{2I}{M\sqrt{2}H}} = 2\pi \sqrt{\frac{I\sqrt{2}}{MH}}$.Given $T_1 = 2^{5/4} \ s$,we have $2^{5/4} = 2\pi \sqrt{\frac{I\sqrt{2}}{MH}} \dots (i)$.When one magnet is removed,the time period $T_2$ for a single magnet is $T_2 = 2\pi \sqrt{\frac{I}{MH}} \dots (ii)$.Dividing equation $(i)$ by equation $(ii)$:$\frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{I\sqrt{2}}{MH}}}{2\pi \sqrt{\frac{I}{MH}}} = \sqrt{\sqrt{2}} = (2^{1/2})^{1/2} = 2^{1/4}$.Therefore,$T_2 = \frac{T_1}{2^{1/4}} = \frac{2^{5/4}}{2^{1/4}} = 2^{5/4 - 1/4} = 2^1 = 2 \ s$.

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