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(B) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic mo

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$2^{7/4}$

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(B) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal magnetic field.For the combination of two identical magnets (each with moment of inertia $I_0$ and magnetic moment $M_0$) placed perpendicularly and bisecting each other:Total moment of inertia $I_1 = I_0 + I_0 = 2I_0$.Resultant magnetic moment $M_1 = \sqrt{M_0^2 + M_0^2} = M_0\sqrt{2}$.Given $T_1 = 4 \, s$,so $4 = 2\pi \sqrt{\frac{2I_0}{M_0\sqrt{2} B_H}} = 2\pi \sqrt{\frac{I_0\sqrt{2}}{M_0 B_H}}$.When one magnet is removed:New moment of inertia $I_2 = I_0$.New magnetic moment $M_2 = M_0$.New time period $T_2 = 2\pi \sqrt{\frac{I_0}{M_0 B_H}}$.Taking the ratio:$\frac{T_1}{T_2} = \frac{2\pi \sqrt{\frac{I_0\sqrt{2}}{M_0 B_H}}}{2\pi \sqrt{\frac{I_0}{M_0 B_H}}} = \sqrt{\sqrt{2}} = 2^{1/4}$.Therefore,$T_2 = \frac{T_1}{2^{1/4}} = \frac{4}{2^{1/4}} = \frac{2^2}{2^{1/4}} = 2^{2 - 1/4} = 2^{7/4} \, s$.

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