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

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$1/3$

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(B) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MH}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $H$ is the magnetic field.Thus,$T \propto \frac{1}{\sqrt{B_{net}}}$,where $B_{net}$ is the net magnetic field.Initially,$T_1 = 2 \ s$ and $B_{net1} = H$.When the magnet is brought near and parallel,the net field becomes $B_{net2} = H + F$,and the time period is $T_2 = 1 \ s$.Taking the ratio: $\frac{T_1}{T_2} = \sqrt{\frac{H+F}{H}}$.Substituting the values: $\frac{2}{1} = \sqrt{1 + \frac{F}{H}}$.Squaring both sides: $4 = 1 + \frac{F}{H} \Rightarrow \frac{F}{H} = 3$.Therefore,the ratio $\frac{H}{F} = \frac{1}{3}$.

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