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(B) The frequency of oscillation $n$ of a vibration magnetometer is given by $n = \frac{1}{2 \pi} \sqrt{\frac{M H}{I}}$,which implies $n \propto \sqrt

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$\sqrt{63}$

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(B) The frequency of oscillation $n$ of a vibration magnetometer is given by $n = \frac{1}{2 \pi} \sqrt{\frac{M H}{I}}$,which implies $n \propto \sqrt{B_{net}}$.In the first case,the net magnetic field is $H$. So,$n_1 = 12 \propto \sqrt{H}$.In the second case,the bar magnet is placed along the axis,so the net field is $H + H_1$. Thus,$n_2 = 15 \propto \sqrt{H + H_1}$.Taking the ratio: $\frac{15}{12} = \sqrt{\frac{H + H_1}{H}} \Rightarrow \frac{5}{4} = \sqrt{1 + \frac{H_1}{H}} \Rightarrow \frac{25}{16} = 1 + \frac{H_1}{H} \Rightarrow \frac{H_1}{H} = \frac{9}{16}$.When the poles are interchanged,the field becomes $H - H_1$. Let the new frequency be $n_3$.$\frac{n_3}{n_1} = \sqrt{\frac{H - H_1}{H}} = \sqrt{1 - \frac{H_1}{H}} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.Therefore,$n_3 = 12 	imes \frac{\sqrt{7}}{4} = 3 \sqrt{7} = \sqrt{9 	imes 7} = \sqrt{63}$ oscillations per minute.

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