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(A) The time period of an oscillating magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ i

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$4:9$

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(A) The time period of an oscillating magnet in a magnetic field 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 component of the Earth's magnetic field.Since the magnets have the same size and mass,their moment of inertia $I$ is the same.Thus,$T \propto \frac{1}{\sqrt{M}}$,which implies $M \propto \frac{1}{T^2}$.Given the frequency $f_1 = 10 	ext{ oscillations/min}$ and $f_2 = 15 	ext{ oscillations/min}$,the time periods are $T_1 = \frac{60}{10} = 6 	ext{ s}$ and $T_2 = \frac{60}{15} = 4 	ext{ s}$.The ratio of the magnetic moments is $\frac{M_1}{M_2} = \frac{T_2^2}{T_1^2} = \left(\frac{4}{6}ight)^2 = \left(\frac{2}{3}ight)^2 = \frac{4}{9}$.

Two magnets of same size and mass make respectively $10$ and $15$ oscillations per minute at a certain place. The ratio of their magnetic moments is

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