The time period of oscillation of a bar magne | vedclass

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

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$\sqrt{2} T_0$

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(C) The time period of a vibration magnetometer is given by the formula $T = 2\pi \sqrt{\frac{I}{MB_H}}$,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 size and pole strength remain the same,$M$ and $B_H$ are constant.The moment of inertia $I$ of a bar magnet is proportional to its mass $m$ (i.e.,$I \propto m$).Given that the new mass $m' = 2m$,the new moment of inertia $I' = 2I$.Therefore,the new time period $T'$ is related to the original time period $T_0$ by the ratio $\frac{T'}{T_0} = \sqrt{\frac{I'}{I}} = \sqrt{\frac{2I}{I}} = \sqrt{2}$.Thus,$T' = \sqrt{2} T_0$.

The time period of oscillation of a bar magnet suspended horizontally along the magnetic meridian is $T_0$. If this magnet is replaced by another magnet of the same size and pole strength but with double the mass,the new time period will be

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