With a standard rectangular bar magnet, the time period in a vibration magneto meter is $4 s e c .$ The bar magnet is cut parallel to its length into $4$ equal pieces. The time period in vibration magnetometer when the piece is used $($ in sec $) ($ bar magnet breadth is small $)$

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Solution

The correct option is C $4$
Time period of vibration: $T \propto \sqrt{\frac{I}{M}}$ where $I$ is the moment of inertia and $M$ is the magnetic moment
$∴$ $\frac{T_{1}}{T_{2}} = \sqrt{\frac{I_{1} M_{2}}{I_{2} M_{1}}}$
When the magnet is cut into 4 pieces parallel to its length, magnetic moment remains same since the breadth is very small..
$∴ M_{1} = M_{2}$
Moment of inertial also does not change.
$I_{2} = I_{1}$
$∴ \frac{T_{1}}{T_{2}} = \sqrt{\frac{I_{1} M_{2}}{I_{2} \times M_{1}}} = 1$
$∴ T_{2} = T_{1} = 4$

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