With a standard rectangular bar magnet 'the time period of a vibration magnetometer is $4 s$ . The bar magnet is cut parallel to its length into four equal pieces. The time period of vibration magnetometer when one piece is used (in second) (bar magnet breadth is, small) is

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Solution

The correct option is C $4$
Time period of magnet, $T = 2 π \sqrt{\frac{I}{M B}}$
When magnet is cut parallel to its length into four equal pieces.
Then new
magnetic moment, $M^{'} = \frac{M}{4}$
New moment of inertia, $I^{'} = \frac{I}{4}$
$∴$ New time period, $T^{'} = 2 π \sqrt{\frac{I^{'}}{M^{'} B^{'}}}$
$\Rightarrow T = T^{'} = 4 s$

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