With a standard rectangular bar magnet of length $l,$ breadth $b (b << l)$ and magnetic moment $M,$ the time period of the magnet in a vibration magnetometer is $4 s$ . If the magnet is cut normal to its length into four equal pieces, the time period (in seconds) with one of the pieces is :

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Solution

The correct option is B $1$
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 broken into 4 pieces, magnetic moment becomes ${\frac{1}{4}}^{t h}$
$∴ M_{1} = 4 M_{2}$
Moment of inertial also changes.
$I_{1} = \frac{m l^{2}}{12}$
$I_{2} = \frac{\frac{m}{4} {(\frac{l}{4})}^{2}}{12}$
$∴ \frac{T_{1}}{T_{2}} = \sqrt{\frac{I_{1} M_{2}}{I_{2} \times 4 M_{2}}} = \sqrt{\frac{1 \times 64}{4}} = 4$
$∴ T_{2} = T_{1} \times \frac{1}{4} = \frac{4}{4} = 1 s$

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With a standard rectangular bar magnet of length l, breadth b(b<<l) and magnetic moment M, the time period of the magnet in a vibration magnetometer is 4 s. If the magnet is cut normal to its length into four equal pieces, the time period (in seconds) with one of the pieces is :

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