When two similar bar magnets are suspended one after the other in stirrup of a vibration magnetometer, make $10$ vibrations and $12$ vibrations in one minute respectively. If the magnetic moment of the first magnet is $2.5 A . m^{2}$ , the magnetic moment of the 2nd magnet is :

3.6 A m^{2}

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2.5 A m^{2}

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3.0 A m^{2}

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5.0 A m^{2}

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Solution

The correct option is A $3.6 A m^{2}$
The time period of oscillation of a bar magnet freely suspended in air is given by
$T = 2 π \sqrt{\frac{I}{M B}}$
$\frac{f_{2}}{f_{1}} = \sqrt{\frac{M_{2}}{M_{1}}}$
$M_{2} = \frac{f_{2}^{2}}{f_{1}^{2}} \times M_{1}$
$M_{2} = 1.44 \times 2.5$
$M_{2} = 3.6$ A m $^{2}$

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Two bar magnets A and B are placed one over the other and are allowed to Vibrate in a vibration magnetometer. They make 20 oscillations per minute when the similar poles of A and B are on the same side, while they make 15 oscillations per minute when their opposite poles lie on the same side. If $M_{A}$ and $M_{B}$ are the magnetic' moments of A and B and if $M_{A}$ > $M_{B}$ , the ratio of $M_{A}$ and $M_{B}$ is