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

4 : 3

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25 : 7

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7 : 5

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25 : 16

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Solution

The correct option is B 25 : 7
Ratio of magnetic moments of two magnets of equal size when in sum and difference position is
$\frac{M_{A}}{M_{B}} = \frac{T_{d}^{2} + T_{s}^{2}}{T_{d}^{2} - T_{s}^{2}} = \frac{v_{s}^{2} + v_{d}^{2}}{v_{s}^{2} - v_{d}^{2}}$
$= \frac{{\frac{1}{20}}^{2} + {\frac{1}{15}}^{2}}{{\frac{1}{15}}^{2} - {\frac{1}{20}}^{2}}$
$= \frac{400 + 225}{400 - 225}$
$\Rightarrow M_{A} : M_{B} = 25 : 7$

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The correct option is B 25 : 7Ratio of magnetic moments of two magnets of equal size when in sum and difference position is MAMB=T2d+T2sT2d−T2s=v2s+v2dv2s−v2d=1202+11521152−1202 =400+225400−225 ⇒MA:MB=25:7