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The frequncy of vibration in a vibration magnetometer of the combination of two bar magnets of magnetic moments $M_1$ and $M_2$ is $6 \mathrm{~Hz}$ when like poles are tied and it is $2 \mathrm{~Hz}$ when the unlike poles are tied together, then the ratio $M_1: M_2$ is
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The correct answer is:
$5: 4$
In the sum and difference method of vibration magnetometer,
$\frac{M_1}{M_2}=\frac{T_2^2+T_1^2}{T_2^2-T_1^2}$
Here,
$T_1=\frac{1}{n_1}=\frac{1}{6} \mathrm{~s}$
$T_2=\frac{1}{n_2}=\frac{1}{2} \mathrm{~s}$
$\frac{M_1}{M_2}=\frac{\frac{1}{4}+\frac{1}{36}}{\frac{1}{4}-\frac{1}{36}}+\frac{\frac{9+1}{36}}{\frac{9-1}{36}}$
$=\frac{10}{8}=\frac{5}{4}$
$M_1: M_2=5: 4$
$\frac{M_1}{M_2}=\frac{T_2^2+T_1^2}{T_2^2-T_1^2}$
Here,
$T_1=\frac{1}{n_1}=\frac{1}{6} \mathrm{~s}$
$T_2=\frac{1}{n_2}=\frac{1}{2} \mathrm{~s}$
$\frac{M_1}{M_2}=\frac{\frac{1}{4}+\frac{1}{36}}{\frac{1}{4}-\frac{1}{36}}+\frac{\frac{9+1}{36}}{\frac{9-1}{36}}$
$=\frac{10}{8}=\frac{5}{4}$
$M_1: M_2=5: 4$
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