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Two magnets are held in a stirrup and allowed to oscillate in the earth’s field. When their poles are pointing in the same direction, they make 12 oscillations /min; when their unlike poles are pointing in the same direction, they make only 4 osiliations/min. The ratio of their magnetic moments is

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5 : 4

2 : 3

5 : 2

3 : 2

answer is A.

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Detailed Solution

The expression for frequency of vibration,

$n^{2} = \frac{1}{4 π^{2}} \frac{M B}{I}$

In the first case, $M = M_{1} + M_{2}$ and $I = I_{1} + I_{2}$

In the second case, $M = M_{1} - M_{2}$ and $I = I_{1} + I_{2}$

$∴ n_{1}^{2} = \frac{1}{4 π^{2}} \frac{(M_{1} + M_{2}) B}{(I_{1} + I_{2})} \to (1)$

$n_{2}^{2} = \frac{1}{4 π^{2}} \frac{(M_{1} - M_{2}) B}{(I_{1} + I_{2})} \to (2)$

$\frac{e q (1)}{e q (2)} \Rightarrow \frac{n_{1}^{2}}{n_{2}^{2}} = \frac{M_{1} + M_{2}}{M_{1} - M_{2}}$

By componendo and dividend,

$\frac{n_{1}^{2} + n_{2}^{2}}{n_{1}^{2} - n_{2}^{2}} = \frac{M_{1}}{M_{2}} \Rightarrow \frac{M_{1}}{M_{2}}$

$= \frac{{(12)}^{2} + {(4)}^{2}}{{(12)}^{2} - {(4)}^{2}} = \frac{160}{128} = \frac{5}{4}$

$∴ M_{1} : M_{2} = 5 : 4$

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