Two magnets are held together in a variation magnetometer and are allowed to oscillate in the earth's magnetic field with like poles together. 12 oscillations per minute are made but for unlike poles together only 4 oscillations per minute are executed. The ratio of their magnetic moments is

3 : 1

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1 : 3

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3 : 5

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5 : 4

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Solution

The correct option is D 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, $T_{1} = \frac{1}{n_{1}} = \frac{60}{12} = 5$ sec
$T_{2} = \frac{1}{n_{2}} \Rightarrow \frac{60}{4} = 15$ sec.
So, $\frac{M_{1}}{M_{2}} = \frac{15^{2} + 5^{2}}{15^{2} - 5^{2}} = \frac{5}{4}$ .

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Two magnets are held together in a variation magnetometer and are allowed to oscillate in the earth's magnetic field with like poles together. 12 oscillations per minute are made but for unlike poles together only 4 oscillations per minute are executed. The ratio of their magnetic moments is

The correct option is D 5 : 4In the sum and difference method of vibration magnetometer:-M1M2=T22+T21T22−T21Here, T1=1n1=6012=5 secT2=1n2⇒604=15 sec.So, M1M2=152+52152−52=54.