Two bar magnets are placed together in a vibration magnetometer vibrates with a time period is $3 s$ . If one magnet is reversed, the combination takes $4 s$ for one vibration. The ratio of their magnetic moments is :

3 : 1

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5 : 18

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18 : 5

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25 : 7

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Solution

The correct option is D $25 : 7$
The time period of oscillation of a bar magnet freely suspended in air is given by
$T = 2 π \sqrt{\frac{I}{m B}}$
$T_{1} = 2 π \sqrt{\frac{I}{(m_{1} + m_{2}) B}}$
$T_{2} = 2 π \sqrt{\frac{I}{(m_{1} - m_{2}) B}}$
$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}}$
$\frac{9}{16} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}}$
$m_{1} : m_{2} = 25 : 7$

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Two bar magnets are placed together in a vibration magnetometer vibrates with a time period is 3s . If one magnet is reversed, the combination takes 4s for one vibration. The ratio of their magnetic moments is :

The correct option is D 25:7The time period of oscillation of a bar magnet freely suspended in air is given byT=2π√ImBT1=2π√I(m1+m2)BT2=2π√I(m1−m2)BT21T22=m1−m2m1+m2916=m1−m2m1+m2m1:m2=25:7

Two bar magnets are placed together in a vibration magnetometer vibrates with a time period is $3 s$ . If one magnet is reversed, the combination takes $4 s$ for one vibration. The ratio of their magnetic moments is :