A magnet freely suspended in a vibration magnetometer makes $30$ vibrations per minute at one place and $20$ vibrations per minute at another place. If the value of $B_{H}$ at first place is $0.27 T$ . The value of $B_{H}$ at the other place is :

0.12 T

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2.1 T

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5.4 T

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0.61 T

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Solution

The correct option is B $0.12 T$
The time period of oscillation is given by
$T = 2 π \sqrt{\frac{I}{M B}}$
So the frequency of oscillation is directly proportional to the square root of magnetic field
$\frac{f_{2}^{2}}{f_{1}^{2}} = \frac{B_{2}}{B_{1}}$
$B_{2} = \frac{400}{900} \times 0.27 T$
$B_{2} = 0.12 T$

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A magnet freely suspended in a vibration magnetometer makes 30 vibrations per minute at one place and 20 vibrations per minute at another place. If the value of BH at first place is 0.27T. The value of BH at the other place is :

The correct option is B 0.12TThe time period of oscillation is given byT=2π√IMBSo the frequency of oscillation is directly proportional to the square root of magnetic fieldf22f21=B2B1B2=400900×0.27TB2=0.12T

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