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A short bar magnet used in a vibration magnetometer is heated so as to reduce its magnetic moment by 36%, then the frequency of magnet is (Neglecting the Changes in the dimensions of the magnet)

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Increase by 20%

Decreases by 36%

Increase by 36%

Decreases by 20%

answer is D.

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Detailed Solution

$Time period, T = 2 π \sqrt{\frac{I}{{MB}_{H}}}$

$Frequency, f = \frac{1}{2 π} \sqrt{\frac{{MB}_{H}}{I}}$

$⇒ \frac{f_{1}}{f_{2}} = \sqrt{\frac{M_{1}}{M_{2}}}$

$If M_{1} = 100 then, M_{2} = (100 - 36) = 64$

$So, \frac{f_{1}}{f_{2}} = \sqrt{\frac{100}{64}} = \frac{10}{8}$

$f_{2} = \frac{8}{10} f_{1} = 0.8 f_{1}$

$So, % decrease in frequency = 20 %$

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