The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by $36$ %. The new periodic time of the magnetometer will be

Increases by

25

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Increases by

36

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Decreases by

25

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Decreases by

36

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Solution

The correct option is B Increases by $25$ %
Time period, $T = 2 π \sqrt{\frac{I}{M B_{H}}}$
$T \propto \frac{1}{\sqrt{M}}$
or $\frac{T_{1}}{T_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$
If $M_{1} = 100$ then, $M_{2} = (100 - 36) = 64$
So, $\frac{T_{1}}{T_{2}} = \sqrt{\frac{64}{100}} = \frac{8}{10}$
$T_{2} = \frac{10}{8} T_{1} = 1.25 T_{1}$
So, % increase in time period $= 25$ %.

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The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by 36%. The new periodic time of the magnetometer will be

The correct option is B Increases by 25%Time period, T=2π√IMBHT∝1√Mor T1T2=√M2M1If M1=100 then, M2=(100−36)=64So, T1T2=√64100=810T2=108T1=1.25T1So, % increase in time period =25%.

The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by $36$ %. The new periodic time of the magnetometer will be