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(C) The time period $T$ of a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{M B_{H}}}$,where $I$ is the moment of inertia,$M$ is the mag

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increase by $11\%$

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(C) The time period $T$ of a vibration magnetometer is given by $T = 2 \pi \sqrt{\frac{I}{M B_{H}}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_{H}$ is the horizontal component of the Earth's magnetic field.From the formula,we see that $T \propto \frac{1}{\sqrt{M}}$.Therefore,the ratio of the time periods is $\frac{T_{1}}{T_{2}} = \sqrt{\frac{M_{2}}{M_{1}}}$.Given that the magnetic moment is reduced by $19\%$,if $M_{1} = 100$,then $M_{2} = 100 - 19 = 81$.Substituting these values,we get $\frac{T_{1}}{T_{2}} = \sqrt{\frac{81}{100}} = \frac{9}{10}$.This implies $T_{2} = \frac{10}{9} T_{1} \approx 1.11 T_{1}$.The percentage increase in the time period is $\frac{T_{2} - T_{1}}{T_{1}} 	imes 100 = (1.11 - 1) 	imes 100 = 11\%$.Thus,the periodic time increases by $11\%$.

The magnet of a vibration magnetometer is heated so as to reduce its magnetic moment by $19\%$. By doing this,the periodic time of the magnetometer will

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