The magnet of a vibration magnetometer is hea | vedclass

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Question

$T=2 \pi \sqrt{\frac{I}{M B_{H}}} \Rightarrow T \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{T_{1}}{T_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}$

If $M_{1}=10

Vedclass · Accepted Answer

increase by $11\%$

Vedclass · Answer

$T=2 \pi \sqrt{\frac{I}{M B_{H}}} \Rightarrow T \propto \frac{1}{\sqrt{M}} \Rightarrow \frac{T_{1}}{T_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}$

If $M_{1}=100$ than $M_{2}(100-19)=81$

So $\frac{T_{1}}{T_{2}}=\sqrt{\frac{81}{100}}=\frac{9}{10} \Rightarrow T_{2}=\frac{10}{9} T_{1}=1.11 T_{1}$

implies Time period 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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