A bar magnet used in a vibration magnetometer is heated so as to reduce its magnetic moment by 19%. The periodic time of the magnetometer will :

increase by 19%

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decrease by 19%

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increase by 11%

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decrease by 11%

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Solution

The correct option is D increase by 11%
The time period of oscillation is given by
$T = 2 π \sqrt{\frac{I}{M B}}$
$m_{2} = m_{1} - 0.19 m_{1}$
$m_{2} = 0.81 m_{1}$

$\frac{T_{2}}{T_{1}} = \sqrt{\frac{m_{1}}{m_{2}}}$

$\frac{T_{2}}{T_{1}} = \frac{1}{0.9}$

$\frac{Δ T}{T_{1}} \times 100 \approx 11 %$

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A bar magnet used in a vibration magnetometer is heated so as to reduce its magnetic moment by 19%. The periodic time of the magnetometer will :

The correct option is D increase by 11%The time period of oscillation is given byT=2π√IMBm2=m1−0.19m1m2=0.81m1T2T1=√m1m2T2T1=10.9ΔTT1×100≈11%