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Q.

A bar magnet of moment of  inertia  I  is vibrated in a magnetic field of induction  0.4×10-4T. The time period of vibration is 12sec. The magnetic moment of the magnet is  120Am2  . The moment of Inertia of the magnet is ( in Kgm2 )

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a

1750×10-2Kgm2

b

175×10-4Kgm2

c

2.1π2Kgm2

d

1.5×10-2Kgm2

answer is B.

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Detailed Solution

Time period of Oscillation of Magnetic needle is                           T=2πIMB ⇒T2=4π2IMB ⇒4π2I=MBT2 ⇒I=MBT24π2Given data :  T = 12 sec ,  B=4×10-5T , M=120Am2                  ∴I=120×4×10-5×1444×(3.14)2=175×10-4Kgm2
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