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