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Question

A suspended bar magnet of moment 3 Am2 is deflected through an angle of 600 in a magnetic field of induction 4×105 T and then released. If its angular velocity, when passing through its equilibrium position is 0.5radians/s, the moment of inertia of the magnet is:

A
24×105kgm2
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B
12×105kgm2
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C
48×105kgm2
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D
64×105kgm2
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Solution

The correct option is C 48×105kgm2
Kinetic Energy = 12Iω2
m=3Am2
ω=12rad/s
θ1=600
θ2=00
U1=Bmcos60=Bm2
U2=Bmcos0=Bm
(12Iω2)=U1U2
=Bm2
12I×(12)2=4×105×32
I=48×105Kgm2

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