alleryStart learning →The time period of small oscillations of a magnet of moment of inertia I in field B is
A2π√(mB/I)
B2π√(B/mI)
C2π√(mI/B)
D2π√(I/mB)
Answer & Solution
Correct answer: D. 2π√(I/mB)
I d2theta/dt2 = -mB theta gives omega squared = mB/I, so T = 2π√(I/mB).
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