Questions
The period of oscillation of a thin magnet at a place is T. When it is stretched to double its length and its pole strength is reduced to of its initial value, then its period of oscillation is
detailed solution
Correct option is D
We know time period of oscillation T=2πIMBGiven : Pole Strength : m1=m, m2=m4;Length l1=l, l2=2l;Moment of Inertia I1=I=ml212, I2=m(2l)212=4I;Dipole Moment M1 = ml = M, M2=m4(2l) = ml2=M2;Now, T1T2=I1I2×M2M1⇒TT2=I4IM2M⇒T2=22TTalk to our academic expert!
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