A thin magnetic iron rod of length 30 cm is suspended in a uniform magnetic field. Its time period of oscillation is 4 s. It is broken into three equal parts. The time period (in seconds) of oscillation of one part when suspended in the same magnetic field, is
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a
43
b
23
c
3
d
43
answer is A.
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Detailed Solution
T=2πIMBHWhere I=mL212 = moment of inertia of thin magnetic rodWhen given rod is broken into three equal parts, thenM'=M3 and I'=I27 = M.I. of each piece. ∴T'=2πI'M'BH=2πI x 327x MBH=T3=43