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A thin magnetic needle oscillates in the horizontal plane with a time period of 2.0 s. If the needle is broken into 4 equal parts perpendicular to its length, then the time period of each part will be

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0.5 s

2.0 s

1.0 s

1.5 s

answer is A.

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Detailed Solution

$T = 2 π \sqrt{\frac{I}{MB}}$

For each part M becomes one-fourth while l becomes (1/64)th.

Hence, $T' = 2 π \sqrt{\frac{I / 64}{(M / 4) B}} = \frac{T}{4} = \frac{2}{4} = 0.5 s$

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