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A thin rectangular magnet suspended freely has a period of oscillation equal to T. Now, it is broken in 1: 2 ratio of length but cross sectional area of each is same as before and smaller piece is made to oscillate freely in same field if its new time period of oscillation is T¹, then T¹/T is equal to :

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$\frac{1}{3}$

$\frac{1}{6}$

answer is B.

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Detailed Solution

$I^{1} = \frac{I}{27} M^{1} = \frac{M}{3}$

$\frac{T^{1}}{T} = \sqrt{\frac{I^{1}}{I} \times \frac{M}{M^{1}}} = \sqrt{\frac{\frac{I}{27} \times M}{I \times \frac{M}{3}}}$

$\frac{T^{1}}{T} = \frac{1}{3}$

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