A long bar magnet of time period T is cut into four equal parts by cutting it perpendicular to both length and breadth. The time period of each part is

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T/16

T/2

T/4

T/32

answer is D.

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Detailed Solution

$\begin{array}{l} T = 2 π \sqrt{\frac{I}{{MB}_{H}}} \Rightarrow T \propto \frac{\sqrt{I}}{\sqrt{M}} \\ \Rightarrow T_{2} = T_{1} \sqrt{\frac{M_{1} I_{2}}{M_{2} I_{1}}} = T \times \sqrt{\frac{M}{M / 4} \frac{I / 16}{I}} = \frac{T}{2} \end{array}$

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