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The time period of a freely suspended magnet is 4 seconds . It is broken in length into two equal parts and one part is suspended in the same way, Then its new time period will be

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answer is 2.

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Detailed Solution

new magnetic moment will be M/2

New moment of inertia, $I' = \frac{(\frac{m}{2}) {(\frac{l}{2})}^{2}}{12} = \frac{1}{8} \frac{m l^{2}}{12}$

$I' = \frac{1}{8} I \{I = \frac{m l^{2}}{12}\}$

Initial time period, $T = 2 π \sqrt{\frac{I}{M B_{H}}} = 4$

New time period, $T' = 2 π \sqrt{\frac{\frac{I}{8}}{(\frac{M}{2}) B_{H}}} = \frac{1}{2} T = \frac{4}{2} = 2. \sec$

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