The time period of oscillation of a freely suspended bar magnet with usual notations is given by:

T = 2 π \sqrt{\frac{I}{M B_{H}}}

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T = 2 π \sqrt{\frac{M B_{H}}{I}}

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T = \sqrt{\frac{I}{M B_{H}}}

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T = 2 π \sqrt{\frac{B_{H}}{M I}}

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Solution

The correct option is A $T = 2 π \sqrt{\frac{I}{M B_{H}}}$
the time period equation of a pendulum is $T = 2 π \sqrt{\frac{l}{g}}$
in case of bar magnet which is suspended g is replaced by $M B_{H}$ & $l$ is replaced by $I$
$∴$ time period of suspended bar magnet $T = 2 π \sqrt{\frac{I}{M B_{H}}}$

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The time period of oscillation of a freely suspended bar magnet with usual notations is given by:

The correct option is A T=2π√IMBHthe time period equation of a pendulum is T=2π√lgin case of bar magnet which is suspended g is replaced by MBH & l is replaced by I∴ time period of suspended bar magnet T=2π√IMBH