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If there is no torsion in the suspension thread then the time period of a magnet executing SHM is

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$T = \frac{1}{2 π} \sqrt{\frac{I}{M B}}$

$T = 2 π \sqrt{\frac{M G}{I}}$

$T = \frac{1}{2 π} \sqrt{\frac{M B}{I}}$

$T = 2 π \sqrt{\frac{I}{M B}}$

answer is D.

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4^th option is right

$T = 2 π \sqrt{\frac{I}{M B}}$

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