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(D) The time period of a magnetic dipole oscillating in a uniform magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment

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$T\sqrt{2}$

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(D) The time period of a magnetic dipole oscillating in a uniform magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the earth's magnetic field.Initially,the moment of inertia is $I$. So,$T = 2\pi \sqrt{\frac{I}{MB_H}}$.When a piece of wood with the same moment of inertia $I$ is attached to the magnet,the new moment of inertia of the system becomes $I' = I + I = 2I$.The magnetic moment $M$ remains unchanged.The new time period $T'$ is given by $T' = 2\pi \sqrt{\frac{I'}{MB_H}} = 2\pi \sqrt{\frac{2I}{MB_H}}$.Comparing this with the initial time period,we get $T' = \sqrt{2} 	imes (2\pi \sqrt{\frac{I}{MB_H}}) = \sqrt{2}T$.

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